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TRIGONOMETRIES BY
PROFESSOR R.
K MORITZ
Text Book on Spherical Trigonometry. 8vo, vi
+ 67
8vo, xiv
+451
xii
+ 518
(With five-place tables.)
pages, 183 figures.
Plane and Spherical Trigonometry. 8vo,
Cloth, $1.25 net.
pages, 41 figures.
Elements of Plane Trigonometry.
Cloth, $2.00 net.
(With five-place tables.)
pages, 224 figures.
Cloth, $2.50 net.
ELEMENTS OF
PLANE TRIGONOMETRY (WITH FIVE-PLACE TABLES)
TEXT-BOOK FOR HIGH SCHOOLS, TECHNICAL SCHOOLS AND COLLEGES
BY
ROBERT
E.
MORITZ
PBJ>. (NEBRASKA), PH.N.D. (STRASSBURG), PROFESSOR OF MATHEMATICS.
UNIVERSITY OF WASHINGTON
NEW YORK & SONS, WILEY JOHN LONDON-
CHAPMAN & HALL,
INC.
LIMITED
COPYRIGHT, 1910 BY
ROBERT
Printed in
.
.
MORITZ
U.S. A.
Stmbope Jprets G1LSON COMPANY BOSTON. U.S.A.
H.
PREFACE TRIGONOMETRY
is
college
mathematics par
excellence.
To
at least
students college mathematics means trigoIt is important, therefore, that the nometry and nothing else. science be presented in as simple and attractive a manner as possible
90 per cent of
all liberal arts
be made more than a mere method of solving triangles. by making the treatment technical than is customary, by introducing considerable
and that
it
The
the author tries to accomplish
less
first
by not presupposing a too ready knowledge of elementary mathematics, and none at all of the topics ordinarily To accomplish the second point the treated in college algebra. the of the science. central is made idea This permits the angle enrichment of the science through the introduction of a variety of historical matter,
concepts and processes ordinarily reserved for advanced courses in mathematics. Since the treatment departs considerably from that current in textbooks on trigonometry, it is fitting that some of the leading characteristics of the present book should be enumerated at the outset: First, as to subject-matter:
The book has been planned to cover five months' work at four Each month's work is followed by a set of review exercises. Where less time must be given to the subject, (1)
lessons per week.
advanced chapters may, of course, be omitted. the graphic method of splving triangles is intended to impress the need of a more accurate method, certain (2)
the
The introductory chapter on method
of trigonometry. For (3) knowledge of logarithms has not been presupposed. this reason a chapter on logarithms and the use of tables has been
A
incorporated at in logarithms
proper place. Classes who are properly prepared of course omit this chapter.
Many of the more important results have been derived by two more independent methods. This has been done,--
(4)
or
its
may
PREFACE
IV
To give the teacher a choice of To offer the ambitious student
(a)
methods.
the advantage which comes from approaching the same truth from two or more directions. To offer an alternative to the student without a teacher
(b)
(c)
who (5) It is
any one
finds
undue
not intended that
class.
with any one given proof.
difficulty all
The problems
the problems should be assigned to
in each set are carefully graded
and
arranged as follows, (a)
The
(b)
The next
first half in each set are very simple applications of the principles and theorems discussed in the preceding
sections.
three or four problems require some originality on the part of the student. The last few problems in each set are for the more ambitious student and frequently give him the opportunity to dis-
(c)
cover for himself results which are discussed in detail in later sections of the
book.
has been bestowed on the applied problems illusthe solution of right and oblique triangles. In each case trating there is given first a set of problems involving miscellaneous heights (6) Special care
and distances. This is followed by separate sets of applied problems from each of the following sciences: Physics, Engineering, NavigaThese tion, Astronomy and Geography, and Elementary Geometry. arc probably the most varied and complete that have been published in America in recent years.
lists
(7)
than
Trigonometric curves have is
usual.
veloped from
The method
received
much
fuller
of representing functions
first principles.
The treatment
by
treatment
curves
is
de-
includes sine curves
of given amplitude and wave length, logarithmic and exponential curves, composition of harmonic curves, the catenary, and the curve of damped vibrations. (8)
A
special section is
devoted to the angle and
its
functions con-
sidered as functions of time. (9) In developing the theorems of Demoivre and Euler no knowledge of imaginaries is presupposed. The chapter on trigonometric series presupposes no knowledge of series. (10) The usual inadequate treatment of hyperbolic functions is
replaced
by a separate chapter
in
which the analogies between the
PREFACE
V
circular and hyperbolic functions are developed both analytically and geometrically. This chapter concludes with the determination
of the area of a hyperbolic sector. (n) Abundant historical matter has
been introduced throughout
the work.
Second, as to method and arrangement: (12) Logical order has been made subsidiary to order of teaching. To illustrate: the study of the functions of an angle is divided into three parts, (a)
The study
(b)
the solution of right triangles. The study of the functions of the obtuse angle, followed the solution of oblique triangles.
by
(c)
The study
by
(13)
of the functions of
an acute angle, followed by
of the functions of the general angle, followed the solution of trigonometric equations.
The
discussion of the general angle, of circular measure, and two or more angles, is postponed until after the solu-
of functions of
tion of the oblique triangle.
This makes
it
possible to complete the
subject through the solution of triangles in half a semester, an important consideration for classes in short summer sessions, and for engi-
neering students who begin surveying and trigonometry the same Furthermore it removes the suspicion, so often felt by the semester. that the solution of triangles is the sole aim of the science. student, (14) Following the plan long since established in Germany and France, only the three principal functions, sine, cosine and tangent, have received detailed treatment. The corresponding results for
the reciprocal functions are left as exercises for the student.
This
plan economizes time and space and leaves the student with a clearer understanding of the entire subject. (15) Every example worked out in the text is followed by one or more checks. Checks are looked upon as an essential part of every solution. In order to cultivate the use of checks, it would seem best to omit the answers to the exercises, yet every teacher knows the importance of "answers" to the beginner in guiding his first un-
certain efforts. In the present text the plan has been adopted of supplying the answers to a part of the exercises only, with explicit directions to the student to check every problem to which no answer is
given.
This leaves the enforcement of the checking habit largely
PREFACE
VI
with the individual teacher, who may assign as without answers as he deems desirable. (16)
No
many problems
pains has been spared to impress the student with the
limitations in the degree of accuracy, in the answers to problems, imposed by the data, as well as with the limitations in the degree of
accuracy due to the use of tables. Superfluous figures and show of accuracy not warranted by the data or the process of computation employed, are nowhere tolerated in the present book. ^17) On the other hand the student is guarded against disregarding figures and remainders without first measuring the effect of the parts neglected on the required results. Thus, in the computation of
logarithms by means of the logarithmic series, or of natural functions by means of the sine and cosine series, the effect of the neglected part of each series on the final result has been carefully considered in each case.
The book embodies
the author's practical experience of seventeen and univer-
years in presenting the subject to beginners in colleges
His experience has convinced him that the subject of trigonometry can be so simplified and enriched that it deserves the fore-
sities.
most place
year of any high school or the first year of any not college curriculum, only because of its intensely practical value, but chiefly because of its unrivaled cultural value. In the mastery in the last
of logarithms, which strips the most complicated and laborious calculations of their difficulties and irksomeness, the student cannot help
becoming conscious
when properly
of the
tremendous power of the human mind
In the application of algebraic processes and symbols to geometrical and physical magnitudes, he is initiated into a cal
directed.
most far-reaching method
geometry.
The study
of
modern
working knowledge of an indispensable tool activity.
The study
research, that of analyti-
of trigonometric curves should give
of the trigonometric
him a
in every field of scientific
and logarithmic
series,
computation of logarithmic and natural functions, opens an entirely new field of thought with its importance for pracThe actual use of tables familiarizes the student with tical ends.
and
their use in the
the principle of interpolation, a knowledge of which
is
demanded
wherever tables are used. Besides these concepts and processes, the importancb of which all, there are an abundance of others which open the
must appeal to
door to higher realms of thought.
The
simplest applications of
PREFACE
VII
trigonometry to imaginary and complex numbers reveals a new conception of addition and multiplication; in the determination of the roots of unity an otherwise unsolvable problem is solved in all generality; imaginary angles lead to the unsuspected region of hyperbolic functions and reveal a new world of symmetry and its
beauty.
Above
all,
while contemplating the lifelong self-sacrificing efforts of who brought the science to its present state of per-
the master minds
fection, of men who spent their lives without either pecuniary compensation or popular applause in order to share in the building of the temple of abstract truth, the student must come to a better
appreciation of truth for its own sake and be helped in part to a realization of the higher objects of human endeavor.
While writing this book the author has received valuable suggestions from several of his colleagues. Special mention is due Mr. George I. Gavett who supplied some of the applied problems from engineering, script
and
and to Mr. Allen Carpenter who read the
verified
many
entire
of the answers to the problems.
GREEK ALPHABET a pronounced "
y 5 e
" " " "
77
r
"
alpha.
v
beta.
o
gamma. p
epsHon.
v
eta.
T
" " 1
\
" ti
rho.
slgma. "
tan.
" i
upsilon. xi
" ^.V7.
\J/
" 4C
"
beta.
iota.
omikron.
"
' '
I
K
"
PI.
delta.
seta.
pronounced nu.
"
kappa. lambda,
mu.
>>//f.
X
w
" r:A/.
" owtfga
manu-
CONTENTS CHAPTER
I
INTRODUCTION PACK
ART. 1
.
2.
3.
4.
Graphic solution of triangles Solution of practical problems by the graphic method Inadequacy of the graphic method Definition of trigonometry
CHAPTER
i
4 7 7
II
TRIGONOMETRIC FUNCTIONS OF AN ACUTE ANGLE '
5.
Definition of function
Definition of reciprocal 7. The six trigonometric functions of an acute angle 8. Trigonometric functions determined approximately
9 9 a
6.
9.
10.
Given one of
its
by measurement ....
functions, to construct the angle
Functions of complementary angles
n. Functions
of
14 17
19
o, 30, 45, 60, 90
21
12.
Fundamental
13.
To
25
14.
express each of the functions in terms of a given one Reduction of trigonometric expressions to their simplest form
15.
Trigonometric identities
33
relations
24
CHAPTER
27
III
SOLUTION OF RIGHT TRIANGLES BY NATURAL FUNCTIONS 16.
Table of natural functions
18.
To To
19.
Accuracy
17.
35
find the natural functions of find the angle less
an angle
less
than 90
than 90 corresponding to a given natural function
of results
35
39 43
Solution of right triangles by natural functions 21. Triangles having a small angle 20.
45
49
22.
Historical note
50
23.
Review
51
CHAPTER
IV
LOGARITHMS 24.
Definition of logarithm
53
25.
Fundamental laws governing logarithms Logarithms of special values
54 55
26.
iz
CONTENTS
X
PAGE
ART. 27.
The common system
28.
Rule
29.
Table of
30.
$\
.
To To
of logarithms
57
for the characteristic
common
58 61
logarithms
number number corresponding to a given logarithm
find the logarithm of a given
62
find the
65
32.
Directions for the use of logarithms
67
33.
Application of logarithms
72
34.
To compute a
table of
common
logarithms
73
N
36.
Relation between loga and log&N Natural or hyperbolic logarithms
37.
Tables of logarithmic trigonometric functions
35.
38. 39.
To To
75
76 78
find the logarithmic trigonometric functions of
an angle
less
than 90
function
81
4o: Logarithmic functions of angles near o or 90 41. Use of 5 and T table 42.
79
find the angle corresponding to a given logarithmic trigonometric
83
84
Historical note
87
CHAPTER V LOGARITHMIC SOLUTION OF RIGHT TRIANGLES 43.:
Logarithmic solution of right triangles
44.
Number
45.
Applied problems involving right triangles
Heights and distances. 47. Problems for engineers
..
46.-
48.
.
.
.
89
.
of significant figures
93
94
.
95
.
.
96
Applications from physics
Problems
98 100
.
.
in navigation Geographical and astronomical problems 51. Geometrical applications 52. Oblique triangles solved by right triangles
49.
.
.
103 106
50.
CHAPTER
no
VI
FUNCTIONS OF AN OBTUSE ANGLE 53.
Rectangular coordinates
54.
Definitions of the trigonometric functions of
55.
The
56.
Fundamental
signs of the functions of
any angle
less
than 180 ....
an obtuse angle
119
relations
Functions of supplementary angles 58. Functions of (90 -f- 0) SQ. Functions of 1 80.
....
57.
.
.
.
.
.
60.
Angles corresponding to a given function
61.
Review
119 120 121
.
.
117 1 18
... .,
121
.
.
122 123
CONTENTS
CHAPTER
xi
VII
PROPERTIES OF TRIANGLES PAGE
ART.
62. 63. 64.
The law of sines ............................................ The projection theorem^ ............................... The law of cosines .^-A? ...... ........... .. .
...
Arithmetic solution of triangles .. 66. The law of tangents. .$*??. ....... i/ 67. Formulas for the area of a triangle .VT. ...... 65.
.
......
.
68.
127 128
130
.132
.
Functions of half the angles in terms of the sides .............
CHAPTER
125 126
134
VIII
SOLUTION OF OBLIQUE TRIANGLES 69. 70. 71.
72. 73.
74.
Solution of oblique triangles
.
I.
them
138 141
....
.
138
144 147
150
Auxiliary geometrical constructions
.
152
.
79.
80.
Geometrical applications
77.
...
150
System of simultaneous equations Miscellaneous heights and distances. $fi Applications from physics Applications from surveying and engineering. Applications from navigation Problems from astronomy and meteorology
78.
of
.
Practical applications (a) System of triangles
(c)
76.
.
.
(6)
75.
...
.
.
Given two angles and one side Case II. Given two sides and the angle opposite one Case III. Given two sides and the included angle Case IV. Given three sides Case
.
.
.
155
.
.
V
.
158 161
164
.
171 .
.
,
.
.
.
172
174
CHAPTER IX THE GENERAL ANGLE AND
ITS
MEASURES
81.
General definition of an angle
82.
Positive and negative angles
83.
Complement and supplement
178
84.
Angles in the four quadrants
178
85.
Sexagesimal measure of angles
86.
Decimal division
of degrees
..
177
178
.
179 180
87.
Centesimal measure of angles
180
88.
The
181
circular or natural
system of angular measures 89. Comparison of sexagesimal and circular measure 90. Relation between angle, arc and radius 900. Area of circular sector 91.
Review
182
185 187
189
CONTENTS
Xll
CHAPTER X FUNCTIONS OF ANY ANGLE PAGB
ART. 92. Definition of the trigonometric functions of any angle 93. Signs of the functions in each of the quadrants
191
Periodicity of the trigonometric functions 95. Changes in the value of the functions 96. Changes in the value of the tangent
193
192
94.
193 195
of results
97.
Summary
98.
Fundamental
195
relations
196
Representation of trigonometric functions by lines TOO. Reduction of the functions to the first quadrant
196
99.
101. 102.
Reductions from the third quadrant Reductions from the fourth quadrant
199 201
104.
Functions of negative angles Table of principal reduction formulas and general rules
105.
Generalization of the preceding reduction formulas
103.
199
203
204 206
CHAPTER XI FUNCTIONS OF
Two
OR MORE ANGLES
106.
Addition theorem for the sine and cosine
107.
Generalization of the addition theorems
108.
Addition theorems.
109.
Subtraction theorems
209 210 211
Second proof f< -r
the sine and cosine
no. Tangent of the .um and difference of two angles in. Functions of double an angle 112. Functions of half an angle 113. Sums and differences of sines or cosines transformed into products.
CHAPTER
212 215 216
216 ...
219
XII
TRIGONOMETRIC EQUATIONS 114.
Angles corresponding to a given function
225
115.
Principal value
225 226
1
16.
1 1 7.
1 18.
Formula for angles having a given sine Formula for angles having a given cosine Formulas for angles having a given tangent of results
i
IQ.
Summary
1
20.
Trigonometric equations involving a single angle
226 227 227 227
121.
Trigonometric equations involving multiple angles
233
122.
Trigonometric equations involving two or more variables
236
123.
Solutions adapted to logarithmic computation Inverse functions
240
124.
125.
Review
250
2 45
CONTENTS
CHAPTER
Xlil
XIII
TRIGONOMETRIC CURVES PAGE
ART. 126.
127. 1
28.
129. 130. 131. 132.
Functions represented by curves
253
The straight line The circle The hyperbola The sine curve The tangent curve The sinusoidal or simple harmonic curves
253
254 255 255 257
260
Angles as functions of time 134. Composition of sinusoidal curves
262
133.
135.
Theorem on composition
264
of sinusoidal
curves having equal
wave
lengths 136.
137. 138. 139.
140. 141. 142.
267 268
Fourier's theorem
The logarithmic curve The exponential curve The general exponential curve The compound interest law The catenary The curve of damped vibrations
26q
270 270 272
274 275
CHAPTER XIV TRIGONOMETRIC REPRESENTATION OF COMPLEX NUMBERS Imaginary numbers Geometric representation of imaginary numbers Geometric representation of complex numbers Trigonometric representation of complex numbers Geometric addition and subtraction of complex numbers Physical applications of complex numbers
278
149.
Historical note
285
^50.
Multiplication and division of complex numbers
151.
Powers of complex numbers Roots of complex numbers
287 288
143. 44.
f45.
146. 147. 148.
152.
153. f 54.
To To
n solve the equation z n solve the equation z
i
=
The cube The cube
292
o
293
157. 158.
The
1
56.
159. [60.
To To
295
number
irreducible case
express sin n$ and cos express cos 6 and sin
angles
282 283
o
roots of unity roots of any real or complex Solution of cubic equations
f 55.
281
289 i
-f-
278 280
295
296
298 iiO
in in
powers of sin 6 and cos terms of sines and cosines of multiple
301
302
CONTENTS
XIV
CHAPTER XV TRIGONOMETRIC SERIES AND CALCULATION OF TABLES ART.
PACE
161.
Definition of infinite series
162.
Convergent and non- convergent
163.
Absolutely convergent series
164.
Sum
165. 1 66.
The limit or r n as n approaches The geometric infinite series
167. 168.
169. 170.
171.
of
an
306 307 308 310 310 310
series
infinite series
Convergency test ... Convergency of special
infinity
31 ! series
313
The number e The exponential series The logarithmic series
316
Calculations of logarithms Errors resulting from the use of logarithms .. sin x tan .v . T f as x approaches zero 174. Limiting values of the ratios
172. 173.
.
.
,
.
,
>
x
175.
Limiting values of cos
w -
/ sm
and
n
\
,
x \ x / n )\
x/n
as H approaches infinity
The
177.
Computation of natural functions table Approximate equality of sine, tangent and radian measure of small
178.
and tangent
326 328
/
176.
sine, cosine
318 320 321 324
series
angles
329 330 333
CHAPTER XVI HYPERBOLIC FUNCTIONS 170. 180. 181.
Series with
complex terms
336
Definition of the trigonometric functions of complex Euler's theorem, ei0 = cos 6 i sin
+
Geometrical representation of Euler's theorem 183. Exponential form of the sine and cosine
182.
Hyperbolic functions defined Duality of circular and hyperbolic functions 186. Table of formulas 187. Inverse hyperbolic functions
184.
.
.
.
185.
1 88.
189. 190. 191.
Geometrical representation of hyperbolic functions Area of hyperbolic sector
Use of hyperbolic functions Review
numbers
337 338
339 340 342 343 345 348 349 350 353 354
PLANE TRIGONOMETRY CHAPTER
I
INTRODUCTION In order to work the exercises in this chapter the student should be provided with a pair of compasses, a protractor, and a graduated ruler divided into tenths of a unit.
In plane geometry it is shown 1. Graphic Solution of Triangles. that the six parts (three sides and three angles) of any plane triangle are so related that any three parts suffice to determine the shape of the triangle, and if one of the known parts is a side, the size of the
Furthermore it is shown how to contriangle is also determined. All struct the triangle when a sufficient number of parts is given. four cases. of or the one another under come cases following possible
To
construct the triangle
One
when
there
is
given,
and two angles. and an angle opposite one III. sides and the included angle. IV. Three sides. I.
Two Two
II.
Usually measures. the actual
side
sides
ment
A
them.
we have given not the actual lines and angles but their From these measures lines and angles corresponding to lines and angles may then be constructed by means of
suitable instruments. 1.
of
Such instruments
are,
graduated straight-edge for the construction and measure-
of straight lines of definite lengths.
The
smallest divisions of
the straight-edge should be decimal, either millimeters or tenths of
an inch.
A
2. pair of compasses for the construction of circles cular arcs.
3.
A
protractor for the construction
angles of definite magnitudes.
and measurement
and
cir-
of plane
PLANE TRIGONOMETRY
EXAMPLE
i.
It
is
[CHAP,
i
required to construct a triangle which has two and 1.75 inches respectively and the included
sides equal to 2.5 inches
angle equal to 36. Solution. By means of the protractor construct an angle (Fig. i) equal On to 36. measure off equal
MAN
AM
to
2.5
inches.
AB
On
AN
measure
off
AC
equal to 1.75 inches. Join B and C by a straight line, ABC is the required Fig.
i.
triangle.
The numerical values of the parts which were not known at the outset may now be found by measurement. BC is thus found to be 1.49 inches, and by means of the protractor, angles B and found to be approximately 43.5 and 100.5 respectively. If it is
not possible or convenient to construct the triangle
C
are
full size,
a similar triangle may be constructed on a reduced scale; that is, any unit or a fraction of a unit on the scale may be taken to represent
Thus lines 3 and 4 inches long in the problem. of a be in solution the triangle whose sides are 30 and may employed 40 miles respectively. The angles of the reduced triangle will of any unit occurring
course be equal to the angles of the triangle represented. Similarly, the unknown parts of a triangle which is too small for actual construction, say some microscopic triangle, may be found by measurement from a similar triangle drawn on an enlarged scale.
EXAMPLE
2.
One
side of a triangle measures
adjacent to this
side
measure 23
angles Find the remaining parts of the triangle. Let J inch represent 100 Solution. drawn 3 inches miles. Then a line
600 miles, and the
and 100
respectively.
AB
long will represent 600 miles. At B draw the angles BAC and
and 100
AC
A and
ABD 23 BD intersect
Let respectively. at E. will represent the re-
ABE
Fig. 2.
quired triangle.
Angle
E
measures 57, which of course could have been found sum of the angles .4 and B from 180.
otherwise by subtracting the
AE
and
BE
are found
inches respectively.
to
measure approximately 3.52 and 1.40
Remembering that each J inch
represents 100
INTRODUCTION
i]
miles, the actual lengths represented
by
3
AE
BE
and
are approxi-
mately 704 and 280 miles
respectively. Solutions, like the foregoing, in which geometrical drawings to a scale are employed instead of numerical calculations, are called
graphic solutions.
EXERCISE
i
1. Review the following propositions in geometry. A, B, C represent the three angles of any triangle and a, 6, c the sides opposite these angles.
c.
Given A, B, c\ to construct the triangle. Given a, &, C; to construct the triangle. Given a, b, c; to construct the triangle.
d.
Given
e.
Under what conditions
a. b.
a, b,
A
To
solutions?
The 2.
to construct the triangle.
;
Given a
=
5,
=
4, c
=
7; find
give rise to two different
Given 6
=
4, c
=
C = 90;
5,
Given
b
=
=
600,
the graphic method.*
B=
30',
34,
find the third side
Ans. a
270, c
by
the angles to the nearest
A = 44
to the nearest 15'. 4.
(d)
following problems are to be solved
Ans. 3.
will
only one solution?
A = 100;
=
3,
5.
=
0.029,
Given a
=
42, b
32
=
51,
136 Ans.
Ans.
of the remaining parts.
c
7.
>1
=
44
30',
B=
57,
C=
= =
700.
45'; find the remaining b 0.081, c 0.104.
find the
or c
Given
and the angles
find the third side correct
15', C=
A = 55;
15'.
30'.
Ans. a
B=
sides. 6.
101
A = 37, B = 53.
to the nearest integer.
Given a
C=
approximate measures
33.6, 24.9,
B= B=
84, 96,
C = 41; C = 29.
78 30'; find the ratios between
the sides opposite these angles.
Ans. Approximately a
:
b
:
c
=
5
:
6
:
7.
*
In order to employ the graphic method successfully the student must pracTwo pencils of medium hardness should be used, one sharpened to a point for marking distances, the other sharpened like a chisel for drawing The pencil points are easily kept sharp with the aid of a piece of fine sandlines. tice
accuracy.
The lines should be drawn sharply and they should bisect the points Through which they are intended to pass. In measuring the required parts, beginners should estimate angles to quarters of a degree and lengths to quarters of
paper.
the smallest division of the scale.
PLANE TRIGONOMETRY
[CHAP,
i
Solution of Practical Problems by the Graphic Method. important practical problems, in which a high degree of accuracy is not essential, can be easily solved by the graphic method. 2.
Many
AB
across a Suppose it is required to find the approximate distance lake or'swamp, without actually measuring it. This may be accomplished in various ways, one of which is as follows: Select
A and B tances
some point
P
from which both
are visible, and measure the dis-
AP
and
BP
and
also
the
angle
APB*
This gives two sides and the included angle of the triangle APB from
which
AB may be
found by the method of
the preceding article. Similarly the heights of towers and trees Fig. 3-
and mountains, of clouds and shooting distances the stars, through impenetrable forests across swamps and through mountains, the widths of rivers, ravines and canyons, may be
Even the
determined.
distances between celestial objects may be method after certain other distances
the graphic
approximated by and angles have been measured.
EXAMPLE
In order to determine the width of a
i.
river,
the dis-
tance between two points A and B close to the bank of the river was measured and was found to be 600 feet. The angles BAP and
ABP, formed
with a point
P
Select a suitable scale, say
Solution.
bank and 36
close to the opposite
were also measured and were found to be 50 Required the approximate width of river. i
of the river,
respectively.
inch
to 100 feet, and construct a triangle ABP, = 6 inches and the adjacent having to 50 and 36 respectively. From angles equal
AB
P
PT
perpendicular to AB. the width of the river. represent
draw
PT.
PT
will
represents 100
be found to measure feet,
DEFINITIONS. observer.
will
2.7 inches,
the width of the river
Let
Through
PT
P
Fig. 4-
Measure is
and 270
since each inch feet.
P
the position of the be any point and and draw a vertical line, draw through
a horizontal line meeting the vertical line in H. * The angle between two visible objects is readily measured by means instrument called a transit.
of
an
INTRODUCTION
P is above H, as in the upper figure, the angle HOP is called the angle of elevation of the point P as seen from O, If P is below H, as in the lower figure, the angle #OP is called the angle of depression of the point P as seen from O. If
It is obvious that the angle
of elevation or
depression of an object depends upon the position of the observer.
of
EXAMPLE
2.
From a
point
P at
M
FiS-
5-
the foot of a mountain, the angle
measured and is found to be 30; after walking two miles toward the summit on an incline averaging 15, the angle of elevation is found to measure 45. Required the
summit
of elevation of the
is
height of the mountain. Solution. Draw a horizontal
PX.
line
angle
XPN =
Construct an
30; then
PN
represents the direction in which the summit of the
mountain
is
from P.
seen
Construct angle XPC = 15, and take PC two units in
. _. Fig. 6.
length. Then, if each unit the one will C mile, position from which the represent represents second observation was made. Through C draw CX' parallel to PX, and construct an angle X'CN'
=
45.
Then CN'
represents the direction in which the
summit
is
seen from C. Since the
summit
is
on each
of the lines
M. PX. Then
PN
and CN',
it
must be
located at their point of intersection
Draw
MF
perpendicular to
MF
will represent
the
mountain on the same scale on which PC represents two miles. Measure MF. If PC was taken equal to 6 inches, MF will measure 5.8 inches. Since 3 inches represents one mile, rep-
height of the
MF
resents 1.933 miles, or 10,200 feet approximately.
EXERCISE
2
The following problems are to be solved graphically. The student is expected to obtain distances correct to three figures and angles correct to nearest
15'-
PLANE TRIGONOMETRY
6
[CHAP,
i
1. At a distance of 400 feet from the foot of a tree, the top of the tree subtends an angle of 20. Find the height of the tree. Ans. 145.6 ft.
A
B
2. 8 miles disstraight road leads from a town A to a town tant; another road leads from A to a third town C 10 miles distant.
The 3.
angle between the roads
What
is
is
How
65.
far is it
from B to C ? Ans. 9.82 mi.
(= angle of elevation) of the sun, when a high casts a shadow 190 feet long on a horizontal
the altitude
building 75 feet
Ans. 21
plane?
30'.
The
great pyramid of Gizeh is 762 feet square at its base and each face makes an angle of 51 51' with the horizontal plane. 4.
Determine the height of the pyramid, assuming that it comes to Ans. 485 feet. an apex. 5. As a matter of fact, the pyramid mentioned in Problem 4 does not come to a point, but terminates in a platform 32 feet square. Find the actual height of the pyramid. Ans. 465 feet. 6.
line
An
observer on board ship sees two headlands in a straight The ship sails northwest for 5 miles, when one of the
N. 35 E.
headlands appears due east and the other due northeast. apart are the headlands? 7.
How
far
Two observers
at the
on opposite sides of a balloon observe the balloon same instant and find its angles of elevation to be 56 and 42
The observers are one mile apart. Find the height of respectively. at the time the observations were taken. the balloon Ans. 0.6 mi. nearly. 8.
AB
In order to determine the distance across a swamp, a distance laid off 100 yards long, and at each extremity of the line AB
was
the angles were measured between the other extremity of the line and each of two stakes P and Q placed at opposite ends of the swamp. At one extremity of the line the angles measured 35 and 85 respec-
end the angles measured 40 and 121 respectively. Find the distance PQ. 9. Find the perimeter of a regular polygon of 7 sides inscribed in Ans. 60.75 fta circle whose radius is 10 feet. = = = 10. The sides of a triangle are a 10, b 12, c 15 respectively. tively, at the other
Find the
radii of the inscribed
the angles of the triangle.
Ans.
and r
of the circumscribed circles
=
3.23,
R =
7.52,
and
INTRODUCTION
3-4]
7
Inadequacy of the Graphic Method. The graphic method of solving triangles, though exceedingly simple and useful, is not suffiFor instance, in the last problem ciently accurate for all purposes. 2 obtained the results of Exercise by the graphic method are: 3.
r
=
3 .2 3
while the
,
#=7-52,
more accurate
A=
=53,
41 30',
results,
C=8 5
3o
/
,
obtained by a method to be de-
scribed later, are: r
=
3.2331,
=52 The
#=
7-5236,
53' 27",
.4
=
41 38' 59", 27' 34".
causes of the inaccuracies are manifold.
First of all the
divisions of the scale used in measuring are not indefinitely small;
they were the eye could not distinguish them. Besides this the graphic method is subject to many other unavoidable errors. The instruments employed in the construction of lines and angles The straight-edge is not perfectly straight, the are imperfect. if
divisions of the scale are not exactly equidistant.
The
points used
in the construction are not true points but dots having dimensions;
likewise the lines of unequal widths.
drawn are not
true lines but pencil or pen marks hand which draws the pencil All perfectly steady, and so on.
Again, neither the
nor the eye which guides it is these sources of error are unavoidable.
By employing better instruments and by using greater care, these errors may be diminished, but they cannot be entirely eliminated. There exists another method of 4. Definition of Trigonometry. mentioned solving triangles which is free from all the errors above and enables us to obtain results correct to any desired degree of the unknown accuracy. This method consists in the computation of parts of a triangle from the numerical values of the given parts. The development of this method has given rise to a separate branch Trigonometry considers the with angles, and properties of angles and certain ratios associated of mathematics, called trigonometry.*
* The word Trigonometry comes from two Greek words, trigonon = triangle, and metron = measure. The method was originated in the second century B.C. by Hipparchus and other early Greek astronomers in their attempts to solve certain The term trigonometry was not used until the close of the spherical triangles.
sixteenth century.
8
PLANE TRIGONOMETRY
[CHAP,
i
applies the knowledge of these properties to the solution of triangles
and various other algebraic and geometric problems.
Incidentally
trigonometry considers also certain time-saving aids in computation such as logarithms, which are generally employed in the solution of triangles.
Briefly stated,
Trigonometry
is the science of
angular magnitudes and the art of apply-
ing the principles of this science to the solution of problems.
H
CHAPTER
TRIGONOMETRIC FUNCTIONS OF AN ACUTE ANGLE When
Definition of Function.
6.
the value of the one depends to be a function of the other.
upon
two variables are so related that
the value of the other, the one is said.
EXAMPLES. The area of a square volume of a sphere is a function of
a function of
is
radius.
its
The
The
its side.
velocity of a began to 'fall.
body is a function of the time elapsed since it The output of a factory is a function of the number of men employed.
falling
In the expression y y
is
a function of
functions of
x.
/
=
\
,
y depends upon x for
>
+
2
x.
2
/
3
x i, x* Similarly x / is a function of t, etc.
3,
its
value, hence
ax H
c,
are
6. Definition of Reciprocal. // the product of two quantities equals unity, each is said to be the reciprocal of the other.
For example, cal of x.
J
4X2=1. xy
=
i
it
is
if
xy
=
i,
x
the reciprocal of y, and y is the reciprofor 2, and 2 is the reciprocal of \,
is
the reciprocal of
In general,
follows that
=
and
x
=
- are reciprocals since 7
,
and y
The reciprocal of any quantity
is
=
,
that
=
i
.
From
is,
unity divided by that quantity.
7. The Six Trigonometric Functions of an Acute Angle. Let A be any acute angle, B any point on either side of the angle, and A BC the right triangle formed by drawing a perpendicular from B to the
tude),
Denote AC, the side adjacent to the angle BC, the side opposite the angle A, by a (for altiand the hypotenuse AB by h.
The
three sides of the right triangle form six different ratios,
other side of the angle.
A by t
b (for base),
namely,
and
a
6
a
It
ti
b'
*
*
a
b'
their reciprocals
9
*.
a
PLANE TRIGONOMETRY
10
[CHAP,
n
a. These ratios do not depend upon the distance of the point B from the vertex of the angle; that is, each of the six ratios will have the same value for every other point B' located on either one of the
sides of the angle.
AB or AB produced, or on A C or and the produced, perpendicular B'C' be drawn to the other f f the AB will be similar to triangle ABC and therefore C side, triangle For
if
B' be any other point on
AC
D7J//V c ~AB
and
~AHi~h' each of
for
similarly
~ a
/" x5C
T
the other
ratios.
The ratios differ
b. lg '
for different angles,
'
for
the ratios were equal, the corre-
if
sponding triangles would be similar (why?), and corresponding angles
which is contrary to the hypothesis that they are different. Since the ratios depend upon the angle for their values, they are functions of the angle according to the general definition of a -funcequal,
tion given in 5. Referring to Fig.
a 7 h
.
,
h
A
side opposite angle & ~~-
.
,
.
side adjacent angle
r
is,'
.
.
..
called the sine of angle
A
,
-
,
AA
;
.
.
.
f
AA
is
called the cosine of angle ft
,
.
side opposite angle ~
is
called the tangent of angle
A
.
,
that
,
the reciprocal of the sine,
,
the reciprocal of the cosine,
is
,
the reciprocal of the tangent,
is called
is,
,--
/.-
.
j
side adjacent angle
AA
is
,
-
,
,
,
.
f
called the cosecant of angle
A
;
called the secant of angle
A
;
h T
,
,
is
;
hypotenuse
h a
^ that ^
, 9
ab
7,
-
is,
name.
of these functions has received a special
hypotenuse
b
7
.
that
Each
.
A
;
the cotangent of angle A.
The six functions just defined are variously known as the trigonometric, circular, or goniometric functions: trigonometric, because they form the basis of the science of trigonometry; circular, because shown presently; goniometric, because of their use in determining angles, from gonia, of their relations to the arc of a circle, as will be
a Greek word meaning
angle.
The
expressions sine of angle A,
TRIGONOMETRIC FUNCTIONS OF AN ACUTE ANGLE
7]
II
cosine of angle A, etc., are abbreviated to sin A, cos A, tan A, esc or cosec A sec A cot A
Besides these six functions, two others are sometimes used,*
A =
versed sine
The
cosine A, abbreviated to vers
i
A =
coversed sine
A
.
,
,
sine
i
viz.,
A\
A, abbreviated to covers
A'.
definitions of the first six trigonometric functions
must be The first three memorized. are thoroughly especially important and should be memorized in the following form: Given an acute angle The sine
in a right triangle,
of the angle is the ratio of the side opposite the angle to
the hypotenuse.
The cosine of
the angle is the ratio of the side adjacent the angle to
the hypotenuse.
The tangent to the
of the angle is the ratio of the side opposite the angle
adjacent side.
The remaining three functions may be remembered most by the aid of the reciprocal relations,
A cos A tan A that
A =
cosec
sin
A =
sec
cot
readily
i,
i,
A
i,
is,
cosec
A= -
-
sin
It will aid the
sec
>
AA
A=
>
cos
memory
A
cot
A=
7
.
tan
A
to observe that only one o appears in each
pair of reciprocal functions. It should be noticed that while 0, 6, and h are lines, the ratio of any two of them is an abstract number; that is, the trigonometric
Again, the expressions sin A cos A are single symbols which cannot be separated, sin has no meaning except as it is associated with some angle, just as the has no meaning_except when 'used in connection with symbol
functions are abstract numbers.
tan Ay
,
,
etc.,
V
some quantity, as
EXAMPLE
i.
in
The
Va, v^,
etc.
sides of a right triangle are 3, 4, 5.
trigonometric functions of the angle *
A
opposite the side
Especially in navigation.
Find 3.
all
the
PLANE TRIGONOMETRY
12
The hypotenuse we have
Solution.
of the triangle equals 5.
ing the definitions,
--
side opposite
A=
.
sin
-:
n
[CHAP,
Hence, apply-
= a6
hypotenuse
The cotangent cot
A =
EXAMPLE
i
-5-
of |
:
The
2.
a
A= .
r
A
opposite the side
A -
A=
=
A
6; find
the
a. 2
ft
;
hence
2
--
= Va? +
b2 >
EXERCISE 1.
+
a and
$.
a
*
.
esc
J.
legs of a right triangle are
The hypotenuse h =Va?
Solution.
A =
tan
J,
A is the reciprocal of the tangent, hence = J, similarly, sec ,4 = i -^ J = |, esc 4 =
functions of the angle
sin
A =
cos
Similarly,
3
right triangle has its sides equal to 5, 12, 13. the angle A opposite the side 5.
Calculate the
six functions of
Ans. sin esc 2.
A = T53 cos A = A = -V-, sec A = ,
i
,
tan
}J, cot
A = fV, A = V-
In the same triangle calculate the functions of the angle
opposite the side 12.
Ans. sin esc
B = }J, cos B = ^, sec
J5
= &, = V,
tan cot
B
B = V, B = TV
the answers in problem 2 with those in problem B = 90, write down six equations, of i, and remembering that A which the following is the first: sin A = cos B = cos (90 A). 3.
By comparing
+
A = tan A = sec A =
Ans. cos
4.
Show
sin
cot esc
B= B = B=
sin (90
.4),
cot (90
A),
esc (90
^4), etc.
that for any acute angle the following equations are true:
sin (90
cos (90
tan (90 (Suggestion. right triangle,
-
A) A) A)
= = =
cos A,
sec (90
sin
A,
esc (90
cot A,
cot (90
A) = esc A, A) = sec A, A) = tan 4.
and 90 A the two acute angles of a and express the functions of each angle in terms of
Consider
^4
the sides of the triangle.)
TRIGONOMETRIC FUNCTIONS OF AN ACUTE ANGLE
7]
5.
The two
a right triangle are 8 and
legs of
13
Write down the
15.
functions of the angle A opposite the side 8; also the functions of the angle B adjacent the side 8.
A = B=
Ans. sin sin 6.
By
using the results of problem sin
A = A .
cos 7.
One
A
A
tan
Compute the
|f, cos
A = -r A
.
.
cot A.
A = |?, cos A = l9 = H, sec A = V
esc ^4 8.
Two
,
legs of a right triangle are sine,
,
side.
Ans. sin
Find the
tan
A = T8y B = V-
9 and the hypotenuse is 41. included between the hypote-
is
A
functions of the angle
,
tan
show that
5,
sin
side of a right triangle
nuse and the given
,
cos -
,
and
A = H, B = fr
8 T T cos
and tangent
cosine
p
2
2
and
,
the angle
of
2
tan cot
.
respectively.
/>
B
A = V> A = ^
opposite
the
side 2 pq.
Ans. 9.
Given
rfna-^.co.*- ^.tana-y^i.
sin
A =
$, find
vers
^i
and covers A
Ans. vers
A =
.
1,
covers
/I
=
J.
EXERCISE 4 1.
Compute
all
the functions of 45.
45= esc 45=
Ans. sin
(Suggestion.
45= sec 45=
cos
N/2
= =
0.707, tan 1.414,
cot
45= 45=
i, i.
Construct a right triangle having an angle 45, and
denote each of the equal sides by 2.
i v/2
Compute
a.)
the functions of 30. Ans. sin J, cos
all
30=
esc 30
=
2,
30=
sec 30
V
\
=
3j
tan 30
x/3, cot
I
30=
v^.
If one angle of a right triangle is double the other, (Suggestion. then the hypotenuse is double the shorter side. Call the shorter
side a.) 3.
By measurement
find the functions of
15.
15= 0.26, cos 15= esc 15= 3.87, sec 15=
Ans. sin
(Suggestion.
By means
0.97, tan 1.03, cot
15= 15=
0.28, 3.73.
of a protractor, or otherwise, construct an
PLANE TRIGONOMETRY
14
Draw
angle of 15.
sin
A =
Given cos
A =
Given
4.
and compute the
sides of the triangle,
,
,
Given tan
7.
Show
A
45
35'.
A =
48
11
50
12'.
; .
f construct and measure the angle. ,
Ans.
A =
that sin
A = A .
cos
A
A =
construct and measure the angle.
Ans. 6.
ratios.)
f construct and measure the angle.
Ans. 5.
n
a perpendicular to one side forming a right
Measure the
triangle.
[CHAP,
cos
.
tan A,
A = A
r T sin
A
cot A*
being any acute angle.
Show
8.
that sin 2 ,4
+cosM =
yl
9.
A =
2
find all the other functions.
./$,
esc
Show
.
2
.}
sin
Ans. sin
10.
2
i
2
(Suggestion.
2
i
2
Given
i,*
+ = sec ^!, = esc A cot A + Remember that a + b = h tan
2
A = / cos .4 = yt = -VS sec A = ff ,
= ?\, A = V-
JJ, tan ,4 |J, cot
that as an angle increases from o toward 90, its sine, increase, while its cosine, cotangent and cose-
tangent and secant cant decrease.
n. Show that every
sine and cosine is a proper fraction, while the and tangent cotangent may have any value large or small. 8.
Trigonometric Functions Determined Approximately by There are various ways of computing the trigono-
Measurement.
metric functions of a given angle. The results of such computations for the sines, cosines, tangents and cotangents of angles between o and 90 have been put together into tables, known as tables of natural functions.
We
shall learn
how to calculate them. know how approximate values of and
later
how
to use such tables
It is of value to the beginner to
the functions
may
be obtained
graphically. *
sin2
A
means (sinX) a tan 2 A means (tan A) 2 A = (tan 4) n etc., except when n =
n tan n (sin A) ,
tan-M,
etc., will
,
,
be explained
later.
,
etc., i.
and generally
sin w ^4
=
The meaning of sin- 1 A,
TRIGONOMETRIC FUNCTIONS OF AN ACUTE ANGLE
8J
EXAMPLE
Find graphically the functions
i.
By means
Solution.
of
25. an angle
of a protractor construct
MON =
25. Take OB any convenient length, say 10 inches, and from O as a center , and with
OB
as a radius describe an arc
ON
in C.
Draw CE and BA
BC
^
cutting
perpendicular
find CE = 4.23 = 4.66 inches, B A 9.06 inches, = 10 inches. Hence = OB OC and by construction
to
15
OB. By measurement we
= inches, OE
sin 25*
cos
=* =
^
=
tan 2 5
25= ^-- -
0.4,3, esc
9.06 - = 0.906, = OE 7^ = 2L10 v OC _ - BA = 4.66 " 466 05 IT -
2
sec 25
Observe that we have used two different
t
=
C0t 2S
'
Fi S- 9-
2.364,
=
1.104, *'
cos 25 i
-
tan
=
2 I4 S'
5
triangles, the triangle
COE
and their reciprocals, and a second triangle AOB to obtain the tangent and cotangent. This was done in order to have in each case TO for a divisor. If for instance we had used the triangle AOB only, we would have had to obtain the sine, cosine
sm
25
= BA- = OA
4.66
=
^ .4.23 -as above. 0.42$, instead of .
I0
EXERCISE 1.
.
.
11.03
5
Obtain by measurement the sine, cosine and tangent of 40. Ans. sin 40= 0.643, cos 40= 0.766, tan 40= 0.839.
2. Find the sine, cosine and tangent of 35. To avoid constructing the triangle and measuring the necessary lines, we may make use of
the diagram facing page 16.
and the measure
Hence
sin
35=
Similarly,
of
AB =
^
=
~
sin 35
57.4
may
= AB
^r
be read
.
Now OA = OR =
off
on the
0.574.
OB
35
-Q3 --a** 81.9
100,
vertical scale.
PLANE TRIGONOMETRY
16
To
find the tangent
it is
better to use the triangle
= 3.
With the
[CHAP,
=-^- =
n
OPR, from which
<
aid of Fig. 10 find each of the results given in the
following table. (Suggestion. Choose your triangle so that the denominator of the fraction equals 100 or some other integer.)
Explanation of table. For angles in the left-hand column the names of the functions appear on top, for angles in the right-hand column the names of the functions appear at the bottom. Thus the number 0.423, which is in the column headed "sin" and has 25 to the left of it, is the sine of 25; the same number being in the column " " cos at the bottom, and having 65 opposite it in the which has right column, is also the cosine of 4.
By
0.707, both as sines
Ans. 0.174 5.
By
65.
use of the table, express the numbers 0.174, 0.866, 0.643,
=
and
sin 10
cosines.
=
cos
80, 0.866
=
sin
60
=
cos
30,
etc.
use of the table, express the numbers 0.364, 2.145 an d i.ooo
both as tangents and cotangents. 6.
Every number
in the second
and third columns
is
the sine of
one angle and the cosine of another angle. What relation do you observe between each pair of angles? Every number in the fourth and fifth columns is the tangent of one angle and the cotangent of another. tween each pair of angles?
What
relation exists be-
TRIGONOMETRIC FUNCTIONS OF AN ACUTE ANGLE
g]
examining the table, verify the following statements and, you can, give reasons for them. a. Every sine and cosine is a fraction less than unity.
By
7.
if
17
Every tangent of an angle less than 45 is a fraction less than i. Every tangent of an angle greater than 45 is some number
6. c.
greater than unity.
The
d.
and tangent
sine
an angle increase as the angle
of
in-
creases.
The
e.
cosine
and cotangent
an angle decrease as the angle
of
increases.
The
/.
sine
and tangent
of a small angle are nearly equal.
Given One of Its Functions, To Construct the Angle. In the last article it was shown how to find by measurement the func9.
We
tions of a given angle.
that
is,
how
now
i.
Solution.
Take
CB
consider the converse problem, of its functions is given.
when one
To
EXAMPLE pendicular
will
to construct the angle
construct an angle whose tangent is f 4 units in length and at C construct a perin units 3 length. Join A and B. Then CAB is the .
AC
CAB =
required angle, for tan
The
and
is
is
the angle whofee tangent
written in short
Fig. ii.
A =
tan' 1
!,
read in either of three ways: a.
b. c.
Similarly,
A A A
is
the angle whose tangent
is
the inverse tangent f
is
the arctangent f
if
y equals the sine of
y
=
sin
x
=
sin" 1
x }',
x,
then
x equals the angle whose sine
or
x equals the inverse sine y x equals the arcsine y.
or
.
.
if
then in words,
i-
expression
A is
=
TT,
y
is y,
is
f
.
is
f
PLANE TRIGONOMETRY
l8
n
[CHAP,
Corresponding meanings are given to the symbols
=
y
cos" 1 *, y
To
=
=
csc" 1 *, y
=
sec" 1 *, y
cot"*
1
*.
construct an angle whose sine is ^. in a line CD of indefinite length, construct
EXAMPLE
2.
Solution.
At a point C
a perpendicular CB 4 units in length. From B as a center, with a radius 7 units in length, draw an arc cutting CD at A. Join A and B. Then the angle
CAB
B
/^
4
the required angle, for
is
sin
CZ?
=
C4J5
=
Another way of constructing sin"^
4 -
is
u
AVv Fig. 12.
.
to construct a semicircle
a having a diameter AB 7 units long. From B as a center, with semicircle at C. the radius 4 units long, describe an arc cutting Join
A and
BAG
Then
C.
(Draw the
the required angle.
is
figure
and
give reasons.)
In the preceding example, find each of the other functions of the angle A.
EXAMPLE
3.
Solution.
Two
third side
is
sides of the right triangle
ABC
being known, the
easily found.
AC = VAB*- Be = V7 -4 =V33. 2
2
2
Hence :
sec
A =
A7
cos
= -=. =
i.
v 33 esc
tan^ =
0.821,
yl
=
218, cot
=
T .
J
s sm^l
^ = -4- =
A = -^ tan
^ 4
=
=
33 4
A
0.696,
=
1.436,
1.750.
Observe that the expressions, -
1 sin- 1 *, cos7
^,
,
tan- 1
7
-^, A
csc'^,
sec- 1 --^,,
4
A/33
V 33
represent the same angle.
all
EXERCISE 6 ^*
Construct an acute angle equal to A, when,
i.
4.
4 = i sec A = |. sin
2,
5.
4 = tan -4 =
tan
$.
3-
cos4
4.
6.
cos
^4
= =
0.5.
0.45.
TRIGONOMETRIC FUNCTIONS OF AN ACUTE ANGLE
IQ]
=
cos 4
7.
Read
10.
A=
-
sin
8.
A =
k.
.
k
in three different
B=
1
sin""
A =
tan
9.
n
19
!,
tan""
ways the
1
=
x
3,
expressions,
esc" 1 3.5,
=
y
cot" 1 \/3.
Construct and measure in degrees each of the following
11.
angles,
Ans.
A = ^4
sin" 1
= 19.5;
Given
12.
=
-|,
sin
cos~ 1
i C=tan~4,
A =
f
;
sec
=
5
A = -7- =
tan
,
VS
B = V;
-= =
A =
$
A =
sin" 1
15.
Show
-J;
=
Ans. sin^l
1 6.
17.
18.
sin"
If
y
^1
=
esc
v5,
|
=
^4
=
= f
^7, cot
ff ,
find the functions of A.
cos
^,
T 7 = cos"
=
Show
A=
i ^3, tan
^4
sec
_
= ^^3,
esc
8
sin #,
1
= jc
cos- 1 f|
=
=
tan- 1
^4
5 = V.
=
2, etc.
A-
tan- y\ find ^ and y. Ans. x 1
show that x
=
cos" 1
1 graphically that tan" \
+
/=
!
tan- 1
1
Construct tan- ^ and tan"
= H,
1
\,
J
B
A
/
A =
,
ABC is,
Now
also cos
B=
,
therefore
A =
2 c
=
Cos
B=
008(90
Then
A and B
C
sin
g
be any
sides.
equals 90, that
and
^
measure each and add
angle plus angle are complementary angles. sin
= A-
= 45.
Functions of Complementary Angles. Let C the right angle, and a, ft, c the three
B
y
tan
right triangle,
Fig. 13.
J
that
(Suggestion. the results.) 10.
$ N/$,
find the other functions of the angle B.
sin- 1 -ft 1
cot
\5,
Ans. sinjB =1?, cosl? 14.
J.
to find the other functions.
3
tan
cot- l
=70.5; C=i8.5; Z)=7i.5; approximately.
Ans. cos A
13.
=
Z)
A).
PLANE TRIGONOMETRY
20
[CHAP,
n
Similarly
cos
4=
tan
A=
=
sin
B
=
cot
B=
cot (90
~ = tan a
B=
tan (90
c
f
=sin
o
cot
A =
sec A
=
esc
=
v4
7 6 -
A),
(90
-
-
A),
A),
=
esc
B=
esc (90
4),
=
sec
B=
sec (90
A).
a
These results may be put in words as follows: The sine of an angle equals the cosine of the complement of the angle. The cosine of an angle equals the sine of the complement of the angle. The tangent of an angle equals the cotangent of the complement of the angle, and similarly for each of the remaining functions. If
we arrange
the six functions in three pairs,
viz., sine, cosine;
tangent, cotangent; secant, cosecant; and call either function of a pair the cofunction of the other, we may express the six rules just
given by a single rule, namely:
Any the
cofunction of
complement
By means pressed as
an angle
of that
EXAMPLE. Solution,
= =
A =
cos (90
A =
tan (90 cot
*
The term
-
sin (90
problem
A =
= = 80) 75)
cos 15, sin
10,
etc.
tan 8 A, to find one value of A. A), and from the condition of the tan 8
A
.4)
=
tan (90
therefore
ex-
of
Given cot cot
corresponding function of
any function of an acute angle can be an angle less than 45. Thus
of this rule,
cos 80
to the
angle*
some function sin 75
is equal
,
tan 8 A.
was not used until the beginning of the lyth century. Beterm sine of the complement (compliment! sinus) was used instead a contraction of which gave rise to the present name cosine. Similarly cotangent is a contraction of complimenti tan gens and cosecant of complimenti secans. The cosine
fore that time the
abbreviations sin, cos, tan, the i8th century.
etc.,
did not
come
into general use until the middle of
TRIGONOMETRIC FUNCTIONS OF AN ACUTE ANGLE
u]
This last equation
21
when
is satisfied
-A=
90
A =
Solving
8-4.
10.
NOTE. This is not the only value that A may have. After the definitions of the functions have been extended to angles greater than 90, it will be seen that 30, 50, 70, etc., are other values of A satisfying the equation cot A tan 8 A*
Functions of o, 30, 45, 60, 90. There are certain values which the values of the functions may be easily
11.
of the angle for
determined exactly. a.
line
The functions and a line
of
OX
center.
o.
OP
A
Let
be the angle formed by a fixed
of constant length h rotating about
as a
Let b be the base and a the altitude of the triangle formed by dropping a perpendicular from P to OX.
As
x lg *
I4
the angle increases, a increases and b deand vice versa, as the angle decreases
creases,
"
a decreases and b increases. As A approaches o, a approaches o and b approaches h, hence in the limit
o= 7=0,
sin
cos
o=
h
b.
A =
7=1, h
tan
o= -
=
o.
n
The functions of jo. When the angle acute the other angle of the right tri30,
a-U
angle equals 60; the right triangle then forms one-half of an equilateral triangle. Each side of this triangle equals h, hence a, the altitude
N
\
N
N*
It
of the right triangle, equals -
*V -
sin
*
/
c.
^^ cos
A
|,
Fig. 16.
^0== 30=
-
x-
is isosceles,
\
V 3,
When A
The functions of 45.
right triangle
v '.A
,
.
|^2= % h >/3, and we have
,,0^ \ h ^ = 30=
*
and the base
that
is,
45 the a and b are equal,
and we have __
A2
=
a2
+a = 2
2
a2
,
a
=
h
Vj = \h \/2,
so that
PLANE TRIGONOMETRY i$.
It
is
[CHAP,
ii
shown in geometry that if the radius of a circle is divided and mean ratio, the greater segment will be equal to the
into extreme
chord subtending an arc of 36; that
AO
is
divided at
B
in Fig. 18
is, if
AO OB = OB
so that
:
:
BA,
then the chord AC, taken equal to OB, subtends
an angle COA equal to 36. Put AC = OB = x, and AO = r, then CDA is a right triangle whose short side is x, whose hypotenuse is 2 r, and the angle CDA = 18.
Hence
sin 18
= 2r
,
where x
is
the equation r
Solving,
we
:
x
the positive root of
=
x
:
r
x.
=
find the positive value of #, x
r
J
(
+
Hence 2 r
Show now
that
=
cos 18
J
VIQ
+ 2V/5, ,
12.
Fundamental Relations. In any
By 6 Fig. 19.
2 ,
2+ (f)
sec 18
=
-
cot 18
=
VS +
finally
2==i (!)
'
>
2
right triangle (Fig. 19)
dividing this equation
and
~T=
by a
2 ,
we
first
2 by A then by ,
obtain in turn
or
(2)
-Y =
or
(3)
^* (i), (2),
and (3) are called the square relations of the trigonometric These together with the three reciprocal relations
functions.
smA csc^l = cos A -sec A = tan A' cot A =
x,
(4)
x,
(5)
i,
(6)
TRIGONOMETRIC FUNCTIONS OF AN ACUTE ANGLE
13 ]
25
constitute the six fundamental relations of the trigonometric funcTo these, two others are usually added, viz., tions.
A=
ton A tan
-
Sfa^-
cos^
cotf
,
A=
COS.! sin
f
,
(7) w>
A
Proof.
-Abo a
sin
cos
A = :
h 7
=
a 7
=
,
A tan A, and .
cos.4sin
/I
=
i
tan
-
A
=
,
,
cot A.
h 13.
To
express each of the Functions in
Terms
of a
Given
One. a.
in
Analytic method.
terms of the
sine.
EXAMPLE
i.
To
express each of the functions
PLANE TRIGONOMETRY
26
EXAMPLE
Geometric method.
b.
To
i.
[CHAP,
n
ex-
of the press each of the functions in terms sine.
In the right triangle
ABC
(Fig. 20), Fig. 20.
sin
A =
sin
A
"DS~i
hence
;
AB
if
chosen for the unit of measure,
is
=^=C, and AC - \/ AB* - J5C
= Vi -
2
sin2 ,4.
i
definitions of the trigonometric functions
Then from the
EXAMPLE
-,
To
2.
ten
A.
-
express each of the functions in terms of the
tangent.
In Fig. tan
A= A= A
.
tan
21,
;
/>G
A C is chosen for the unit of
hence if
measure,
T>,
=J5(
i
Then by
Fig. 21.
sia
A
definition
BC
tan
A
.etc.
AB EXERCISE 8
i.
From
(i)
sin
show that
A =
cos
4)
(i
4)
sin
(i
(i
cos2
N/I
(2)
y
+
A = Vi sin2 2 sin A) = cos A,
cos
sin2
A
cos
i
From
(i
+ cos ^4) = sin
2.
A
cos
A,
A
A, sin
i
A
+ sin A cos
+ cos A
i
A
sn
show that
secyl
=
v/i
+ tan
(sec A
tan
2
/!,
tan
.4
.4
+
.4) (sec
= VsecM tan A)
=
i.
i,
A
TRIGONOMETRIC FUNCTIONS OF AN ACUTE ANGLE
14]
3.
Show
that
= Vi + cotM,
cscA
(esc A 4.
27
From
(7)
cos sin
cot
= VcscM + cot A) = i.
cot4
(esc A
A)
i,
show that
A A
tan sec
A = A =
sin
tan
A A
A
cot
cos A
esc
sin
,
,
A = -4 =
cos A, cot A.
5.
Express in words the relations given by formulas
6.
Use the analytic method to express the tangent
(i) to (7).
in
terms of
sine
and the
the cosecant; in terms of the cosine. 7.
Use the geometric method to express each the
cosine in terms of the tangent.
In the following exercises compare your results with those given in the table 8. 9.
on page
35.
Express each of the functions in terms of the cosine. Express each of the functions in terms of the cosecant.
10.
Express each of the functions in terms of the secant.
11.
Use
12.
sin 15
sin
Ans. cos
=
30
=
\
\ to find each of the remaining functions of 30.
^2
N/3; find the other functions of 15.
15=! ^2+^3, CSC 15
tan
15= 2-^3,
=
\/2
2
+ \/3,
sec
is=2 ^-V^, = 2 + V3
COt 15
.
13//_
IA /~
-\
where
prove that
14.
cos*
Reduction
= of
s
=
(
and
Trigonometric Expressions to their Simplest
Form.
Like algebraic expressions, expressions involving trigonomay frequently be reduced to a simpler form. As a of the rule the reduction is most easily effected by expressing each the and sine of terms in the functions which occur in expression
metric functions
resulting expression like any algebraic cosine as two separate quantities. and sine the cpcpression, treating be result In the end the expressed in terms of whatever may again the simplest form. Of course result the function or functions give of a single function, say the terms in one might express everything
cosine
and by reducing the
PLANE TRIGONOMETRY
28 sine,
[CHAP.
11
but that would introduce radicals for every function except the By using both the sine and cosine radicals are avoided,
cosecant.
unless, of course, the expression involves radicals to begin with.
EXAMPLE and
cosine,
Reduce the expression
i.
Substituting for cot
Solution.
A
+
A
sin
A
cot
cos A.
value in terms of the sine
its
we have
4
sin
cos 4. +2^4 sin A
Reducing to a common denominator,
A
sin
+ cos A cos A
A
sin
sin
Now by
sin
EXAMPLE
sin
A
sin
and
_
+ cos
A
sin
A =
2
tan
A
+ cos
2
A
A
+ cos
4
A
.
+ 2 sin A cos A.
A and
cot
A
their values in terms
we have
+ 2 sin
A
A
cos
A
2
A
A ^ (sinM
cos2
A
sin
= cos
4
esc
A
sec
+ cosM) A
2
cos A
A.
Reduce the expression
3.
(cos x cos y
sin
#
sin y)
+
z
# cos y
(sin
+ cos # sin y)
2 .
Squaring the expressions enclosed in parentheses,
Solution.
x cos y
sin
x
sin y)
=
2
cos2 x cos2 y
+ (sin
finally,
sm A
A
sin
(cos
hence
;
-
cot
cosine, (7), Art. 12,
sin
EXAMPLE
i
A
A
= esc A + cot A cos A = sm A
A
cos 4
2
sin
Substituting for tan
Solution.
+ cos
A
Reduce the expression
2. 2
of the sine
2
A 2
(i), Art. 12,
__ sin
# cos y
+ cos # sin y) = 2
sin2
x
sin2 * cos2
+ cos
2
Adding the right-hand members 2
2
cos y (cos x
+ =
cos2 ?
+ + sin
+
two
sin2 y (sin2 2
cos x cos y sin x sin y
2 sin # cos y cos # sin y y x sin 2 y.
of the last
sin2 x)
2
sin2 T,
? =i.
x
expressions,
+
cos2 x)
we have
TRIGONOMETRIC FUNCTIONS OF AN ACUTE ANGLE
i4]
2Q
Complicated expressions, involving a single angle, may be most easily reduced by putting for each function its value in terms of the The resulting expression may then sides a, b, h of a right triangle. be reduced like any other algebraic expression. Of course, the re= a2 b2 may be made use of whenever an advantage is
lation
W
+
gained by
it.
EXAMPLE
Reduce the expression
4. 2
sin
Solution. sin
A
+
A
tan
cos2
A
cot
.4
+
2 sin
A
A
cos
.
Putting
A =
~
cos
,
A =
h
-
tan
,
A =
h
~ b
cot
,
A =
,
a
we have a_?^
ab
+ ^4Va
b
=
+ * + 2 aW = 4
a*
hh
( a2
tan
+ cot
^4
The preceding methods
vl
'
2
*
2
6
)
aM2
abh*
=
+ &2 = a + b = a
-|
6
~""afc
a
.
are perfectly general, but frequently
it is
of
advantage to use other expedients. Any one of the seven relations in Sometimes the denomiArt. 12 may be employed in the reduction. nator
may be
the fraction
by a binominal factor
tan A, esc pression
A
+
cot
A
cos
A
,
i
;
+ sin form
A
,
sec
of
A
of the ex-
operation which does not change Radical expressions should be avoided
may be changed by any
whenever
possible.
EXAMPLE
5.
Reduce the expression
sin 2
A.
Solution.
i
= EXAMPLE
6.
i
sin 2
x
i
Greek
-
+ cos x = + cos x
--
sin 2
i
since
i
cos2 x
-+ -
x(i *
;
cos x
+ cos Xj
x
cos x
i
=
cos ;
cos^
i
a)L
x
sin 2 x.
cos0*
Reduce sec
*
like i
in short, as in algebra, the
the value of the result.
,
removed by multiplying both terms
simplified or
letters are frequently
+ tan
used to represent angles.
For the benefit of
the Greek letters and their names, the Greek alphabet has been printed in the front part of this book. The Greek letters are written as they are printed.
those students
who do not
already
know
PLANE TRIGONOMETRY
30
[CHAP,
n
tan
0,
Multiplying both terms of the fraction by sec
Solution.
we have cos0
= =
cos0
^
+ tan
sec
cos
(sec
sec0 =
since cos0
sin0,
0- tan 0) _ 0- tan 0)
since sec2
tan 0),
(sec
i
(sec
* t
+ tan 0)
(sec
cos
tan
=
2
2
i
tan0
cos0
i,
0- tan0) - tan
(sec
sec2
=
sin0.
EXERCISE 9
Reduce the following
*A A
IB
_JL_.
.
3.
esc 2
tan 2
x
,
Am
*HU1_.
+ cot 0) sin cos cos x tan s + sin x cot #.
,
-4$. sin x
8.
cot0H
9.
sin 4 J5
^ 4
2
cos2
5 sin2 5. Ans.
.
A
sin
i
x
(sec
4
+ csc4)
cos
.4
(sec
4
(x
sec
|8
4. sin^l)
|8
esc
(secA
2
-
16.
17.
+ tan^, + cot 5 cos #) # + cos x) + (sin (sin 2 (sin + cos 0) 3 (sin + cos 6
6
2
sin #.
esc A.
Ans
t
cos A.
an^tan5.
A
2
15.
A
^4w5.
tan^l).
tan^ cot
+
i
4*w.
2 sin 0).
(i
csc0.
csc.4).
sec
cot
i.
cos #.
4ws.
>T
sin
+
^4fw.
--.
+ cos B +
cos i
I3>
.
Ans. cos a.
tan a sin a.
a
7.
12.
x.
+ cos:y -4i.
0.
(tan
I0
sin>
x>
tan y
sec
n.
4ws. tan2
i.
jw*2
6.
Am
sin^i+cos^ sec A esc -4
i
5.
Ans. tan A.
cot2 A
sin
2
expressions,
vers
A
vers2 A.
2
a;
4
Ans.
.
4
0)
+
i.
2.
Ans.
o.
Ans. sin 2
A
TRIGONOMETRIC FUNCTIONS OF AN ACUTE ANGLE
IS]
A
esc .
i
+ cosJS
(cos
22.
(x cos
2 esc
5 2 cot2 #.
+ sin # sin y)*+ (sin # cos y cos sin a) + (x sin a + y cos a) sinM - cscM + cscM cofA
a;
sin y)
2
2
a
secM
2 .
<4ws.
.
;y
4s.
i.
-4 ns.
i.
#2
+/
.
'
A
sec2 /i
-
sin
i
+
sin
V
sin
# cos y
21.
6
A
cos 4 #)
esc 4 x (i
20.
sec
31
sin 2
X A
4 - esc A 2
cos2 4
4^
.
sec
4-
tan A.
Trigonometric Identities. Equations which express general relations, that is, equations which remain true no matter what values be given to the quantities which are considered variable, are 2 = 2 x + i is an identity, because it is x2 called identities, i) (x 2 6 = oisnot an be value true no matter what $x given to x. x 15.
+
+
+
identity, since 2
sin
x
+ cos
2
it is
x
=
not true unless x has the value
and tan
i
A = ^S_ cos A
2
or 3.
Similarly,
a re identities, for they ex-
.
which hold true no matter what x = J is not an \ sin x = sin" 1 \. = x when that x sin when it for is true is, only identity, To distinguish equations which are not identities from those which press general relations, relations
or
A may
On
be.
the other hand, sin2 x
,
former are sometimes called equations of condition, since they express conditions to which the variables are restricted and not general relations as do identities. are, the
The fundamental
12. trigonometric identities are given in Art. may be derived from these by
All other trigonometric identities
it is sufficient properly combining them. To prove a given identity to reduce both sides to the same form. If no shorter way suggests each side to its itself, this may always be accomplished by reducing
simplest form.
EXAMPLE
i.
(i
Solution.
tan
4 cos
Prove the identity
-
tan4)
(i
cot A)
Putting
A
sin
A
+ sec A esc A
=
2.
PLANE TRIGONOMETRY
32 the left-hand
[CHAP,
n
member becomes
(cos
A
cos
A/
sin
cos4/\
A
sin ^4) (sin
cos A sin
cos
A
2 sin
cos
A
A
sin
+
cos A)
A
i
A
+
cos2 A
sinM
cos .4
A
i
A'
sin
__
baud
cos^l sin^l
If
we had put
tan
A =
~
,
cot
A
.= -
A =
-
A=
-
the
re*
of the equation separately,
we
sec
,
esc
,
,
duction would have been as follows, /
_a\/
1
b\
^A
V
(&-a)(a-6)
fe&_r
.
1
ba
a)
= 2ab- dt-tf +
sin
2. 2
x tan #
Solution.
Show
k?
=
2jo6
2
#
=
2
tan # sec #
Reducing each member 9
a-
sec 2
Likewise the second
2
that
+ esc x sec
esc
^
ab
have x+ sm x Atan201
2
ab
a6
EXAMPLE
+A
x
= sm
2
o:
sin -jc cos2 x
+ esc x
-i
+ -sini x cosr,
2
sin #.
-
sin 4
*" -:
sin
oc
+1
#
2 x cosT" #
'
member becomes
-~-- sin* cos x
cos x 2 sin2
x
+ cos
2
sin
jg
#
sin2
x cos2 a
x cos2 #
2sin2 #+(i sin
sin
sin2 x) cos2
sin2 a) (i
sin
^ X+ 4
sin
I
x cos2 *
^
&
x cos2 #
x cos2 x
sin* s)
TRIGONOMETRIC FUNCTIONS OF AN ACUTE ANGLE
IS]
EXAMPLE
Is sec
3.
A
2
2
esc
A =
tan
2
+ cot
A
2
A
33
+ 2?
Solution. sec2
A
esc2
Hence the
A = = = =
+ tan A) (i + cot A) + tan 4 + cot A + tan 4 cot 4 + tan A+ cot A + since tan A cot = tan .4 +cotM + 2
2
(i
2
2
i
2
i
2
.4
i,
2
identity
2
2
1.
2.
is true.
EXERCISE 10 Prove the following sin
i.
2 3.
.
2
A
5+
(sin
sin 4
sec
A =
cos
cot .4 cos
7.
=
cos 0)
A
2
_
A
i.
=1 + sin 2
=
cos2 #
4 sin
i
cot
B.
2 sin J5 cos
x
A
2
cos
=
cos
2 cos2
i 2
#
=
2 sin2
i.
4
yl
^4
3
#
0.
cot A cos + cos sin # cos x). sin x + cos # = (sin x + cos x) (i = + cos A) (tan + cot A) sec 4 + esc 4 (sin + cot sec0+ cscfl _ tan +
cot A
6.
sec 2
jB)
2
(sin e
identities:
2
cos 4 x
# 2
4.
2
3
.4
4.
.
i _. i
sec
tan
esc
9.
10. 11.
i
cot
i
sec 4 + tan 4 = i + 2 sec2 4^ tan2 4 tan a + tan = tan a tan (cot a + cot 0). 2 = 2 (i + sin*) (i -f cos*). (i + sin^c+ cos*) 4
4
.
/3
4
sec
.4
+
4 13. cos x
sin 4
*
=
tan
2
+ sin
2
i
cos2 x (i 2
cos *
15.
(cos a cos 0+sin a i
2
*sin y 4" sin *cos ^
14.
=
tan *) 2
sin
ft
sin2 ft sin2 7.
cos?)
2
+
(i
=
+ tan *). i.
(sin a
cos ft
cos a sin
ft
cos
2
34
PLANE TRIGONOMETRY
8
^
+
1
+ M
>
2
[CHAP,
+
s
-f
10 I
'55
55
c
'55
'55
> II
II
II
11
II
8 35
o u
ii
CHAPTER
III
SOLUTION OF RIGHT TRIANGLES BY NATURAL FUNCTIONS 16. Tables of Natural Functions. In the preceding chapter we computed the functions of 30, 45 and 60. Later we shall learn how to compute the functions of any given angle. Now we cannot remember the values of the functions for all angles, and the computation by which the values are determined is too laborious to be
repeated each time the value of a particular function is needed. For this reason the values of the functions for every degree, minute
and second from o
to 90 have been computed once for all and the tabulated in tables, known as tables of natural functions. Usually such tables contain the sines, cosines, tangents and cotan-
results
gents of angles differing by i minute; in some tables the angles From such differ by 10 seconds, and in still others by only i second. tables, the value of the sine, cosine, tangent and cotangent may be
found whenever needed, and conversely, when the value of a function is known, the corresponding angle may be found by use of the tables.
Secants and cosecants are not given directly by the tables,
may be found ^ since sec A =
but
cos 17. a.
indirectly from the cosines
A
,
and
esc
A
.
sin
A
and
sines respectively,
.
To Find the Natural Functions of an Angle Less than 90. When the angle is less than 45, the degrees are found at the
The head of the column and the minutes in the left-hand column. number, in the same horizontal line as the minutes and in the same column as the degrees, is that function of the angle whose name appears at the head of the column.
EXAMPLE
i.
To
We
find the sine of 26 31'. column in the table of natural sines
and follow 26 and headed is which cosines (specimen "page, p. 40) by down the column marked sine on top until we come to the number 4465 which is in the line beginning with 31'. The number 4465 Solution.
find the
considered a decimal
is
the required sine, that sin 26 31' = 0.4465. 35
is,
PLANE TRIGONOMETRY
36
[CHAP, in
When
the angle is greater than 45, the degrees are found at the column and the minutes in the right-hand column. The number in the same horizontal line as the minutes in the right-hand column and in the same column as the degrees at the bottom, is that function of the angle whose name appears at the foot of the column. b.
foot of the
EXAMPLE
To
2.
We
find cos 63
29'.
column in the table of natural sines and cosines (see specimen page) which has 63 at its foot, and follow up the column marked cos at the bottom until we come to the number 4465 which is in the horizontal line ending with 29'. The number Solution.
find that
4465 considered a decimal
the required cosine, that
is
cos 63
When
c.
the
angle
29'
=
is,
0.4465.
and seconds, the and minutes must be corrected
of degrees, minutes
consists
result obtained for the given degrees for the additional seconds.
EXAMPLE From
To
3.
find sin 28 46' 36".
28 46' 36"
Solution.
the table
we
=
28 46
.6'.
find sin 28
46'= 0.4812
sin 28
47'= 0.4815
difference for
i'
=
0.0003
The
angle 28 46.6' whose sine we seek is /fr the way between the two angles 28 46' and 28 47', hence we take for its sine the sine of 28 46' increased fi
! tf
by
of 0.0003
/fr
of the difference
~ 0.00018,
or 0.0002
between
sin 28 46'
and
we carry 4 decimal
if
sin 28 47'.
places only.
Hence sin 28 46'
EXAMPLE From
To
4.
61
Solution.
the table
13'
we
36"
=
find cos 61
24"
=
61
+ 0.0002 =
0.4812
0.4814.
13' 24". 13.4'.
find
cos6ii3 =
0.4815
14'
=
0.4812
difference for i'
=
0.0003
/
cos6i
iV of 0.0003
= o.oooi 2, or o.oooi if we carry four places only, and since
the cosine of 61 cos 61
13.4' 13'
must be between cos 61
24"
=
0.4815
o.oooi
=
13'
and cos 61
0.4814.
14',
SOLUTION OF RIGHT TRIANGLES BY NAT. FUNCTIONS
17]
37
Observe that in Example 3 the difference was added to the sine of the smaller angle, while in Example 4 the difference was subtracted from This
the cosine of the smaller angle.
is
because as the angle in-
creases the sine increases while the cosine decreases.
reason the difference must be added
to the
tangent
and
For the same subtracted from
the cotangent of the smaller angle.
The tangent or cotangent of an angle is found from the table of natural tangents and cotangents in exactly the same way that the sine or cosine is found from the table of natural sines and cosines.
EXAMPLE
5.
To
63
16'
we
find
Solution.
find tan 63
i6'32".
=
32"
63 16.5!'. the table of natural tangents and cotangents (see specimen
From
page, p. 41)
tan 63 tan 63
1
difference for
SJ times 0.0014
=
6.
From
To
1 6'
1.9868
=
0.0014
i'
32"
=
1.9854
^ our
places.
=
+ 0.0007
Hence
1.9861.
=
26 43-4!'-
the table (specimen page) cot 26
43'=
1.9868
=
1.9854
=
0.0014
cot 26 44' difference for
4
t
find cot 26 43' 28".
26 43' 28"
Solution.
1.9854
7'=
0.000747, or 0.0007
tan 63
EXAMPLE
=
1 6'
=
of 0.0014
i'
0.00065 or 0.0007 to four places.
cot 26 43' 28"
=
1.9868
-
0.0007
=
Hence 1.9861.
function of an angle interprocess of finding the value of a mediate to two consecutive angles whose functions are given directly Thus in Example 4, cot 36 43' in the is called
The
interpolation. while cot 36 43' 28", 44' are given directly in the table,
table,
and cot 36
which is not given in the table, was found by interpolation. In interpolating we assumed that the increase or decrease of the to the increase of the angle, that is, we assumed function is
what
is
proportional as the principle of proportional parts.
known
it is this,
Briefly stated
PLANE TRIGONOMETRY
38
For small changes in is
*
nearly
proportional
EXAMPLE hypotenuse
One
7. is
235.0.
[CHAP, in
change in the function of an angle change in the angle.
the angle the to the
angle of a right triangle is 25 48.5' and the Find the remaining sides of the triangle.
In the adjacent figure, let the given angle and c the hypotenuse. To find a we have, Solution.
A
represent
^
-
=
sin
A, or a
=c
sin
Flg 22 '
A.
c
=
c
2 3 5>
and from the
hence
To
a find
by
we
=
we have
235
X
0.4354
cos
A
or b
=
A =
sin
0.4354,
102.3.
have, c
^
=
table
235,
=
c cos
A
,
and from the table we have cos b
hence,
,
=
235
X
0.9003
=
A =
0.9003,
211.6.
use the relation 7 = tan A, that is, if our b results are correct, the quotient of a by b must agree with the value of tan A as found from the table.
To
check our
?
*?^3
b
211. 6
=
results,
we
0.4835, while tan
A
as given in the table
is
0.4836.
difference of i in the last decimal place arises from the neglected parts of the decimals. If we had carried out the work to five significant figures instead of four, the quotient of a divided by b would
The
have been 0.4836.
EXERCISE
n
All the functions called for in this exercise are found
men
on the
speci-
pages of natural functions on pp. 39, 40.
1.
Find
2.
Find cos 64,
sin
26, cos 27 sin 62
30',
tan 25 45', cot 29 59'. Ans. 0.4384, 0.8870, 0.4823, 1.7332.
30', cot
64
15',
tan 60
01'.
Ans. 0.4384, 0.8870, 0.4823, 1.7332. *
no say nearly, for no exact proportion exists, All we can say is, that when the in the angle.
We
matter how small the
change in the angle is in most cases are suffiwhich results of the gives small, proportional parts principle
change
ciently exact for practical purposes.
i7l
SOLUTION OF RIGHT TRIANGLES BY NAT. FUNCTIONS SPECIMEN PAGE
39
PLANE TRIGONOMETRY SPECIMEN PAGE
[CHAP, ui
SOLUTION OF RIGHT TRIANGLES BY NAT. FUNCTIONS
i8]
3.
Find
sin 28 30' 24", tan 61
41
60 30' 25".
24' 36", cos
Ans. 0.4773, 1.8349, 0.4923. 4.
5.
Find cos 25
50' 10", cot 28
Find sec 2535',
esc 25
25' 50", tan
27oo'3o".
Ans. 0.9001, 1.8471, 0.5097. Ans. 1.1087, 2.3256.
28'.
Solve the following right triangles: 6.
A=
Given an acute angle
26 15' and the hypotenuse c
= 35.0;
to find the remaining parts.
Ans. 7.
8.
Given Given
a and
c
A =
63
a
45',
=
15.5, b
=
63 40' 30" and the side adjacent, b
and check your
To Find
18.
B=
28 50' and the side opposite, a = 150. Ans. B = 61 10', b = 272, c
A =
= =
31.4.
311.
363.
Find
results.
the Angle Less than 90
Corresponding to a
Given Natural Function. a.
When is
angle
the function is one of the numbers given in the table, the found by taking the number of degrees which stands at the
head or foot
of the
column according
name
as the
of the function
appears at the head or the foot of the column in which the number is found, and the number of minutes at the left or right end of the
which the number is found according as the degrees have been taken from the top or the bottom of the column.
line in
EXAMPLE written at
Find the angle whose sine
i.
We
Solution.
find the
number 4633
head and 62
its
at
its foot,
is
0.4633.
column which has 27 which begins
in the
and
in the line
with 36' and ends with 24' (see specimen page, p. 39). Since the " appears at the head of given number is a sine, and the name "sin the column in which 4633 is found, we take the degrees from the top of the column and the minutes from the left of the line in which 4633 is
found, that
is,
the angle whose sine
EXAMPLE Solution.
bottom
2.
is
Find the angle whose cosine
This time the
of the
1 0.4633, or sin" 0.4633
column
in
name
which 4633
the angle whose cosine
is
is
=
27
36'.
0.4633.
of the function appears at the is
found, hence 1
0.4633, or cos" 0.4633
=
62
24'.
PLANE TRIGONOMETRY
\2
[CHAP. HI
6. When the junction is not given in the table, we find the corresponding angle by reversing the process by means of which we find the function when the angle is given.
EXAMPLE
3.
=
tan x
0.5492, to find x.
The number 5492
Solution.
not found in the table of natural
is
tangents and cotangents, but two other numbers are found (see specimen page), namely, 5490 and 5494, one of which is a little smaller, the other a little larger than the given
0.5490 0.5494
= =
number.
tan 28 46' tan 28 47'.
whose tangent is 0.5492, 47'. Applying the principle
It is plain, therefore, that x, the angle
somewhere between 28 46' and 28
we
proportional parts
x-
have,
46'
-
tan x
_
28 46'
~ 28" 2847' that
is
of
tan 28 46' - tan 28 46'
tan 28 47'
is,
#
2846'
_
i'or6o"
and solving
O.549O
0.5494
0.5490
_.
O.OOO2
__ 2
0.0004
4
_
I
2
for x,
x It is
0.5492
=
28 46'
+
of
60"
=
28 46' 30".
not necessary to go through all this work each time. All we is that the smaller of the two angles between
need to remember
which x
lies
must be increased by of 60",
where
D
is the difference (without regard to the decimal point) between the functions of the two angles between which x lies, d is the difference (without regard to the decimal point) between the function of the smaller angle and the given function.
EXAMPLE Solution.
4.
tan x
From
=
1.9887, to find x.
the table (specimen page)
= tan63i8 = tan 63 19' 1.9897 = D 19897 19883 = 14, d = 19887 - 19883 = 4, x = 63 18' + A of 60" = /
1.9883
63 18'
if.
SOLUTION OF RIGHT TRIANGLES BY NAT. FUNCTIONS
IQ]
EXAMPLE
cos x
5.
From
Solution.
=
0.4767, to find the table
0.4769 0.4766
D= d x
= =
4769
4769
-
6i 3 i'
43
x.
= cos6i3i' = cos6i32' 4766 4767
+
=*
=
of
3, 2,
60"
=
61*31' 40".
19. Accuracy of Results. When an angle has been obtained from a four-place table (a table giving four places ol decimals only), the number of seconds in the angle found cannot be relied upon with
This
certainty.
best
is
shown by considering a
special example, as
Example 4 above. Since the tangents are given to four places only, the actual value 1 8' is not necessarily 1.9883, but may have any value
of tan 63
between 1.98825 and 1.98835. Similarly the actual value of tan 63 19' may be any number between 1.98965 and 1.98975. Hence D, the difference between tan 63 19' and tan 63 18', is not necessarily 0.0014, but may be any number less
than 1.98975
1.98825 =0.00150,
and greater than 1.98965 Again, tan
x,
1.98835
=
0.00130.
four places only being given, may have any value
between 1.98865 and 1.98875, so that and tan 63 18', may be any number than 1.98875
1.98825
and greater than 1.98865
1.98835
less
= =
0.00050, 0.00030.
Hence, r:
D
of
than -3- of 60" 130
=
23"
and greater than -3- of 60"
=
12".
60" must be some number
less
150
The
conclusion
uncertain
by
is
that the result
at most 6".
63i8'i7"
previously given
is
PLANE TRIGONOMETRY
44
[CHAP, in
From the example just given, it appears that the amount by which the result obtained from a table is uncertain depends upon the difference D, which varies not only for different functions of the
No angle, but also for the same function of different angles. general rule can be laid down to cover the amount of uncertainty in all cases. If absolute certainty in the number of seconds is required, same
a seven-place table should be used, giving the values of the functions for small angles, and for intervals of 10" for
from second to second larger angles.
When a given
to the
four-place table is used, and no special consideration is nature of the differences involved, the number of seconds of
an angle obtained by
The
interpolation cannot be relied upon.
some aid to beginners: angle less than 45 can be obtained more accurately from a sine than from a cosine, while an angle greater than 45 can be obtained more accurately from a cosine than from a sine. 1.
following rules will be of
An
This
is
because the sines of angles
less
than 45 vary more than the
and at the same time the
principle of proportional parts is more nearly true for sines of small angles than for cosines, while the opposite is true for angles greater than 45.
cosines,
2.
Very small angles can be obtained with greater accuracy by interfrom tangents than from cotangents, while angles near po can
polation
be obtained with greater accuracy
The
from cotangents than from
tangents.
the fact that the principle of proportional parts ceases to apply to cotangents of small angles and to
reason for this
lies in
tangents of angles near 90.
EXERCISE 12 Find the following angles: 1.
sin~ l 0.4904, cos~ l 0.4904, tan" 1 1.8940, cot"" 1 1.8940.
2.
1 1 1 sin" 1 0.4267, cos" 0.4900, tan" 2.1036, cot" 0.5644.
Ans.
Ans. 3.
4. 5.
cot"
1
Show Show
25
15' 30",
29 22', 60 38', 62
10',
27 50'.
60 39' 30", 64 34' 30", 60 33' 30".
1 1 2.1441, tan" 0.4737, tan" 1.7611, cot" 1.7611. Ans. 25 oo' 15", 25 20' 45", 60 24' 40", 29 35'
1
+ sin"" 0.4488 + sin"
that sin" 1 0.4250
1
that sin" 1
1
0.9052
0.4746
= =
90. 55.
*<>''.
SOLUTION OF RIGHT TRIANGLES BY NAT. FUNCTIONS
20]
If
6.
tan" 1 0.5000
+
=
tan" 1 x
87 34', find
45
x.
Ans. x =1.8040.
In a right triangle one side b Find the included angle A. 7.
A =
Ans.
c
= 10.0.
62
25'.
In a right triangle the hypotenuse c = 35.00 and the side Find the angle opposite a, and the third side b. 31.29. Ans. A = 63 23', b = 15.68.
8.
a
and the hypotenuse
A = b/c.)
cos
(Suggestion,
= 4.63
=
Two
9.
the angles
sides of a right triangle are a
A and B and
the hypotenuse
A =
Ans.
63
Determine the uncertainty
10.
=
475.0, b
26',
B =
237.5.
26 34', c
number
the
in
=
Find
c.
of
=
531.1.
seconds
of
tan" 1 2.1211 as given by a four-place table, assuming the principle 2". Ans. tan" 1 2.1211 = 64 45' 30" of parts. proportional
20.
Solution of Right Triangles by Natural Functions. In order to solve a right triangle, two parts must be given besides the right angle, and one of these parts must be a side.
Let
ABC
be any right
the angles, and sides.
Fig. 23-
?=sin,4,
=
-
(i),
a,
cos^,
~=
(2),
C
tanyl, (3).
o
sufficient for the solution of
These three equations are
A, B,
corresponding
Then
c
c
triangle,
c the
b,
any
right
triangle, for, If
one of the given parts is an angle, we may must be either a, b, or c\
call this
angle A, and
the other given part I.
II.
III.
Given Given Given
If the
A and a; A and b\ A and c\
then
(3) gives b,
then (2) gives then (i) gives
c,
a,
and and and
(i) gives
c.
(3) gives a. (2) gives b.
given parts are both sides, there are three more cases,
Given a and b\ then (3) gives A, and c is found as in I or II. V. Given b and c\ then (2) gives A, and a is found as in II or III.
IV.
VI. Since
Given
c
and
A and B
the relation
a;
then
(i) gives
A, and
are complementary,
A
+
B=
B
B may 90.
is
found as in III or
I.
always be found from
PLANE TRIGONOMETRY
46
[CHAP,
m
It is not necessary to consider the various cases of right triangles separately, for the following simple rule governs all cases:
Employ that trigonometric function of the angle which involves the two sides under consideration. The two sides may both be given, or one may be given and the other required to be found.
EXAMPLE i. Given one side of a right triangle equal to 418.5, and the hypotenuse equal to 614.0; required the other parts. Solution. Denote the given side by a or b, let us say by b, and the hypotenuse by c. We then have * Given b = 418.5, Required A c
To
i.
We
=
=47
/
614.0.
02',
B =42 58', a = 449-3-
,
find A.
/
h *ve
b_
-
ms
Jf
cos
A =
'
-
= 4^-5 =
c
614.0
.68i6,
and from the table cos- 1 0.6816 2.
To
find B.
3.
To
find a.
From From
=
47 02'.
A+B = 90, hence B = 90
(3),
the table
tan
a = A =
-
47 02'
42
58'.
6tan^l. tan 47 02'
=
1.0736
6=
418.5
a
Multiplying,
Check.
=
=
449.3
To guard against possible
mistakes, the answers should be tested by some formula which has not already been used in obNow in solving a right triangle we never use taining the answers. 4.
more than two
may
of the three formulas
(i), (2), (3),
hence the third
always be used as a check.
used in finding A, and (3) by using (i), that is,
In the present problem, (2) was in finding a, hence we may test our re-
sults
if
our results are correct they should
satisfy the relation (i), or
= A = c = a
sin
Multiplying,
we
get
c sin
A
sin 47
02'
=
0.7318
614 a
=
449.3,
which agrees with the value of a as determined above. NOTE. We might have used the relation aa -f 62 =c2 as a check, but have required more work.
this
would
SOLUTION OF RIGHT TRIANGLES BY NAT. FUNCTIONS
20]
47
EXAMPLE 2. Given one angle of a right triangle equal to 36 50', and the hypotenuse equal to 3.12; to solve the triangle. Solution. Denote the angle by A, and the hypotenuse by c, then B
we have, Given
To
i.
A = c =
50',
3.12.
find a. Fig. 25.
From From
36
= = c =
a
(i),
the table,
sin ^4
c sin
A.
sin 36
50'= 0.5995 3.12
a
Multiplying,
To find b. From (2),
2.
From
the tables,
cos
b = A = c =
c cos
To
find B.
A
+B=
3.12
B=
90
-
36 50'
2.497
53
10'.
Having already used (i) and (2), we use a = b tan A
Check.
4.
.
b
90, hence
1.870
cos 36 50'= 0.8004
Multiplying, 3.
A
=
(3).
.
From
the tables,
A =
tan
tan 36 50'= 0.7490
b=
2.497
a = 1.870, get which agrees with the value of a as obtained above. Instead of referring to equations (i), (2), and (3), it
we
Multiplying,
is better to the triangle under consideration that ratio which needed to solve for a particular side or angle. The method will be
write is
down from
sufficiently clear
EXAMPLE
3.
from an example.
Given
= =
15
Required
25',
345-
Solution. "i.
To find 9.
= 2.
of
6
90
+ = 90,
0=
74^5',
n=
1251,
p
1298.
-15
25'
=74
35'.
and m are given, we must use that function which involves n and the given side m.
To
find n.
=
tan0,
Since
or
n=*mcot
= 345X3.6264=
1251.1.
PLANE TRIGONOMETRY
48 3.
To
find p.
We
use that function of
<
[CHAP, in
which involves p and
the given side m.
__
or sin<
p 4.
Check.
-
=
sin
If
6,
our results for or
n
=
0, n,
p sin 6
and p are
=
.
0.2658
1298
X
correct,
0.9640
we should have
=
1251.3.
In this case there is a slight discrepancy between the result of the check and the value of n as found in 2. This is due to the fact that
we have given the value The values
four places.
n to five places while the table gives but to four places agree exactly. of
EXERCISE Solve the following right triangles.
13
Check your
results
when the
answers are not given. 1.
= 10.00, A = 25; find b = 21.45, c = 23.66. b = 256, A =36 30'; find a, B and c. c = 350,^ = 56 45'; find B = 33i5', a = 293, b = 192. = = a 346, B 50; find the other parts. c = 45.7, B = 44 50'; find a = = 32.2. 32.4, b = =* b 28 40'; solve the triangle. 13.5, B a = 170, b = 350; find ,4 = 25 54', B = 64 06', c = 380, a = 0.81, c = 2.54; solve the triangle. 6= 6.57, c= 10.6; find yl = 51 42', = 38i8',a = 8.32.
Given a
2.
Given
3.
Given
4.
Given
5.
Given
6.
Given
7.
Given
8.
Given
9.
Given
t
Find the altitude of an isosceles triangle whose base is 368, and whose equal sides make an angle of 64. Ans. 294. n. Find the perimeter and area of a regular pentagon inscribed in a circle whose radius is 10. Ans. 58.78, 237.76. 10.
12. Show that the area of any right triangle is equal to either one of the expressions \ be sin or \ ac sin B.
A
13.
angle
In the right triangle
B = 40; show how
median angle
.4,
ABC, to
Fig.
27,
AB =
338,
the length of of the bisector of
find
the the length AS the and the angle included between these two.
AM,
Fig. 27.
SOLUTION OF RIGHT TRIANGLES BY NAT. FUNCTIONS
21]
14.
angle
ABC
An
B =
has oblique triangle 673o'; solve the triangle.
Ans.
152,
120,
BC =
and the
150,
A = 654i', C = 46 49'. A or C to the opposite
Drop a perpendicular from
(Suggestion. side, dividing
AC =
AB=
49
the triangle into two right triangles.)
15. Given one side of an oblique triangle equal to 57.3, and the adjacent angles equal to 35 45' and 753 0/ respectively; find the Ans. 59.5, 35.9, 68 45'. remaining parts.
Draw
(Suggestion.
given
the altitude from a vertex adjacent to the
side.)
21. Right Triangles Having a Small Angle. Given the hypotenuse c and a side b of a right triangle triangle in case b
and
The angles A and B cos
A =
Fig. 28.
,
to solve the
c are nearly equal.
are given -
=
by the
relation
sin B.
c
is apparent from the figure that if b is nearly equal to c, angle A must be very small and angle B must be nearly equal to 90. By examining the table of natural sines and cosines it will be seen that for small angles the cosines are so nearly equal that there is no difference at all in the first four places, and similarly the sines of
It
angles near 90 all
are nearly equal. Thus, so far as the table shows, to o 34' have the same cosine, and likewise all
angles from o
angles between o34' and o 60', between ioo and follows that a small angle cannot be accurately found
from
nor an angle near 90 from its sine. To avoid using the cosine, the following formula
used whenever
/
is
i
16', etc. its
It
cosine
the given parts b and c are nearly equal,
Proof. c the
Let
ABC
hypotenuse.
be any right triangle, b the base, a the Produce CA to O, making
AO = c. Join O and B. Now angle ,4 (= CAB) = angle AOB + angle ABO, hence since triangle AOB is isosceles, angle A = twice the angle AOB, that is, angle AOB = ^, and
c+b
c+b
Fig * 29 '
-*/#- = t/c ~* V(C + J)2 \c + b'
altitude,
PLANE TRIGONOMETRY
O
EXAMPLE
Given b
i.
Solution.
b
c
= =
c+b=
+A _
=
25.7, c
26.8; to find
26.8
25.7
26.8+
25.7
= =
V
-=
8
y
B, and
a.
52.5.
3.020952
2
A
m
i.i,
/i.i _
4
[CHAP,
=
0.1447,
52.5
4 =
14',
16
28',
2
= 90 - 16 28' = 73 32'. a = 6 tan ^4 = 25.7 X 0.2956 = a2 = c - b = (c - b) (c + b). 2
Check.
7.6.
2
(c- b)(c+b)^
X
i.i
5 2 -5= 57-75-
EXERCISE 14 1.
Given
6
=
A =
Ans. 2.
Given a
=
c
9.6,
=
Show
that tan
^ = 2
4.
From
22.
Fig. 29
c
~
13',
B=
64 47', a
=
2.24.
and check your
b .
a
show that
Historical Note.
25
solve the triangle
10.4;
results. 3.
solve the triangle.
=5.25;
4. 7 5,
The
sin
= 2T2C
use of
\/
^,
cos
tables of
2
= V V
'
.
2C
natural functions
In the second century B.C., Hipparchus, Greek astronomer and mathematician, constructed tables of chords (double sines) which answered the same purpose as a table of sines. The tables of Hipparchus have been lost. The oldest table now dates back to antiquity.
j
is that of Ptolemaus (second century A.D.), giving double from minute to minute with an accuracy which in our system of numeration would be expressed by five places of decimals. Hindu mathematicians as early as the fifth century A.D. were in much as we possession of a small table which they memorized, very
extant sines
memorize our multiplication table. Values not given in the table were computed from memory as occasion required, by means of a formula put in verse. The first table which approached in extent and arrangement the " " Canon doctrinae triangulorum of Rhetitables now in use is the
SOLUTION OF RIGHT TRIANGLES BY NAT. FUNCTIONS
23 1
51
cus (1551). This table gives each of the six functions for intervals of 10" from o to 45. Like present tables, the degrees and seconds proceed from top to bottom in the left marginal columns, while the
complementary angles proceed from bottom to top
in the right Later, Rheticus prepared a second table for which the sines of angles for intervals of 10" were computed to 15 " places of decimals, though only 10 places were retained in the Opus Palatinum," the name under which the table was published.
columns.
marginal
Rheticus' table contained numerous errors, which were largely rePitiscus, an indefatigable arithmetician of the seventeenth
moved by
In addition to revising the existing tables he computed from o to 7 for intervals of i' to from 20
century.
anew the
sines of angles
to 25 places of decimals.
under the
in 1613
title
Pitiscus's improved tables were published Thesaurus mathematicus." These tables
"
formed the basis of all subsequent tables, until the discovery of improved methods of computation in recent times has made it comparatively easy to check old tables or to compute new ones.
Review.
23.
Define trigonometry, explain the etymology of the word the science originated, (b) Define in words the sine, cosine, and tangent of an acute angle, (c) Define the secant, cosecant, and cotangent of an angle, (d) What is meant by the versine 1.
and
(a)
how
tell
and coversine
of
an angle
?
Name
three pairs of functions such that in each pair either (a) is the cofunction of the other. (6) Explain the origin of the terms and cosine, cosecant, cotangent, (c) Prove that sin A = cos (90 A) 2.
y
cos
A=
sin (90
A), tan
A=
cot (90
as a function of an angle less than
A),
(d)
Express sin 76 40'
45.
Construct the following angles: sin- 1 J, cos~ l 0.4, tan" 0.5, (b) Give from memory the values of the sine, cosine, and 1
(a)
3.
cot'
1
3.
tangent of each of the following angles: o, 30, 45, 60, (c) Draw a figure and deduce the functions of 30 and 45.
Name
(a)
4.
function sin*
A
three pairs of functions such that in each pair either
the reciprocal of the other, (b) Prove the relations 2 2 tan 2 A i = sec A, cos2 A = i, cot 2 A i = esc A, is
+
+
tan A = ^~ cos tions of
90.
A
.
.
.4
(d)
(c}
Given
sin
A =
+
^
,
find each of the other func-
5
Express each of the functions of
A
in terms of tan
A
.
PLANK TRIGONOMETRY
52 (a)
5.
Reduce
cos
(b)
6.
(a)
Prove the
sin
.
X
X X
cot
i
identities,
A
+ esc A
sec
A
esc
is
X
tan
sec
What
form the expression
to its simplest
i
[CHAP, in
__
A A
tan tan
^4
meant by a
+
i __ i i
+ cot A cot
i
^4
table of natural functions?
(b)
Which
functions increase and which decrease as the angle increases from o to 90 ? (c) What is meant by the term " interpolation" ? by " the (
accurately from
its sine
or from
its
cosine
angle be found more accurately from
gent 7.
II.
8.
?
Why ?
(/)
Can a
tangent or from
its
small
cotan-
Why ?
?
(a)
There are four
different cases of right-triangle
according as the given parts are: IV.
its
One
Two
side
and an
angle.
Show how
sides.
formula for
III.
to solve each case.
=
Prove the formula tan
B tan -
problems
The hypotenuse and an angle. One side and the hypotenuse.
I.
2
c ""
\/
C
+
,
and write down a
similar
.
2
To determine its flagstaff AB stands on top of a tower. was feet measured in a horizontal off a distance OP long 500 height, At P the angles OPA anc." direction from the foot of the tower. OPB were measured, and were found to be 22 and 28 respectively. 9.
A
Find the length of the flagstaff by natural functions, also by the graphic method, and compare your results.
CHAPTER
IV
LOGARITHMS Definition of Logarithm. The numbers representing the given parts of a triangle and other numbers obtained by careful measurement usually contain three or four significant figures and 24.
sometimes
five,
six,
or
it
may be even
seven significant
figures.
Multiplication, division, the extraction of roots and raising to powers of such numbers by the ordinary methods, require a great deal of
tedious labor, most of which
be avoided by using another method of logarithms. This requires the use of tables, by the aid of which multiplication of two or more numbers is accomplished by adding certain other numbers found in the tables. Similarly division is reduced to a mere subtraction of two numbers found in the tables. To raise to a power or to extract a root of a number, the corresponding number in
method method
of computation,
the table
is
may
known
as the
multiplied or divided
by the index
of the
power or
root.*
of logarithms presupposes that when some positive different from unity, has been chosen, every other positive can be expressed as some power (integral, fractional, nega-
The method number,
number
incommensurable) of this number. Thus, if a is some posinumber not equal to one, and N any positive number, we assume that a number x can always be found such that
tive, or
tive
ax This number x
is
=
N.
called the logarithm of
N
to the base a,
and
is
usually written
# in words: * The usefulness of the method of logarithms may be anticipated from the testimony of Laplace, the great French astronomer, who said the method of log" arithms by reducing to a few days the labors of many months, doubles, as it were, the life of an astronomer, besides freeing him from the errors and disgust inseparable from long calculation." The advantages which the use of logarithms offers to the astronomer are shared, of course, by all others who deal much with numerical calculation.
S3
PLANE TRIGONOMETRY
54
[CHAP, iv
The logarithm * of a number to a given base is the exponent of power to which the base must be raised to produce the number.
For example, since
= 100, 2 = 1000, 3 io* = 3.1623, \ i lo" = o.i, = 0.25, 2 (0.5)2 = 9, ! 2? 2 = =8 6 = Iog 2 64, 64, 10 2 103
1
2e
=
a
4
= = = = = = 3
logio
100
;
logio 1000; logio 3. 1623; logio o.i
;
logo.50.25;
log2 79;
=
Iog 4 64,
2
=
Fundamental Laws Governing Logarithms.
25.
the
logs 64.
Since loga-
rithms are exponents, the laws of logarithms are the same as the laws of exponents. Now the laws of exponents are,
a
(a)
=
av
00
x
is
y x
is
+
a"**,
the logarithm (exponent) of the first factor, the logarithm (exponent) of the second factor, y is the logarithm (exponent) of the product;
hence,
The logarithm of
the product of two factors is equal to the
sum
of the
logarithms of the factors, or
P = M*N,
If
log
P=
log
M H- log N.
Similarly, If
Thus
log 15
(b)
a*
-*
P = L*M*N = log 3 + log 5, a
17
x
is
v
is
.v
=
,
logP=logL+ logM + \ogN+ log
30=
log 2
+ log 3 + log
.
5.
a"-*,
the logarithm (exponent) of the dividend, the logarithm (exponent) of the divisor, y is the logarithm (exponent) of the quotient;
hence, *
From
logos
=
ratio,
and arithmos
=
number, so called by the Scotch mathe-
matician John Napier, one of the inventors of logarithms, because, as originally conceived of, logarithms were a set of numbers, as k,
//,
f,
/,
n,
etc.,
h a k a1 a m a n etc., corresponding to a second set a the ratio between any consecutive two that so chosen second set the being ,
same.
,
,
,
,
is
the
LOGARITHMS
26]
55
The logarithm of the quotient of two numbers is equal to the logarithm of the dividend diminished by the logarithm of the divisor, or,
Q = M + N, = log 5 -log 3. log $
If
Thus (a-r
(
x
=
log()
= logM-log#.
a,
the logarithm of the quantity which is to be raised to the wth power, nx is the logarithm of the resulting power; hence, The logarithm of any power of a number is equal to the logarithm of the number multiplied by the index of the power to which it is to be raised, is
or
P = Nn
If
Thus
Iog3
^P
(d)
x
=
is
5
logP
,
=
fllogtf.
=5lg3*
a",
the logarithm of the quantity whose nth root
is
to be extracted,
-is the logarithm of the resulting root; hence,
n
The logarithm of any root of a number is equal to the logarithm of the number divided by the index of the root to be extracted, or
P
If
_=
Thus
log N/I7
Logarithms of Special Values.
26. (a)
a1
=
a,
hence log a
The logarithm of (b)
-f-
a
= 0,
The logarithm of
By
Art. 25
article, log i
that
=
o,
=
that
i,
is,
the base is i.
Any number
nents a
(c)
log 17.
divided
by
so that a
=
i to
(6),
itself is i,
any base
logT;^
i,
but by the law of expo-
and therefore
log a
i
=
o,
that
log
i
log TV,
and by
(b)
of
this
hence
is,
The logarithm of logarithm of the
the reciprocal of
number.
is,
is o.
any number
is
equal to minus
PLANE TRIGONOMETRY
56
[CHAP, iv
Definition. The logarithm of the reciprocal of a number the COLOGARITHM * of the number.
is called
Thus logio
10=
logio 100
=
hence logio
2,
=
logio 3. 1623
A=
hence logio
i,
T&TF
-1 =
=
=
hence logio
,
cologio 10;
=
2
-"1
cologio 100;
*
00^103.1623.
3.1623
N N =M-~,
Since
we have
= logM + =logM + log^
log
that
colog
N,
equal
to the
is,
The logarithm of
numbers
the quotient of two
is
logarithm
of the dividend plus the cologarithm of the divisor.
EXERCISE 15 3
Given
1.
the base
5,
5
=
down
=
2
,
1
25, 5
=
5;
write
down
the logarithms to
rb-
^V,
0.5, Iog3 8 1
=
4, logio 3. 1623
equivalent expressions free
Find Iog3
3.
125, s
of 125, 25, 5,
Given Iog 4
2.
write
=
2
=
27, logio 10,000, log a a, Iog3
Express
4.
terms of log
log 15, log
+ log
Ans. log 3 2 (log 2
+ log 5),
Given
5.
,
=
*
log 2
5,
(log 3
logio 2
log
2,
,
and
log 3,
log
log 100, log
+
-
log 5
\^
log 3,
=
,
*
Whenever x
+y=
specifically arithmetic
=
i
0.60206,
+ 0.47712,! constant, ^
i
is
c\
etc.
-3,
f.
.
log
-
log 2
log 3
0.47712, Iog 10 5
^-2
-
log 3,
+ A log 5=
0.69897;
.
729
+ 0.87506, 0.23299, 0.12224.
and y are said to be complements (more
complements) of each other.
constant, hence log
3, 4, i,
log 5
find log] 4, logio 0.3, logio 0.75, logio ^S> logioV T yln5.
=
the following:
5,
+ log 5-log 2), A log 2 + J
0.30103, logio 3
a
Jr, loga a*.
Ans. in
0.5, log b
from the symbol log. Ans. 4' 5 = 2, 3 4 = 81,
Now
log
x
-{-log
= log i
the complement of log x, which on contraction becomes
colog x. t
part
When is
10
is
positive.
the base, the logarithm
is
always written so that the fractional
LOGARITHMS
27]
b2
=
If b
6.
=
184, c
57
59; find the logarithm of
\/ T
and
,
also of
C
Ans. 0.14434, 4.48251.
Show
that the fractional part of a logarithm, to the base 10, is 7. not changed if the number is multiplied by 10, or by a power of 10. 8.
Compute
to four places of decimals the
rithms to the base 10 are
J, i, J,
numbers whose
loga-
.
Ans. 3-1623, 1.7783, 2.1544, 1.4678. 27.
The Common System
of Logarithms.
Each
different base
determines a different system of logarithms. The system in common use has 10 for its base, and is called the common * system of logarithms. Common logarithms have been carefully computed and tabulated, so that the logarithm of any number can be readily found
by referring to the table, and, vice versa, if the logarithm of a number is known, the number itself can be found from the table. The advantage of the common system over other systems of logarithms consists in this: the fractional part of any common logarithm remains unchanged when the number is multiplied or divided by 10 or a power of 10. To see this, let us consider a special case,
say the number whose logarithm I0
-
=
5
10*
Multiplying by 10
X
1<5
Similarly
I02.s I03.r>
I0 4.5
= _ _ =
X
3.16228,
therefore log 13.16228
therefore log 31.6228
=
^6.228,
therefore log 316.228 therefore log 3162. 28 therefore log 31622.8
= 2.5; = 3.5; = 4.5;
3!62.28, 3x622.8,
we
io~ 1+0 6 = 0.316228,
-i
1.5.
etc.
X .oM-i*22 10
-
is
0.5.
have,
e. 10 10
* This system
=
3.16228,
31.6228,
Again, dividing by 10,
or
is 0.5.
10,
io- 5 = 10 io
or
= Vio =
log 0.316228
=
1+
0.5
=
9.5
10.
known as the Briggsian system, in honor of Henry Briggs who was the first one to compute and publish a table of
also
of Oxford (1556-1630),
logarithms to the base 10. t In the remainder of this chapter, and always when computations are concerned, if no base is expressed, the base 10 is understood.
PLANE TRIGONOMETRY
$8
[CHAP, iv
Similarly
= = log 0.00316228
.5
0.0316228,
0.00316228,
and so on, the form i and sign between
2
log 0.0316228
3
+ 0.5 = + 0.5 =
8.5
10;
7.5
10;
10 being introduced to avoid the plus
9.5 0.5.
Now
the logarithms of each of the numbers 3.16228, 31.6228, 316.228, 3162.28, 31622.8, etc., 0.316228, 0.0316228, 0.00316228, etc., have the same fractional part, namely, 0.5, while the integral parts *> 2 , 3> 4> etc.,
i,
-2,
3, etc., plainly
depend upon the position
of the decimal point.
For convenience the integral part of the logarithm characteristic of the logarithm,
and the
is
called the
fractional part is called the
mantissa.
Thus, the logarithm of 316.228
and the mantissa
is
composed
of the characteristic 2
0.5.
The logarithm of 0.00316228 is composed of the characteristic 3 and the mantissa 0.5. We have seen that the mantissa is independent of the position of the decimal point, that is, it is the same for all numbers composed of the same figures taken in the same order. 28.
Rule for the Characteristic. We shall now learn how the number may be determined before
characteristic of a logarithm of a
the logarithm
is
known.
10 10
1
102
103
= = = =
Since
i,
or
10,
or
100,
or
or
1000, log
i
=
o
and
= o; = i; = 2; log TOO = 3; log 1000 = i, log io log i log 10
etc.
every number between or,
i and io has a logarithm between o and i,* every number whose integral part has one digit has a logarithm
whose characteristic
is zero.
Since
log 10
=
i
and log 100
=
2,
every number between io and 100 has a logarithm between *
i
and
2,
This statement assumes that to the greater of two numbers corresponds the > N, log > log N, a theorem which can be is, if
greater logarithm, that easily
M
proven by elementary algebra.
M
LOGARITHMS
28]
59
every number whose integral part has two digits has a logarithm whose characteristic is one. Similarly, every number between 100 and 1000 has a logarithm between 2 and 3, or, every number whose integral part has three digits has a logarithm whose characteristic is two. Every number whose integral part has four digits has a logarithm whose characteristic is three, and so on.
or,
This gives us the following rule: I.
is
The
one
characteristic of the logarithm of
less
any number
greater than i
than the number of digits in the integral part of the number.
Thus, the integral part of the number 31622.8 consists of 5 hence the characteristic of its logarithm is 4.
The
integral part of 3.16228 consists of
teristic of its
logarithm
i digit,
digits,
hence the charac-
is o.
Again,
=
10
lo"
1
or log
i,
=
= 10
io~ 2
=
= io2
io~ 3
= -~ =
io~ 4
=
= io 4
o.i,
i
=
or log o.i
o;
=
i;
=
o.oi,
or logo.oi
o.ooi,
or logo.ooi
o.oooi,
or logo.oooi
2;
=
=
3;
4; etc.
Hence, every number between
and
i, or,
i and o.i has a logarithm between o every fraction greater than o.i has a logarithm whose
characteristic is
i.
Similarly, every
tween
i
and
number between
2, or,
o.i and o.oi has a logarithm beevery fraction greater than o.oi but less than
2. has a logarithm whose characteristic is like manner, every fraction greater than o.ooi but has a logarithm whose characteristic is 3, and so on.
o.i
In
less
than o.oi
less
than i
This gives us a second rule: II.
The
characteristic of the logarithm of
any number
is
a negative number one more than the number of ciphers between the decimal point and the first significant figure of the fraction.
PLANE TRIGONOMETRY
60
[CHAP, iv
Thus, the fraction 0.00316228 has two ciphers between the decimal first significant figure, hence the characteristic of its
point and the
is 10. 3 or 7 fraction 0.316228 has
logarithm
The
and the rithm
no cipher between the decimal point
first significant figure,
is
or 9
i
hence the characteristic of
its
loga-
10.
EXERCISE 16 Write down the characteristics of the
1.
common
logarithms of
367, 36.7, 3 6 7<>, 3-^7, 0.000367, 0.367.
Ans. 2.
=
log 635
4 or 6
2, i, 3, o,
2.80277; write
down
i
10,
or 9
10.
the logarithms of 6.35, 63500,
0.635, 0.0000635.
Ans. 0.80277, 4.80277, 9.80277
How many
3.
whose logarithm 4.
To
digits are there in the integral part of the is
number
3.1567; 1.6533; 0.6831?
number
the
10.
10, 5.80277
57 corresponds the mantissa 75587; what is i.75S 8 7; 2 -755 8 7; -755 8 7; 9-755 8 7
number whose logarithm
Ans. 57; 57o; 5.
is
-
the i
?
5-7J 0-57-
To the number 673 corresponds the mantissa 82802;
find the
logarithm of 673; of (673)2; of ^67^; of ^(673)2. Ans. 2.82802; 5.65604; 1.41401; 1.88535. 6.
=
log 3
0.47712;
how many
25 in 3 100 ; in digits are there in 3 ;
5
Ans.
in 27 ? 7.
8-
15 ;
12; 48; 23; 8.
Write down the cologarithms of the numbers in problem 2. Ans. 9-19723 - 10; 5- I 97 2 3 - 10; -i97 2 3; 4-i97 2 3log 5
=
Vs; ofVi;
0.69897; find the logarithm of i; of
of
10; 8.60206
9.65052
To
number
the
-
s > of
0-05; of
V^s; ofVj.
Ans. 9.30103
9.
30
10; 8.69897
10; 9.89966
10; 0.34948;
10; 9.76701
number 3 corresponds the mantissa 47712, and
10.
to the
7 corresponds the mantissa 84510; find the logarithm of 21;
of *; of Vo.3
X
49; of V}.
Ans.
1.32222; 9.63202
10; 0.58366; 9.96362
10.
LOGARITHMS
2 9]
10.
The formula
for the
compound interest at compounded annually,
amount (A)
(R) per cent for
61
of a principal (P) (/)
put out on
years, the interest being
is
-, loo/
whence log
A = IOO/
Find the number
of digits in the
amount
of $i at
compound
interest at 6 per cent for 100 years, the mantissa corresponding to
Ans. 3
106 being .02531.
digits.
Common
Tables of Logarithms. Our next step is to learn to use a table of logarithms in actual computation. Common logarithms, except those belonging to numbers which 29.
how
are integral powers of 10, cannot be exactly expressed in decimals. must therefore omit all the figures after a certain decimal place.
We
Just where to stop depends upon the accuracy desired. If some of the numbers which enter a given problem are the results of measure-
ment, as is frequently the case, their accuracy will not ordinarily exceed four or five figures, consequently it will be useless to retain more than five figures in the decimal part of their logarithms. In other words, a table of logarithms which contains the mantissas to five places will answer for the solution of most practical problems
which approximate answers are all that is necessary or possible. Such a table is known as a five-place table of logarithms. The explanations which follow, and all the answers in this book obtained from logarithmic computation, are based upon a five-place
in
table.
When a more than
five-place table is used, it is not worth while to retain five significant figures in the answers to the problems,
and even then the last figure is not always exact. When more than five-place accuracy is required, and even when the fifth place must be known with certainty, larger tables must be used.
Tables containing
places, eleven places,
to
two hundred and
six places,
seven places, eight places, ten
twenty places, and partial tables containing up There are sixty places, have been published.
also smaller tables containing three
and four places
only.
Tables containing more than seven places are seldom used, for seven-
PLANE TRIGONOMETRY
62
demand
place tables meet practically every
The
[CHAP, iv
of present-day science.
by means of a seven-place table are as exact as the most careful measurements obtained by the most skillful observers by means of the most precise instruments under the most favorable results obtained
conditions.
To Find
30. (a)
When
Number. The characteristic figures.
the Logarithm of a Given
number has four
the given
is
found by the rule in Art. 28. To find the mantissa, enter that line of the table which begins with the number made up of the first three figures of the given number and take out that number of the line
which
is
found in the column headed by the fourth figure of the
given number.
The number thus found
constitutes the third, fourth
and
fifth
The first and second figures are figures of the required mantissa. found in the column headed by o either in or above the line in which the third, fourth and fifth figures are found, except when these figures as given in the table are preceded by a star (*) in which case the first two figures are found in the next following line.
EXAMPLE
i.
From
the specimen page of logarithms, page 63,
we
find
mantissa log 6315 mantissa log 65.24 mantissa log 6.608
= = =
hence log 6315 hence log 65.24 hence log 6.608
80037,
81451, 82007,
= = =
3.80037; 1.81451;
0.82007.
We know (b) When the given number has less than four figures. that the mantissa of the logarithm of a number is not changed if the number is multiplied or divided by 10 or by some power of 10. After we have determined annex as
many
four places
and
the characteristic of a logarithm
EXAMPLE 2. Find the logarithm of 64. The characteristic of log 64 is i. The mantissa as the mantissa of log 6400, which
Hence, log 64 (c)
When
the
we may then
ciphers to the given number as we need to make find the mantissa of this new number by case (a).
by case
=
(a) is
i. 80618.
number has more than four figures.
EXAMPLE 3. Find the logarithm of 6425.4. The characteristic of log 6425.4 = 3.
up
of log 64 is the same found to be 80618.
30]
COMMON LOGARITHMS OF NUMBERS (SPECIMEN PAGE)
PLANE TRIGONOMETRY
64
By
(a),
difference for
Now
= = =
the mantissa of log 6425 the mantissa of log 6426 i
[CHAP, iv
80787
80794 7
evidently larger than 80787 and than 80794, and since 6425.4 lies fa the way between 6425 and 6426, we will assume that the mantissa of log 6425.4 lies fa the way between 80787 and 80794. We must therefore increase the smaller
the mantissa of log 6425.4
is
less
of the
two mantissas by fa
by fa of
of the difference
between the two, that
is,
7.
fa of
and the mantissa Hence
7
=
2.8 or 3 to the nearest integer,
of log 6425.4
=
80787
log 6425.4
=
=
+3
80790.
3.80790.
We
have here assumed the principle of proportional parts for logarithms of numbers, namely, that for small changes in the number, the change in the logarithm is proportional to the change in the number.
is
EXAMPLE 4. Find the logarithm of 6.5487. The characteristic of log 6.5487 = o. The mantissa the same as the mantissa of log 6548.7. mantissa log 6548 mantissa log 6549 difference f or
=
difference for 0.7
Hence
0.7 of 6
mantissa log 6 548. 7
and
log 6.5487
=
= =
i
= = =
81611
81617 6
4.2 or 4 to the nearest integer.
81611
+4=
81615,
0.81615.
EXAMPLE 5. Find log 0.000635945. The characteristic of log 0.000635945 = 4 The mantissa of log 0.000635945 is the same 6359-45-
mantissa log 6359 mantissa log 6360 difference for difference for 0.45
-
Hence
and
0.45 of 7
of log 6.5487
=
mantissa log 6359.45 log 0.000635945
i
= = =
or 6
10.
as the mantissa of
80339 80346 7
3.15 or 3 to the nearest integer.
= =
80339
+3
6.80342
80342, 10.
loj
LOGARITHMS
31]
EXAMPLE
By
65
Find the cologarithm of 65.021. a number the reciprocal of that number, hence 6.
definition, Art. 26, the cologarithm of
rithm of
colog 65.021
=
log
i
31. (a)
To Find When the
acteristic is
the
the loga-
log 65.021.
=
= 10 = 1.81306 log 65.021 = 8.18694* colog 65.021 log
is
i
o
Number Corresponding
to
10
10.
a Given Logarithm.
The chargiven mantissa can be found in the table. used only to determine the position of the decimal point
number has been found. Find the given mantissa in the The first three figures of the number sought are found in
after the table.
the
same
line
fourth figure mantissa.
EXAMPLE
with the mantissa, in the column on the left, and the found at the top of the column containing the given
is
i.
Find the number whose logarithm
is
5.82158.
Find the given mantissa 82158 in the table (see specimen page, p. 63). The number in the left-hand column and in the same line with the given mantissa
is
663,
and the number at the top
of the
column containing
hence the significant figures of the required The characteristic is 5, hence the integral part of the required number has 6 places, that is, the required number the mantissa 82158 number are 6631.
is
is i,
663100.
2. Find the number whose logarithm is 8.81043 IO Corresponding to the mantissa 81043 we find in the table the num10 or ber 6463. The characteristic is 8 2, hence by the rule
EXAMPLE
-
number
for the characteristic the required
one cipher preceding the
Therefore the required number
When
is
a decimal fraction with
first significant figure. is
0.06463.
mantissa cannot be found in the table, two other (6) consecutive mantissas can always be found in the table, one of which the given
smaller and the other a
little larger than the given manfour figures corresponding to the smaller of these mantissas will be the first four figures of the required number; the fifth
is
a
little
tissa.
The
* The subtraction is performed from left to right by subtracting each figure from 9 except the last one, which is subtracted from 10, thus: i from 9 = 8, 8 from = 6, o from 9 = 9, 6 from 10 = 4. 9 = 1,1 from 9 = 8, 3 from 9
PLANE TRIGONOMETRY
66
[CHAP, iv
and sometimes the sixth,* can then be found by interpolafrom the principle of proportional parts.
figure,
tion
EXAMPLE 3. log N = 1.80395, to find N. The table does not contain the mantissa
80395, but it contains mantissas 80393 and 80400, one of which is smaller and the other larger than the given mantissa. The numbers corresponding to these mantissas are 6367 and 6368
the two consecutive
respectively.
= =
mantissa log 6367 mantissa log 6368 difference for
80393 80400
=
i
7
Since the given mantissa lies between 80393 and 80400, we infer that the number corresponding to the given mantissa lies between 6367 and 6368; let it be denoted by 6367 x, we then have
+
=
mantissa log 6367 mantissa log 6367 -f- x
80395
=
difference for x
and the
80393 2
principle of proportional parts gives i
:
7
=
#
:
2,
that
is,
x
=
?,
and the number corresponding
to the given mantissa is 6367$. It remains to determine the decimal point. The characteristic is hence the required number is
still
i,
N=
63. 67^
=
63.673 to five figures.
EXERCISE
17
(In this exercise the 1.
specimen page of logarithms may be used.) Find the logarithms of the following numbers: 6315, 632.5
6.454, 0.0655, 0.0065.
Ans. 3.80037, 2.80106, 0.80983, 8.81624 2.
10, 7.81291
10.
Find the cologarithms of each of the numbers in i. Ans. 6.19963 10, 7.1989410, 9.1901710, 1.18376, 2.18709.
* In the
first
part of the table, the sixth significant figure of a
number may be
found by interpolation. In the latter part of the table, where the difference in the mantissas corresponding to a difference of i in the numbers is much smaller than in the first part of the table, the sixth figure of the number obtained the
by
principle of proportional parts cannot be
depended on.
LOGARITHMS
32]
67
Find log 63.454, log 65.061, log 6.6095,
3.
Ans.
1(
>g 0.0064159.
1.80246, 1.81332, 0.82017, 7.80725
-
10.
Find the numbers whose logarithms are: 1.80277, 2.80584,
4.
0.81003, 9.81351
-
10, 8.80017
~
I0 3-81184. >
Ans. 63.50, 639.5, 6.457, 0.6509, 0.06312, 6484.
Find the numbers whose logarithms are: 1.80958, 2.81922,
5.
0.81006, 9.80002
10,
8.80022
10, 3.82506.
Ans. 64.503, 659.51, 6.4574, 0.63099, 0.063127, 6684.3. 6.
log 63.275
=
x, log
y
=
1.81864; find x and
Ans. x
=
v.
1.80124,
Without multiplying or dividing the numbers,
7.
rithms of 66.027
X
0.65034, 6301
Ans.
-r-
6.454, (6535.4)2,
y
65.863.
find the loga-
^63. 275.
1.63287, 2.98958, 7.63055, 0.90062.
N = V64550; find log N and then N. Ans. log N = 0.80165, ^ =
8.
Find (6.3096) 5 by means of logarithms.
9.
Ans.
6.3336.
10000.
Find the logarithms of sin 40, cos 48 30', cot 8 50'. Ans. 9.80808 10, 9.82125 10, 0.80854.
10.
(Suggestion.
and then
First find the natural functions of the given, angles
find the logarithms of the resulting numbers.)
Find the
first three significant figures and the number of in the figures integral part of the twenty-fifth power of 6.2.
11.
Ans. 645, 20 places. Directions for the
32.
Use
directions will aid the student in
The
of
Logarithms.
an
intelligent use of logarithmic
following
tables. (a)
In finding the logarithm of a given number, or in finding the to a given logarithm, the interpolation should
number corresponding
be performed mentally and only the complete result
set
down
in
writing. (6) In writing down the cologarithm of a number, the subtraction from 10 should be performed mentally and from left to right. (c)
By
The
results obtained
neglecting the sixth
by logarithms are approximations only. and following significant figures of a number
the inaccuracy introduced can never exceed one-half a unit in the
PLANE TRIGONOMETRY
68 fifth place,
that
is,
[CHAP, iv
the error cannot exceed the %fo part of
i
per
cent. (d)
When
the sixth figure of a
number
is 5,
and we wish
to retain
we we
increase
only five significant figures, it is immaterial whether the fifth figure by i or leave it unchanged. In case
increase
the fifth figure by i the resulting number will be too large by half a unit in the fifth place; if we leave the fifth figure unchanged, the number will be too small by half a unit. In order to cause the inaccuracies arising from this source to offset one another, it is customary, on dropping a final 5, to increase the preceding figure by i
when
odd, but to leave
it is
it
unchanged when
it is
but
0.154755 becomes 0.15476, 0.154745 becomes 0.15474;
one-half of
4-23453 becomes 2.11726,
but one-half of
4.23463 becomes 2.11732,
Thus,
when the
even.
results are abridged to five places.
Sometimes the nature of the problem is such that a four-place table would give all the accuracy required. In that case the fifth figure of the mantissa in the table may be omitted and the fourth (e)
figure increased
by
i
if
the omitted figure exceeds
figure of the mantissa given in the table is
5.
If the final
marked with a
stroke,
thus ^, on omitting it the preceding figure is left unchanged; but if the final 5 is unmarked, the preceding figure is increased by i. The reason for this
is
that the final 5 of a mantissa
is itself
the result of
approximation, that is, it is either in defect (5 plus something les^ than J), or it is in excess (4 plus something greater than J), and the latter case is distinguished from the former by printing a stroke
over the
5.
Thus from the
table,
log 2.078
but
when abridged
log 2.079
= =
0.31765 0.31785
= =
0.3176, 0.3179,
to four places.
the mantissa, which Every logarithm consists of two parts, is always positive, and the characteristic, which is always integral When the character(or o), but maybe negative as well as positive. istic is negative, it is customary to change the form of the logarithm by adding and subtracting 10. (/)
LOGARITHMS
32]
Thus,
This
log 0.4562
=
log 0.0032
=
done
is
69
+ 0.65916 is written 9.65916 10, written 7.50515 - 10. 3 + 0.50515 i
is
in part to avoid mistakes
which might
arise
from con-
fusing the positive and the negative parts of logarithms. For similar reasons, if a logarithm whose characteristic
is negative as in extracting a square root, we first modify its form by adding and subtracting 20; if the logarithm is to be divided by 3, we first add and subtract 30; and generally, if the
is
to be divided
by
2,
logarithm whose characteristic
we
first
Thus,
modify
its
is negative is to be divided by form by adding and subtracting ion.
= - 2 + 0.51242 \ of log 0.03254 = \ of (18.51242 i of log 0.03254 = i of (28.51242 = i of (38.51242 i of log 0.03254 \ of log 0.03254 = \ of (48.51242
The
40)
= 8.51242 = 9.25621 = 9.50414 = 9.62811 -
50)
=
log 0.03254
20)
30)
n,
10, 10,
10, 10,
10, etc.
9.70248
between two consecutive mantissas is called the is printed under D in the last column on each page of logarithms. At the bottom of each of the first three pages of logarithms the tabular differences which occur on the page are (g)
difference
tabular difference
and
by each of the nine known as tables
multiplied
sulting tables,
The
digits expressed as tenths.
re-
of proportional parts, are used as an
aid in interpolation.
EXAMPLE Solution.
From
Find by logarithms the product Denote the product by x.
i.
table
By Art. By Art. EXAMPLE by
2.
= = log 0.85734 = log x = x log 37.543
I,
25, (a),
31,
of 37.543
and the
table,
by
1-57453
9.93315
0.85734.
-
IP
1.50768 32.187,
Find by logarithms the quotient of 6.3725 divided
82.756.
Solution.
From
By
Denote the quotient by the table,
Art
From
26, (c),
and the
the table,
x.
= 82.756 = = log x = x
log 6.3725 table,
colog
0.80431 8.08220
10
8.88651
10
0.077003.
PLANE TRIGONOMETRY
70
EXAMPLE
Find the square root
3.
of 0.89355.
= = = log ^0.89355 ^0-89355 =
Solution.
log 0.89355
Dividing by
2,
Hence
EXAMPLE
Find
4.
=
3
x, if
jc
(o.5824) log
Multiplying by
Dividing by
EXAMPLE
5.
x
-
By log x
0.5824=
= = = logo; x =
log x?
3,
8.35
(1.256)'
(62.5)* X X 623.7 XV5-736
2 log 62. 5
log 5-673
\ log 5.673 colog 1.256 3 colog 1.256
i colog 5.736
9.97556
10
0.94528.
.
9.76522
10
19.53044
20
-
29.53044
30 10
9.84348 0.69740.
find
^
the rules of Art. 25, log 8.35 + 2 log 62.5 + 1 tog 5-673 + 3 colog 1.256 + colog 623.7 + I colog 5-736-
= = = = = = = =
=
0.92169
=
3-59176
=
0.37690
1.79588 2
X
I-795 88
0-75381 \
X
3
X
0.75381 10 9.90101 (9.90101
colog 623. 7 colog 5.736
19.95112
=
log 8.35 log 62. 5
IO
20
.
2,
X
~~
9-95 112
2
Solution.
Solution.
[CHAP. IV
9.24139 I
X
-
-
10)
= 9-733 = 7.20502
10 10
10
(29.24139
-
30)
= 9.74713 - 10 log* = 31.54553 -30 = 1-54553 * = 35.118.
EXERCISE 18 Solve 1.
2.
by
logarithms:
3.784 X 7.843. 67.845 X 0.03457.
3-
0.67375 -^ 3.468.
4.
92.57 -h 1.3785.
Ans.
29.677.
Ans. 0.19428.
LOGARITHMS
32]
X
5.
(684.7)*
6.
(o.8oo3 )
7.
V3284-5.
8.
3
71 Ans. 27.236.
(0.03873)3.
-5-
V5
.
73
.
Ans. 5.05. 10
(O.I2S62)
x
1-56
.
3
67-574 600.45;
I0
2
(37-5)
X X
1.0025
i/
j.x^u/xuxx
V
1.056
"
*/-S.
X 2.738 X X 23.7" V_(27-3) V(28. 9 2) X 16.5
Ans. 4.2188.
10
2
12
.
'
3
13.
The
area of a triangle
is
given
by the formula
A = ^s(s-a)(s-b)(s-c), where a, 6, c represent the three sum. Find A when a = 617.34,
sides of the triangle
b
=
345-65, c
=
and
5 half their
467.75.
Ans. 80127. 14.
In Art. 21
it
was shown that 2
where b is a side of a right triangle, c the hypotenuse, and A the angle between b and c. Find A when c = 325.76, b = 324.13. ^W5. 5 44'. 15.
X Solve the equation 2
16.
=
Find x in (3.1416)*
=
Ans. 1.8371.
3-573-*
9.8697.
17. Solve the equation
3 18.
2X
12
X
3*
+
ii
=
=
o,
Ans. x
=
Ans. x
o.
or 2.1827.
Solve X
&
_3*
+ e = 4, 2
where
e
*
=
2.7183. d= 2.6339.
Such an equation as this, in which the unknown quantity appears as an exponent, is called an exponential equation. It is solved by first taking logarithms of both sides of the equation, thus: X log that
is,
from which
2
0.30103 x
x
= =
log 3.573,
0.55303, 1.8371.
PLANE TRIGONOMETRY
72 33.
[CHAP, iv
The solution of triangles, which most important application of logarithms, will be considered later on. At present we give a few miscellaneous Application of Logarithms.
furnishes the fully
problems which are especially adapted to solution by logarithms.
EXERCISE 19
Many the
of the following list of
compound
1.
interest formula,
problems require for their solution Problem 10, Exercise 16.
%
Find the amount on $100 for 100 years at 4 compound = 0.0170333.) Ans. $5050.4. (log 1.04
interest, 2.
In what time
pound
a
will
sum
of
money double
itself
at 8
%
Ans. 9.007
interest ?
com-
yrs.
A man
bequeaths $500, which is to accumulate at compound interest until the interest for one year at 5 will amount to at least 3.
%
$300, after which the yearly interest
is
to be
awarded as a scholar-
How many
years must elapse before the scholarship becomes available, assuming that the original bequest is made to earn 5 ship.
%
compound 4.
Ans. 51
interest ?
At what
rate of interest
invested in order that
40
yrs.
must the bequest in Problem 3 be may become available in
the scholarship
Ans. 6.4
yrs. ?
%.
5. In 1624 the Dutch bought Manhattan Island from the Indians for about $24. Suppose that the Indians had put their money oat
at
compound
interest at 7
% and had added the interest to the prin-
cipal each year, how large would be the accumulated amount in 1910 ? (From White's Scrap Book of Mathematics.) Ans. In round numbers $6,000,000,000. The actual valuation of
Manhattan and Bronx
6.
The population
real
and personal property
of the state of
Washington
in
in 1890
1908 was
was 349,400
in 1900 it was 518,100. What was the average yearly rate of increase ? Assuming the rate of increase to remain the same, what
and
should be the population in 1910 7.
The founder
of a
new
faith
?
Ans. 4
%;
767,970, nearly.
makes one convert each
year,
and
each new convert makes another convert each year, and so on.
LOGARITHMS
34]
73
How long would it require to convert the whole earth to the new faith, assuming that the population of the world
is
1,500,000,000
Am. Between
?
30 and 31
yrs.
The combined wealth of the United States and Europe was estimated (1908) to amount to about $450,000,000,000. Let us 8.
assume that the entire wealth of the world amounts to $io 12 How to equal long would it take $i put out at compound interest at 3 .
%
or exceed this
amount ?
Ans. 935
34. To Compute a Table of
A
Common
yrs.
Logarithms. from the successive
logarithms may be computed square-roots of 10 and multiplications. of
table
io*=Vio = io*=^/io*=V3. 16228 iot = \/ ioi=\/i. 77828
3.16228,
(i)
1.77828,
(2)
i.3335 2 >
(3)
1.15478.
(4)
= = =
logarithm of a number is the power to By which 10 must be raised to produce that number, hence from definition, the
10*
=
common
1.15478,
10* = io^ = 1.33352, + 10* = io* * = 10*
log 1-1548*=
-h
=
A
log x -3335
io* =
I.3335 2
X
1.15478
=
10* =
Check.
10* =
io
+ l4*
10*
=
* = 10*
10* = io*
*=
10*
10* = io* + * = 10* + 10* = io* * = 10*
1.53992,
1.53992
A = 0-1875; 1.77828, 1.15478 = A = 0.2500. log 1.7783
by (2). 10* = 1.77828
X
1.15478
10* =
=
=
1.77828
=
2.05352,
A = 0.3125; 2.37136, 2.05352 X 1.15478 = A = 0.3750; log 2.3714 10* = 2.37136 X 1.15478 = 2.73840, = A = 0.4375; log 2.7384 = = 10* 16225, 2.73840 X 1.15478 = A = 0.5000. log 3.1623 log 2.0535
+
0.0625; 0.1250;
X
log 1.5399
+ 10* = io* * = 10*
= =
=
10* =
10* = 10* = 3.16228 by (i). + 10* = io* * = 10* 10* = 3.16228
=
3-
Check.
X
1.15478
=
log 3.65 1 7 *
Only
3.65174,
=
A = 0.5625;
in order to secure accuracy in the last figure of all 5 figures are carried
the numbers in the
list.
PLANE TRIGONOMETRY
74 iot$
- io* + * = io*
10* = 3.65174
X
[CHAP, iv
1.15478
=
4.21696,
=
log 4.2170
iott
=
i& +
i
*
I0 tt
.
IO A =
4.21696
X
1.15478
=
log 4.8697 io*3
=
io**
+
*=
10* = 4.86966
ioi*
X
=
}
0.6250;
4.86966,
=
U = 0.6875;
=
5.62339, 1.15478 = 0.7500. !g 5- 62 34 =
H
Check.
IO
10**
and
(i)
W=
=
10^
=
10*
10*
=
3.16228
X
1.77828
io^i
=
iot*
+
A=
10* =
10^
5.62342
X
1.15478
=
+
*=
ioW
.
I0 * =
6.49382
X
1.15478
=
log 7.4989
IO W =
i
ii +
* = io& 10* =
7.49892
X
1.15478
=
log 8.6596
IOH =
io^
+*
=
10^
lo^6
=
8.65960
X
1.15478
=
log 10.000
Check.
5.62342
by
(2).
log 6.4938
iolt
=
10**
=
io
l
=
6.49382,
H = 0.8125;
=
7.49892,
H = 0.8750;
=
8.65960,
=
if
=
0.9375;
9.99993,
=
i
=
i.oooo.
10.
is on the four lines just preceding, and in each case exact agreement to five figures between the results of the check and the result of the line just preceding. The discrepancy in the sixth figure, where it occurs, is due to the fact that all our re-
Each check
there
is
sults are
approximations only.
We
have thus computed the numbers whose mantissas are tributed between o and i at intervals of TV.
dis-
Similarly, with the aid of
longer
lists of
IO A
=
io^f
= v 10^ = v
x/ IO *
= Vi. 15478 = 1.07486
=
1.07468, 1.03663,
.
etc.,
numbers could be computed whose mantissas
differ
continuing this process the interval between sucWhen the difference cessive mantissas can be made small at will.
by
sV> ffV etc.
By
between successive mantissas has been made
sufficiently small, the
mantissa corresponding to any number, intermediate to two numbers in the list already found, may be found by interpolation. If
LOGARITHMS
35]
now
the numbers on the
75
are selected at equal intervals
left
and tabu-
with their mantissas, we shall have a table of
lated, together
common
logarithms.*
The characteristics need not be tabulated, for they can always be determined from memory by the two rules of the preceding article. In most of the tables the decimal points are also omitted; for instance,
we
find in
most tables corresponding to
number
the
the mantissa
4753
meaning that .67697
67697, is
the fractional part of the logarithm of any
number whose significant figures are 4753. The method just explained, simple as it which the existing tables
We
lated.
shall learn later
which logarithms
may
is,
is
not the method by
have actually been calcuthat other methods exist by means of
of logarithms
be calculated with much greater ease and
speed.
N
35. Relation between loga and log6 N. We will now show that the logarithms of the same numbers to two different bases are proportional, so that when the logarithm of a number to a given is known, the logarithm of the same number base other may be obtained from it by multiplying the known any constant. It is therefore unnecessary to a certain logarithm by
base, as the base 10, to
actually construct logarithmic tables to To prove that log & ]V
Proof.
=
Let
Now
(i)
and
logaAr,
logb^
and
From
u.
logo-/V
(2)
where
= =
\i
x,
y,
V=
is
more than one
base.
the constant
from which
b*
=
N,
(i)
from which
av
=
N.
(2)
a*.
(3)
let
*
=<*c
,
*
(4)
The student will observe that in the original arrangement the numbers on the right (mantissas) are at equal intervals. In this form the table is known as a table of antilogarithms. A table of antilogarithms would serve the main purposes Tables of antilogarithms have of computation as well as a table of logarithms. been published and are used by some computers.
PLANE TRIGONOMETRY
?6
and
[CHAP, rv
substitute (4) in (3), then
(a*y
=
from which
= a, ex = y,
acx
*-Z
or
(5)
o>
where from Putting in
c
(4),
(5) for
=
loga&.
x and y their values from
and
(i)
(2),
and
M forc
we have Iog 6
^=
|UogJ
(6)
where
may
(6)
If in (7)
that
also be written
we put
N=
a,
we obtain
(since loga#
=
i)
is,
logbd
and
logab are reciprocals.
The
constant multiplier /* is called the modulus of the system of logarithms whose base is b with reference to the system whose base is a.
36. Natural or Hyperbolic Logarithms. Theoretically any posinumber different from i may serve as the base of a system of
tive
logarithms, but in practice only two systems are used. The first is the common system, used exclusively in numerical computations; the other is known as the system of natural or hyperbolic logarithms,
which
is
used extensively in theoretical investigations.
The base *
of the natural
system
*
is
an incommensurable number,
Natural or hyperbolic logarithms are known also as Napierian logarithms in this name is a misnomer, since neither
honor of John Napier (1550-1617), though
Napier nor any of his contemporaries had any conception of the number e or the system of logarithms which has e for its base. Napier's base is the number 0.367879, which happens to be nearly equal to
rithms as well as the
name
is
.
The
discovery of natural loga-
due to Nicolas Mercator (1620-1687).
LOGARITHMS
36]
known
as
e,
the twin sister of the
remarkable properties,
number
2
It will be
2-3-4
3
from which the approximate value of e
By
the preceding article
of logarithms
which
?r,
e is defined as the limit
proaches, as x increases indefinitely. limit is equivalent to the series
2
77
=
we
many
+ iV apthis
..,
found to be
.
modulus
of the natural system
common system
= log 2.71828
has
shown later that
e is readily
find the
IT
which (i
2-3 -4.5
2.71828
with reference to the
like
2 3
is
259
'
0.434294
EXERCISE 20 1. Use the results of Art. 34 to compute to four places the number whose common logarithm is -fa = 0.15625. Ans. 1.4330.
a method similar to that of Art. 34 compute to four places common logarithm is J; also to four places the number whose common logarithm is J. Ans. 2.1544, 1.2913. 2.
By
the 'number whose
3.
Similarly
compute the number whose common logarithm Ans.
4.
Given log
5.
Show
6.
102
=
=
-9ge
=
0.434294
X
Find log e 100, loge 1000, loge o.oi, loge 2. Ans. 4-60518, 6.90777,
7.
Use the
8.
Given Iogio3;
results of Art.
log e N.
- 4-60518,
36 to compute loge 3.65174. Ans.
1.29521. i
r
'
, '
2 log
Show how
0.69315.
fin d logs 10, Iog27 1000.
Ans. 9.
1.4678.
Ans. 3.3219.
0.30103; compute Iog2 10.
that logioN
is J.
3
logio 3
to obtain:
logarithms to the base 8 from logarithms to the base 2; logarithms to the base 3 from logarithms to the base 9. Ans. Divide by 3; multiply by
2.
PLANE TRIGONOMETRY
78
//
Prove that
10.
log&a
Given log 4
11.
[CHAP, iv
=
Iog c 6
Iog c
=
0.60206, log 9
=
i.
y///
0.95424; find log 6. Ans. log 6 = 0.77815.
Tables of Logarithmic Trigonometric Functions.
37.
In the
solution of triangles and elsewhere we constantly encounter expressions involving trigonometric functions. Suppose it were required to
compute the value
of
325-6 sin 23 45' tan 18 24' cos 37 30' It
is
plain that
we might
first
find the values of the several trigono-
metric functions from the table of natural functions; thus, sin
With
2345 = ;
0.4027, tan 18
24'
=
0.3327, cos 37 30'
=
0.7934.
these values the expression above written becomes
x
=
325.6X0.4027
X
0.3327
f
0-7934 Solving
by logarithms we have log 325.6 log 0.4027
log 0-3327
colog 0.7934
It will
= = =
= = log sin 23 45' = log tan 18 24' = colog cos 37 30' = log x x =
be observed that
sulted
two
sin 23
45'
=
10
9.52205
10
0.10051 1.74022 54-983-
in order to obtain log sin 23
different tables: first,
=
2.51268
9.60498
we con1
45'
the table of natural sines, to find
0.4027; second, the table of logarithms, to find log 0.4027 10. To avoid the necessity of referring to two tables,
9.60498 the logarithms of the sines, cosines, tangents and cotangents have been separately calculated and arranged in a new table, known as the table of logarithmic trigonometric functions *. The table contains the values of the logarithms of the sines, cosines, tangents of angles between o and 90. that the sine and cosine of every angle and the tangent of every angle less than 45, as well as the cotangent of any angle be-
and cotangents
We know
* The choice of the term is much better. It is of interest
Trigonometric logarithms would be that Napier's tables, the first tables ever
unfortunate. to
know
published, were tables of logarithmic trigonometric functions.
LOGARITHMS
38)
79
tween 45 and 90, is less than unity, consequently the characteristics of the logarithms of these functions will be negative. To avoid negative characteristics, each negative characteristic has been replaced by a positive characteristic by adding 10. For example, while the true value of log sin 23
45'
= -
i
+ 0.60503 *
the table gives log sin 23
This latter value
To find l
is
45'
=
9.60503
-
10,
9.60503.
called the tabular logarithmic sine, hence
the true value of
a logarithmic sine
the corresponding tabular
logarithmic sine must be diminished by 10.
Since the secant
is
the reciprocal of the cosine, and the cosecant
the reciprocal of the sine, we have The logarithmic secant or cosecant
may be
obtained by taking the
cologarithm of the cosine or sine respectively, that ing the tabular logarithmic cosine or sine from 10.
is,
by
subtract-
38. To Find the Logarithmic Trigonometric Functions of an Angle less than 90. (a) When the angle is less than 45, we find the number of degrees at the top of the page and the number of minutes in the left-hand
column.
The
values of the log sin, log tan, log cot, log cos, as indicated at the head of the column, are then found in the same line as the minutes. If seconds are given, they may be reduced to a fractional part of a minute, and the value of the logarithmic function may be found by interpolation, just as was done in finding logarithms of numbers.
EXAMPLE 11
i.
Find the log
sin,
log cos, log tan
and
log cot of
21'.
Remembering that table are too large by
certain of the characteristics as given in the 10,
we
find
from table II
log sin 11
21'
log tan 11
21'
n
2 1'
log cos 11
2 1'
log cot
= = = =
9.29403
10,
9.30261
10,
0.69739,
9.99142
10.
PLANE TRIGONOMETRY
8o
EXAMPLE From the
2.
Find log
table
we
sin 15
; i524 =
0.6',
log sin 15
EXAMPLE From the
3.
9.42416
= 9.42461 difference for i' = 45 difference for 0.6' = 0.6 X 45 = log sin 15
hence
24' $6".
find,
log sin
36"=
(CHAP. IV
24'
=
36"
To find log we find
25'
+
9.42416
27
=
27,
9.42443.
cos 27 36' 40".
table
= 9.94753 = 9.94747 6 difference for i' = = difference for o.6f 4, log cos 27 36' log cos 27 37'
40"=
0.6$',
hence
log cos 27
Observe that in
this
36'
40"
=
9-94753
~
4
==
9-94749-
example the difference was subtracted from the
mantissa of the logarithm of the smaller angle, because the logarithm of the cosine decreases as the angle increases.
When
(b)
more than 45
the angle is
and
less
than
po, we
find the
degrees at the bottom of the page and the minutes in the righthand column. The values of the log sine, log tangent, log cotangent and log cosine as indicated at the foot of the column are then found in the
same
EXAMPLE From the
line as the minutes.
4.
Find log tan 61
table
we
10' 27".
find
log tan 61 10' log tan 61 n' difference for
27"
=
0.4^',
hence
The
i'
difference for 0.4^'
log tan 61
;
io 27"
=
= 0.25923 = 0.25953 = 30 = 4! X 30 =
0.25923
+
13.5
=
13.5,
0.25936.
difference 13.5 is additive because the log tangent increases with
the angle.
LOGARITHMS
391
EXAMPLE
Find log cot 75 51' 15".
5.
log cot 75 51' log cot 75 52' difference for i'
15"
=
difference for 0.25'
0.25',
hence
The
log cot 75
difference 13
log cot 75
81
is
51'
15"
= = = = =
9-40159 9.40106 53 13,
9.40146
10.
subtracted because log cot 75 52'
is
less
than
51'.
In general when interpolating:
For
sines
and
tangents the difference
of the smaller angle, because the sine
must be added to the function and tangent of an angle in-
creases as the angle increases.
For
cosines
and
cotangents the difference
must be
subtracted
from
the function of the smaller angle, because the cosine and cotangent of the angle decreases as the angle increases. 39. To Find the Angle Corresponding to a Given Logarithmic Trigonometric Function. When the given number can be found in the table, the number of degrees is found at the top or bottom of the page, according as the name of the function appears at the top or bottom of the column. The number of minutes is found in the same line with the given in the left column if the degrees are taken from the top function, of the page, in the right column if from the bottom.
When secutive smaller,
number cannot be found in the table, two connumbers can always be found, one of which is slightly the other slightly larger than the given number. The rethe given
quired angle can then be found
by interpolation. EXAMPLE i. Given log sin x = 8.73997; to find x. The number 8.73997 is found in the first column on page 30
of the
This column has log sin written at the top and log cos at the bottom. Since the given number is to be a log sine, the degrees
tables.
are taken from the top of the page and the number of minutes from thus find the left-hand column in line with the number 8.73997.
We
x
=
3 09'.
EXAMPLE 2. Given log cos y The given number is the same represents a log cosine.
8.73997; to find y. as in Example i, but this time
Therefore
we take
it
for the required angle
PLANE TRIGONOMETRY
82
[CHAP, iv
number of degrees at the bottom of the page and the number minutes found in the right-hand column, opposite the number in the table. Result, y = 86 51'. the of
It should be observed that the given mantissa 73997 appears a second time in the table, namely, on page 60, first column. But the numbers in that column have the characteristic 9, while the
given characteristic is 8. Beginners sometimes overlook the characteristic and are thus led into mistakes.
EXAMPLE 3. The number
Given log tan x = 0.08685; to 0.08685 cannot be found in the
but on page
table,
66,
^
e third column, are found the next smaller number 0.08673 anc* next larger number 0.08699. The difference between these two is
and
26,
that
two corresponds the angle 50
to the smaller of the
is,
mantissa log tan 50 41' mantissa log tan x
but
=
08673 08685
=
difference
41',
12.
we denote by d"
the difference between 50 41' and x the principle of proportional parts gives us
If
d"
:
12
=
9
60"
:
26, that
is,
and the required angle is
EXAMPLE
4.
= =
d" x
Given log cos x
=
i|
X
60"
=
9.87561
10; to find x.
mantissa log cos 41 19' = 87568 mantissa log cos 41 20' = 87557 n. difference for i' or 60" = mantissa log cos 41 19' = 87568 = 87561 mantissa log cos x = difference for
Also
and
:
:
hence the required angle
x
is
=
41
28",
50 41' 28".
60"
=
38",
19' 38".
EXERCISE 21
From 1.
the table of logarithmic trigonometric functions find:
log sin 13
24', log
cos 25
12', log
Ans. 9-3 6 5 2.
log sin 72
tan
ioo2
/ ,
log cot
17^00'.
2 > 9-95 6 57> 9- 2 4779> 0.51466.
n', log tan 46 17', log cot 65 13', log cos 67 59'. Ans. 9.97866, 0.01946, 9.66437, 9-57389-
LOGARITHMS
4 oj
83
tan 36 30' 15", log sin 70 15' 40". Ans. 9.62932, 9.86928, 9.97370.
3.
log sin 25
12' 30", log
4.
log cos 16
26' 45", log cot
9 27' 42", log cot 54 25' 09".
Ans. 9.98186, 0.77818, 9.85456. 5.
log sin 42
15' 10", log cos 17
10' 54", log
tan 51
18' 57".
Ans. 9.82763, 9.98017, 0.09653.
By means 6.
of the table find x,
x
log sin
=
when
there
=
9.77980, log tan x
is
given:
0.40017, log cos x
Ans. 37 7.
8.
log sin x
=
= log tan x
9.68931, log cos x
log sin x
=
68
Am.
9.89609, log tan x 29 16' 30", 38 04' 30", 55
0.30562, log
=
sin#=
8.97296, log sin x
+ log sin 32
log 2
10'
15"
23' 30",
5
Verify the relation
log cos 27
io'4o"
=
log (cos 13 35' 20"
+ log (cos 13
35' 20"
-
=
sin 13
+ sin
9.79558. 21'.
51
=
0.16999.
5'
15"-
9.97296.
69 59' 24".
+ log cos 32 Ans. x
10.
18',
=
Ans. 63 40' 36", 9.
02',
=
10' 15".
64 20' 30". 35' 20")
13 35' 20").
Use the natural function table to find the quantities (Suggestion. within the parentheses.) 11.
Given tan x
=
4.6525; find x without using a table of natural
functions. 12.
Given
(sin x)%
=
0.253; find x.
Ans. x
=
7
i8'4o".
or 90. When Logarithmic Functions of Angles near is very small, say less than 2, the logarithms of the sine, tangent and cotangent vary so rapidly that the principle of proportional parts fails to apply with sufficient accuracy, causing the results obtained by interpolation to be unreliable. The same remark applies to the logarithms of the cosine, cotangent and tangent 40.
the angle
of
an angle near 90, say between 88 and 90. In such cases tables used which give the required functions for sufficiently small
may be
intervals of the angle, say for each second. When such tables are not available the following rules may be employed. If the
sidered,
angle
is less
than
2
and only tenths
of minutes are con-
PLANE TRIGONOMETRY
84
[CHAP, iv
The log sin or log tan of the angle may be found by increasing the logarithm of the number of minutes in the angle by 6.46373-10. Conversely,
The angle corresponding to a given log sin or log tan may be found by The resulting diminishing the given log sin or log tan by 6.46373-10. number ^s the logarithm of the number of minutes in the required angle.
EXAMPLE
To
i.
find log sin o log 50.7
50.7'.
=
1.70501
6.46373
= = Given log tan x = log tan x log sin 50. 7'
EXAMPLE
2.
8.
16874
8.
50724
8.50724 6.46373
log* x
= =
-
10
10 10; to find x.
to
-
10
2.04351 110.5'
=
I
5-5
/ -
rules are based on the theorem, which will be proved later, that the sine and tangent of a small angle are each approximately equal to the length of the arc subtended by the angle at the center of a
These
circle
whose radius
is
the unit of measure, so that for very small angles may be substituted for either the sine or the
the measure of the arc
tangent of the corresponding angle. Now the measure of an arc in terms of the radius is found by multiplying the number of minutes in the arc
by 0.0002909,
for since the semi-circumference (the arc sub-
= 10800') measures 3.14159 tending an angle of 180 radius is i, each minute of arc measures 3.14159 -5- 10800 .
Log 0.0002909
=
of a small angle
.
.
when
the
= 0.0007,909.
10, consequently if the log sin or log tan 6.46373 increased by this amount the result will be the loga-
is
rithm of the number of minutes in the corresponding angle. When the angle is less than 2 and the exact number of seconds In such cases a special table, are considered the foregoing rule fails. known as the S and T table, is commonly employed.
For small angles 41. Use of the S and T Table (Table III). the ratio of either the sine or tangent to the length of the arc of unit radius is nearly constant, that is * ,
x
x being the
as well as .
5L, x
is
nearly constant,
ratio of the length of the arc to the radius.
LOGARITHMS
4i]
85
On
taking logarithms we have log sin x log x, as well as log tan x Let us put
-
log sin x
log x
then on transposing log x
=
5,
and
-
log x,
log tan x
is
nearly constant.
log x
=
T,
we obtain
log sin
a?
=
log
&
+
S
log tan
so
=
log
a?
+
T.
(T)
and
(2)
The values of 5 and T, corresponding to various values of x expressed in seconds, have been carefully calculated and assembled in III. The given angle x is first reduced to seconds, and the corresponding value of S or T is then taken from the table. This value increased by log x gives log sin x or log tan x, as the case
Table
may
be.
Conversely, if log sin x or log tan x is given and x is to be found, or T from the table, subtract it from log sin x or log tan x,
we take 5
may be, and obtain log x, from which x is found. cosine or log cotangent of an angle near 90 may be found log from the same table by substituting for the cosine of the angle the as the case
The
sine of its
complement, and
gent of its
for the cotangent of the angle the tan-
All this will be better understood
complement.
by
fol-
lowing through the examples which are worked out below. The characteristics of S and T are negative, so that 10 must be to each value taken from the table. appended
Given x
=
EXAMPLE
i
o 56' 26"; to find log sin
x.
x
=
o
56' 26"
=
3386".
Applying (i), From Table III,
=
3.52969
S =
4.68556
=
8.21525
log*
log sin x
EXAMPLE Given log
sin
x
=
8.21524
By
III,
(i),
=
10.
2
= S= log* =
x
10
10; to find x.
log sin x
From Table
-
3386"
=
o
56' 26".
8.21524 4.68555
3.52969
10
-
10
PLANE TRIGONOMETRY
86
[CHAP, iv
EXAMPLE 3
To find log tan
10' 51".
i
10'
i
51"=
4251".
= T= 51" =
log 4251
Applying From Table III, (2),
log tan
10'
i
3.62849
4-68564
-
10 10.
8.31413
EXAMPLE 4 Given log tan x
=
10; to find x.
8.31413
= T= = log x
log tan x
From Table
III,
Subtracting,
x
=
4251"
=
find log cos 89
log cos 89
log sin o
o 34' 49"
10.
log 2089
8.00551
34' 25'
log sin (90
log cot
i
i
x).
x)
7
o' 29". o'
10
6
= S= - *) = log (90 = 90-* 2089" =
III,
find log cot
4-68557
10; to find x.
EXAMPLE
To
3.31994
49"
log sin (90
From Table
=
S=
EXAMPLE
=
= 8.00551 n" = 8.00551 -
2089".
log cos 89
log cos x
10
49"
34'
=
log sino
=
-
3.62849
5
Applying (i), From Table III,
log cos x
10
n".
25'
n" =
25'
10
-
4-68564
10' 51".
i
EXAMPLE
To
8.31413
29"
-
log
- -
-^o
t
V^
log tan
i
o' 29".
8.00551
4-68557
10
-
3 .3 1 994
34'
49"
10
LOGARITHMS
42]
Then, as in Example
3,
log tan
i
o'
29"
log cot
i
o'
29"
= = =
10 8.24541 10 8.24541 1-75459-
EXAMPLE log cot x
=
1.75459; to find x. = log log cot x
-
log tan x
Then, as in
Example
-7
=
8
=
tan*
i. 75459,
10.
8.24541
4,
*= io'2 9 ". EXERCISE 22 Find log
1.
sin
o 45' 45", log cos 88
20' 02", log tan
i
Ans. 8.36696
10, 8.12414
54' 36". 10.
10, 8.27928
Find log sin o 50' 50", log cos 89 01' 55", log tan i 15' 17". 3. What would have been the error in each of the functions in 2, their values had been found from table II by interpolation ? 2.
if
Find each
4.
III,
of log tan 88
and check your
5.
x and
Given log
sin
result
x
=
05' 20", log cot 89
by Table
8.22925
50" from Table
tan y 10; find 8.43340 o 58' 17", y = i 33' 14".
10, log
Ans. x
y.
16'
II.
=
42. Historical Note. With the awakening of science in the sixteenth century, measurement became more precise, the resulting-
numbers more complex, and computation more and more tedious and time-consuming. The demand for shorter methods of computation than were then known led to the invention of loga ithms. It is therefore not very strange that the method of logarithms should have been developed independently and almost simultaneously by two mathematicians, John Napier, a Scotchman, and Jost Buergi, a tables Napier's tables were published in 1614. Buergi's Both 1620were computed before 161 1 but not published till Napier's
German.
and Buergi's
tables were soon superseded
by
Briggs' tables.
tables contained fourteen place logarithms of all numbers 20,000 and from 90,000 to 100,000. Briggs' tables were
by Adrian Vlacq
(1628),
who
Briggs' i to
from
completed shortened Briggs' tables to ten places,
PLANE TRIGONOMETRY
88
and computed the logarithms
of the remaining
[CHAP,
w
numbers from 20,000
to 90,000. Briggs' and Vlacq's tables are substantially the same as the tables in use to-day, though the tables have been checked and parts of them recomputed many times. The most complete check was under-
taken by the French authorities in 1784. It required the work of nearly one hundred mathematicians and computers for over two years.
The
resulting tables, giving fourteen place logarithms of all i to 200,000, besides natural sines and logarithmic
integers from sines
and tangents, have never been published.
Two
manuscript
copies are preserved, one at the Observatory, the other at the Institute in Paris.
CHAPTER V LOGARITHMIC SOLUTION OF
RIGHT TRIANGLES AND
APPLICATIONS 43. Logarithmic Solution of Right Triangles. In Article 20 it was shown how to solve right triangles by means of natural functions. Now we shall employ logarithms, which as a rule shortens the work. We shall illustrate each of the cases which may arise by an example.
EXAMPLE c
o
i.
= =
a
Given
521.2.
^
A =
sin
-
(b)
To
cos
.
find
A =
(c)
log a
Check.
+b = 2
+ colog
If
10
log
10
+
~ log
(c
log (c
c cos
+
ft
ft
ft)
ft)
= log c + log cos = 2.71700 log 10 cos A = 9.90009 = 2.61709 log b = 414-1-
-
62
=
(c
correct, it
is
+
ft)
(c
= = = =
-
ft)
must
=
,4
may
a2
= + b) + log = 935-3> 5 21 2 + 4H.i (c
ft)
^4
.
A.
satisfy the rela-
,
2 log a.
-
5 2 *' 2
"~
4J4- 1
107.1.
log a
2.97095
=
2.50037 2
2.02979
5.00074 checks,
=
ft
log (c c
ft
ft
log b
c.
our result for b
2 c2 , or c
from which
Since
or
,
c
= 2.50037 log a = 7.28300 colog c = 9.78337 log sin A A -37 23.5'. -52 36.5'-
tion a2
b.
=-
^4
c
log sin
'
_. Fig. 30.
To find A and B.
(a)
23.5',
52 4I4<1
A
Solution.
37
*>
-
-
i
= =
Required A
316.5,
74
5-
be assumed to be correct
from A. 89
also, for
ft
was found
PLANE TRIGONOMETRY EXAMPLE
v
2.
= =
a
Given
[CHAP,
b
40
Required
6.325,
47-9'>
49 I2
'
7.328. =
-
l7 >
9.680.
Solution.
A and B. = tan A T
To
(a)
Fig. 31-
find
To
(b)
A = A
sin
-
find
c.
Of
-
or c
,
sin
= log a + colog sin A. = log a + colog b. log c log tan 4 = 0.80106 a = 0.80106 log log a = A sin = io 0.18483 colog 9.13501 colog b c = 0.98589 = I0 6 log A 7 9-93 log tan c = 9.680. A = 40 47-9'. B = 49 12.1'. - b) 2 log a. 2 c 2 - & = a\ log (c + b) + log (c (c) Check, = = c + b 17.008, 9.680 + 7.328 - 7.328 = 2.352. c - 6 = 9.680 = 1.23065 loga = 0.80106 log(c + 6) 2 = b) Q.37H4 log (c 1.60212
1.60209
The discrepancy
is a slight discrepancy in the check. results final the of mantissas in the last figure only; that is, if the is 1.6021. Agreement in the are cut down to four figures, each side final logarithms of the check of the mantissas of the
Here there
first
is
four figures
is all
that can be expected
EXAMPLE Given
c
when a
=
35.145* 25
(b)
A =
-
= = logc sin A = log a
= log a
a
=
log c
To cos
.
3I-74S.
64 35' 30"
find
A
b.
== -.
c
c
log
= =
Fig. 32.
find a.
sin
used.
b
24' 30".
Solution.
To
is
Required a = 15-079,
3.
A =
(a)
five-place table
+ log sin A
I.54S 86
9.63252 1.17838 15-079-
io
= log c + log cos A = 1.54586 c log "" 1 cos A = g.955 82 L = 1.50168 log 6 b = 3I-745. log b
log
.
.
RIGHT TRIANGLES AND APPLICATIONS
43l
(c)
91
Check. log
ft)
log (c
ft)
+
+b=
c
log (c +
(c
= =
ft)
+ log (c -
b
c
66.890,
= =
ft)
2 log a.
3.400.
log a
1.82536
=
_
1.17838 2
0.53148
2.35676
2.35684
EXAMPLE Given
4.
=
ft
Required a
^65W
25.01,
=
=
^
Fig. 33-
11.57,
2*
S
-
Solution. (a)
To
find a.
(ft)
=
tan
-
To
find
sin
.
c.
B=
a
= = log cot B = = log a = a log a
log
log
(c)
.
c
+ log cot B.
ft
= log + colog = 1.39811 log sin B = 0.04214 = 1.44025 log c = 2 7-56c log c
1.39811
ft
-
ft
sin J5.
ft
10
9.66537
colog
1.06348 11.574-
Check. log (c
log (c
c
+
+
ft)
log (c
EXAMPLE Given
c
6
= =
+
ft
ft)
= =
=
ft)
+ log (c -
52.568,
ft
=
2 log a.
2.548.
log a
1.72072
=
1.06348
0.40620
2
2.12692
2.12696
5.
B Required
34.57, 34.04.
A
^^^S^\ Fig. 34.
Solution.
ft)
= c -
a
A = ^= a =
10 02.7' 79
57-3'
6.030.
In this problem the given side and hypotenuse are method employed in Example 3 does not
so nearly equal that the
give sufficiently accurate results. Art. 21.
We
therefore use the formulas of
92
PLANE TRIGONOMETRY
.
To
(a)
A and
find
J3.
+ =!-
(6)
logtan^
X
l
2
2
"
V
'
~
'
find a.
og* =
""
=
34.57
+b=
34.57
b
-
34.04
+
34.04
log (c
colog
6)
+
(c
ft)
= = = =
0.53,
68.61.
\A = .4 = B =
5
A =
79
-
8.94394
57' 16", or 79
2.7'
c+ log sin A
1.53870
10
9.24163
0.78033 6.030.
10
8.16361
10 02' 44", or 10
log
10
9.72428
2)17.88789 log tan \ 01' 22"
= sin A = = log a a = log C
log c
To
V c
2
-
[CHAP, v
20
10
to the nearest tenth minute.
57.3' to the nearest tenth minute.
Check.
(c)
+ b) + log (c b) = 34.57 + 34.04 = 68.61,
log c + b = c-b = = log (c + b) = b) log (c
(c
34.57
-
34.04
=
2 log a.
0.53.
1.83639
log
a
=
0.78033
10
9.72428
2
1.56066
1.56067
EXERCISE 23 Solve the following problems to the nearest second
and
by
all sides
logarithms, computing all angles to five significant figures. It is
when no answers
expected that the student check his results
are
given. 1.
a
=
=
168.92, c
= = = = =
A = 35 4o' 33", B = = Find4 = 43.148, b 84.107. = 3.71322.5346, c = 0.28634. 0.37640, b = 32 15' 24". Find 547.5, ^ = ^ 27 30'. 6700, Find
2.
a
3.
6
4.
a
5.
a
6.
a
289.64.
54 19' 27", b 27 09' 29", c
Find Find b
=
867.5, c
= = a,
235.28.
94-53
A, B.
A and B.
=
1025.8.
Find b and
c.
-
RIGHT TRIANGLES AND APPLICATIONS
44]
c
7.
=
672.34,
A =
8.
c
g.
c
10. c
= = =
i. ooi,
369.27,
ft
5464-35*
Number
44.
B*
two kinds: (a) Those (b) Those
16' 25".
35
Find 45
93
B =
54 43' 35", a
=
=
388.26, 6
548.90.
Find a and
45' 45".
= 235.64. Find A = = 545 2 'i3-
50
Find
<*
of Significant Figures.
20' 52",
a
B =
= 3
6.
284.31. 49' 57".
Numerical problems are of
which the given numbers are exact numbers. which the given numbers are approximate only. For example, when we say that each of the sides of a hexagon inscribed in a circle with unit radius is i, and each angle 120, i and 120 are exact in
in
numbers, that
more nor
less
On
120.
is,
the sides in question are to be considered neither i, and the angles neither more nor less than
than
the other hand,
when we say
that the side of a field meas-
ures 631.7 feet and the angle at a corner 73 37', the numbers 631.7 and 73 37' are mere approximations. So far as we know the exact
length of the measured side may be any number between 631.65 and 631.75 and the measured angle may have any value between 73
36.5'
results
the
and 73
When
I.
37.5'.
numbers of a problem are exact numbers, the asked for can be carried out to as many significant figures as
number
the given
of figures in the mantissas of the logarithms used in the
where the computations are based on a fivebe limited to five significant figures and must place table,* lengths Even then to seconds. the fifth figure and the number of angles solution.
In
this book,
seconds cannot always be relied upon. 11. When the given numbers of a problem are the results of measureneed answers the not more contain ment, significant figures than the least accurate of the given parts. Thus, if one side of a triangle measured to the nearest inch and another to the nearest tenth of
is
an inch, the answer for the third side need only be given to the nearest inch.
make
The
following directions will assist the student to
consistent measurements
and
to avoid useless calculations.
Distances expressed to three figures to the nearest five minutes, and vice versa. i.
*
Five-place tables answer most of the
demands
call for
angles
expressed
of applied science.
The
struments ordinarily used by engineers read angles to the nearest minute only.
in-
PLANE TRIGONOMETRY
94
[CHAP,
v
2. Distances expressed to four figures call for angles expressed to the nearest tenth of a minute? and vice versa.
3. Distances expressed to Jive figures call for angles expressed to the nearest second, and vice versa. 4. Distances expressed to six figures call for angles expressed to
the nearest tenth of a second, and vice versa. A six-place table must be used to obtain like accuracy in the answers.
Distances expressed to seven figures call for angles expressed to the nearest hundredth of a second, and vice versa. A seven-place table is necessary to obtain like accuracy in the answers. 5.
In this connection the student should observe that, whenever a is the result of measurement or other approximation, a cipher
number
to the right of a decimal fraction has a distinct significance and cannot be dropped at. will, as is customary in dealing with exact numbers. For example, the square root of 3 is approximately represented by
each of the numbers
1.7, 1.73, 1.732,
proximation being closer the
more
1.7320, 1.73205, etc., the ap-
figures
we
write; but 1.7,
when
used as an approximation for V3,_has not the same meaning as 1.70, for the former means that \/3 has some value between 1.65
and 1.75, while the latter means that the number represented has some value between 1.695 an(^ I -75 which is not true. Simithe numbers when measures of 62, 62.0, 62.00, larly they represent distances or other quantities, are not equivalent. The first implies that the measurement has been carried out to the nearest unit, the second to the nearest tenth, and the third, 62.00, that the measurement has been made to the nearest hundredth of a unit. 46.
Applied Problems Involving Right Triangles.
The
follow-
ing six sections deal with applied problems involving the solution of The problems are grouped with reference to the right triangles.
questions dealt with, and the problems in each set are so arranged that the more difficult come last. It is not expected that any one
student work every problem, but only as many as may be necessary to make him reasonably familiar with the method of solving right triangles
by means
After that an additional hour or
of logarithms.
two may profitably be spent in the analysis of the more difficult problems involving two or more right triangles. In problems where no answer is given the result must be checked by the student. *
The
nearest 10"
is
somewhat
closer.
RIGHT TRIANGLES AND APPLICATIONS
46]
95
Heights and Distances
46.
EXERCISE 24 1. From a point 185 feet from the foot of a wireless telegraph mast, the top of the mast was found to form an angle of 52. Find the height of the mast. Ans. 237 ft. 2.
A man
walking along a straight road observes a church in a making an angle of 50 with the road. After walking
direction
another mile, he comes to the crossroad on which the church is The roads cross at right angles. How far is the church
located.
from the intersection of the roads?
Ans.
1.19 miles.
3. The summit of a mountain, known to be 14,450 feet high, is seen at an angle of elevation of 29 15' from a camp located at an altitude of 6935 ^eet Compute the air-line distance from the camp -
summit
to the
The
Ans.
of the mountain.
2.9 miles.
the height of a roof to its span is one-fourth 4. is the inclination of the roof to the horizontal what (quarter pitch), line? Ans. 26 34'. ratio of
5. During a storm a tree was broken into two parts which remained connected. The broken part made an angle of 35 with the ground and its top reached a mark 165 feet from the foot of the tree. Required the height of the remaining stump and the height of the tree
before
it
Ans.
broke.
116
ft.,
317
ft.
ft. long are to be an angle of 25 and supported by four uprights, one at each end and two at equal distances between the two ends. How far apart must the uprights be placed and what are their lengths,
6.
In constructing a grand-stand, timbers 28
inclined at
the shortest being 6
A
ft.
long?
long stands on a building whose height is a point at the same level as the foot of the building the angles of elevation of the top and bottom of the flagpole are measured and are found to be 57 and 53 respectively. Required 7.
flagpole 25
unknown.
ft.
From
the height of the building.
Ans.
By
natural functions, 156
ft.
8. From the top of a tower the angle of depression of a point in the same horizontal plane with the base of the tower is observed to
be 47 as seen
What
be the angle of depression of the same point Ans. 28 23'. from a position halfway up the tower?
13'.
will
PLANE TRIGONOMETRY
96
[CHAP, v
9. A spherical balloon whose radius is 10 ft. subtends an angle of i46', while from the same position and at the same time the angle of elevation of the center of the balloon is 54. Determine the
height of the center of the balloon.
Ans. 525
ft.
An
observer finds that the top of a spire due south of him has an angle of elevation of 25 36'. He goes to a point 650 ft. east of his first position and now finds that the spire bears 40 12' south10.
west.
Find the height of the spire. It was found that the shadow
of a tall factory chimney lengthwhile the sun's elevation changed from 59 to 42. ReAns. 167 ft. quired the height of the chimney. 11.
ened 85
ft.
47. Problems for Engineers. It is suggested that the student use the graphic method in checking the problems in this set to which no answers are given.
EXERCISE 25
A
from a point A to a miles east and which is 5.95 second point 9.36 miles north of the first. What will be the direction of the road, assuming that it fol1.
branch railroad
is
to be constructed
B
N. 3227' E.
Ans.
lows a straight line?
To determine
the width of a stream a surveyor measures a long along one bank. At B he turns a right angle and his assistant places a stake in the line of sight at C on the oppo2.
line
AB
site
bank
wide
ft.
of the stream.
The
BAG
angle
the stream?
On
3.
are
is
375
map on which
a
drawn
i
measures 64 42'. How Ans. 793 A.
inch represents 1060
for differences of 100
ft.
in altitude.
contour lines
ft.,
What
is
the actual'
by that portion of the at which the contour lines are one-fourth inch apart?
inclination of the surface represented
Ans. 4. is
A
bolt 2 inches in diameter has
map
21 48'.
six threads to the inch.
What
the inclination of the thread to a cross section of the bolt ?
Ans. 5.
A
car track runs from
at an incline of 7 45',
1585
ft.
C
is
known
A
to
JB,
and then from
B
.
a horizontal distance of 1275 ft. B to C a horizontal distance of
to be 509 ft. above A. to C? from
inclination of the track
/
i3i.2
What
is
the average
RIGHT TRIANGLES AND APPLICATIONS
471
Two towns A and
6.
B, of which
by a new
to be connected
B
Ten
road.
97
is 25 miles northeast of A, are miles of the road is constructed
A
from
in the direction N. 23 E., what must be the length and direction of the remainder of the road, assuming that it follows a straight
line?
Ans.
A
7.
16.17 miles >
N. 58
23.8' E.
surveyor wishes to ascertain the distance between two inA and B. He starts from a point C in a straight
accessible objects line
with
A
and
tended by
B
and measures
in a direction at right angles to
AB
At D he measures the angles suband EC and finds them to be 75 35' and 34 46' Find the distance between A and B on the supposi-
CD AC
a distance
respectively.
equal to 500
ft.
tion, (a) (b)
C is
That That
between
A and
B is between A
B,
and C. Ans.
One end
8.
crank BO,
AB
of a connecting rod
i ft.
long, while the other
which
How
is
far
(a) (b)
A
OC
9.
and
Two
AB
to
They ft.
radius.
P of
Find how
the
be,
to
AB?
=
1.099
ft-
*V^+J-4=
1424
ft.
(a)
V26
4
from (b) drop a perpendicular J5C find AC.) ABC the from then triangle C, at
are to be connected
A
move along AO.
In case
railroad tracks intersect
of 54 1 6'. of 100
ft.
fastened to a crosshead
When OB is perpendicular When angle BOA = 6o? (&)
(Suggestion.
long, is fastened to a
5
from the extreme position
Ans.
find
2292, (b) 1598.
is
constrained
crosshead will
Fig- 35-
y
end
(a)
B
to
AO,
an angle
by a curve
far
from the
intersection point O of the tracks the curve begins and the length of the curve.
Ans.
OB =
51.25
ft.
Arc
AB =
94,71
ft-
Fig. 36.
18 inches and 30 inches respecpulleys whose radii are Find the length of the ft. apart from center to center. are 8 tively belt connecting the pulleys, 10.
Two
the pulleys are to turn in the same direction, directions. the If (b) pulleys are to turn in opposite (a) If
PLANE TRIGONOMETRY
98
[CHAP,
v
Applications from Physics.
48.
EXERCISE 26 1.
The
horizontal distance between the two extreme positions of
a pendulum 39.1 inches long it swing? 2.
Two
forces of 10
3.
Through what angle does Ans. 8
and 24
makes with the
What
platform 6
5.73 inches.
first of
the two given forces. Ans. 26 Ibs., 67
A
23'.
necessary to roll a barrel weighing 500 Ibs. onto a high along an inclined ladder 12 ft. long?
force ft.
Ibs.
is
Ans. 4.
24'.
respectively act at right angles to Find the resultant force, and also the angle which the
each other. resultant
is
ball weighing
300
Ibs. rests
250
Ibs.
on a smooth plane inclined at an is necessary to keep
angle of 12 30' to the horizontal. What force the ball from rolling down the plane, (a) If the force acts parallel to the inclined
If the force acts in
(ft)
plane a horizontal direction ?
Ans. 5.
A
block of
wood
rests
An
64.93 l^s., (b) 66.51 Ibs.
As the
on an adjustable inclined plane.
inclination of the plane reaches 29 Find the coefficient of friction.* 6.
(a)
?
37' the block begins to slide.
automobile moving at the rate of 45 'miles per hour is overAs seen from the automobile the raindrops
taken by a shower.
seem to come down at an angle velocity of
of 30
with the vertical.
Find the
the raindrops, assuming that their actual direct %on
Ans. 38.1
vertical.
ft.
is
per sec.
in 7. Through what angle must a fir log 30 ft. long and 54 inches diameter, standing on end, be tilted before it begins to fall? The Ans. 8 37' 40". log is assumed to be cylindrical in shape.
According to Wollaston the intensity of sunlight is equal to 61,000 standard candles acting at a distance of i meter. What is 8.
the intensity of sunlight striking a surface at an angle of 31
Ans. 9.
The
which *
The
is
52210
27"?
c.p.
fans of a windmill are inclined 25 to the plane of the wheel at right angles to the direction from which the wind
coefficient of friction is
of the plane
08'
on which the block
equal to the tangent of the angle of inclination rests.
RIGHT TRIANGLES AND APPLICATIONS
48]
What
blows.
fraction of the wind's force
is
99
effective in turning the
wheel?
Ans. 0.383.
A weight
of 437 Ibs. is
suspended and pushed
17 30' out of the a the horizontal force. horizontal force necesRequired by the in to hold this body position. sary 10.
vertical
n. What
is
the displacement
light
AB
from
air to glass
CM
of a ray of
in passing through a glass plate PQ, 0.215 inches thick, at an angle of 55 47' with the perpendicular EB, the index of refraction *
being approximately 1. Ans. 0.098
in.
Fig. 37.
A sail ship sails
against the wind at an angle of 60. are set so as to make an angle of 15 with the direction 12.
ship.
What
The
OF
part of the wind's force
sails
of the
is effec-
motion of the
tive in producing the forward
Ans. 0.183. Let be the direction of (Suggestion. the wind, OS the direction in which the sail ship?
WO
Decompose a unit two components,
is set.
RO Fig. 38-
sail
into
and
of the wind's force
RP
parallel to the
PO perpendicular to the
sail.
PR
has
and may therefore be disregarded. The other component PO may again be resolved into two components, namely, PQ perpendicular to the direction of the ship and QO in the direcno
effect
on the
tion of the ship.
sail,
PQ
is
neutral so far as the forward motion of the
concerned, leaving QO as the only part of the wind's force effective in the direction OF.) is
ship
10 ft. above the level of the water A person whose eye is at observes at I the image of the foot of E
13.
PI
y
,
a pile driven in the water.
The
horizontal
the observer, from the place where the image is formed, is 20 ft., his
distance
of
What is distance from the pile is 65 ft. the length PF of the pile below the surface of the water, the refractive index
to water being approximately $. *
Index of Refraction
from
air Fig. 39-
Ans. 49.7 sine of angle of incidence sine of angle of refraction
'
ft.
100
PLANE TRIGONOMETRY
[CHAP,
v
Two
bodies A, weighing 2 Ibs., and B, weighing 3 Ibs., are so placed that B is exactly 10 ft. west of A. A moves north and B west, 14.
each at the rate of 12
ft. per second. What the direction and the velocity of their common center of gravity?
EB
is
c
Fig- 40.
Ans. 33 41'; 8.65 ft. per sec. Locate the center of gravity, C, in two positions, as Find and EC, then solve the triangle CEC".)
(Suggestion.
EC
C and C.
The arms of a lever At the extremity A tively.
/ F
Six
16.
F =
may
10, act
2.34
and
of the first arm, acts in a direction
a=
63
FA
45' with
FB =
5.27 respec-
a force of 5.34 units
making an angle
What
produced.
of
force
must be applied at J5, the extremity of the second arm, in a direction making an angle ~ 5 l I f with FB ft S produced, in order
s/4\
p.
that there
FA =
are
15.
be equilibrium?
forces,
A =
B=
15,
6,
on the same point and
C=
in the
5.7,
D=
7.9,
E=
12.3,
same
plane.
The angle between A and B is 12 the angle between A and C is 31 the angle between A and D is 47 the angle between A and E is 58 the angle between A and F is 72
30', 21',
46', 10',
18'.
.
F Required the resultant force and the direction it makes with A. Ans. 50.51, 36 34'. (Suggestion. Resolve each force into two components, one along the other in a direction perpendicular to PA. Sum the components along each of these directions separately. The sums are the
PA,
rectangular components of the required force.)
49.
Problems in Navigation.
In the following problems it is acquainted with the divisions of the the mariner's compass the total angular divided equally into 32 divisions, each of
assumed that the student mariner's compass.
space about a point
which
is
On is
called a point, that
is
is,
a point
is
3-5equivalent to
32
=
n
15'.
RIGHT TRIANGLES AND APPLICATIONS
49]
101
Each point is divided into two half-points, each half-point into two In the figure below, the names of the 32 points quarter-points. are indicated by their abbreviations. Between north and east the points read:
North by northeast
by
east,
north northeast, northeast by north, northeast,
east, east northeast, east
by
north,
and
similarly for
each of the other quadrants.
Fig- 43-
In the following problems the surface of the earth is considered a plane and the distances straight lines, not arcs. By a mile is understood a sea mile or knot, which is the length of a minute of arc measured on the earth's equator so that the earth's circumference measures X 60 = 21,600 sea miles. A sea mile is approximately i J
exactly 360
common
miles.
Definitions.
The
and west component of a course, or disis called the DEPARTURE of the course or distance, the north and south component is called east
tance between two points,
WN
the difference in LATITUDE, that is, if repreany course or distance, and a right triangle is formed, by drawing through a line east
sents
W
WSN
and west, and through
WS Fig-
44-
is
N
a
line
north and south,
called the departure,
and
SN
in latitude of the course or distance
In nautical problems difference degrees and minutes
(i
mile
=
i')
the difference
WN.
in latitude is usually expressed in
PLANE TRIGONOMETRY
102
EXERCISE 1.
A
2.
A
[CHAP.V
27
N.E. by E. at the rate ship the rate at which it is moving due north. sails
Find
of 8 knots per hour.
Ans. 4$ knots per hour. ship
sails S.E.
by
S.
parture.
A
3.
vessel sails
W. by
a distance of 578 miles. Find its deAns. 321.1 miles.
S. until
is 315 miles. Find Ans. 321.2 miles.
the departure
the actual distance sailed.
A
4.
473o' N. on a course N.W. by N.
ship sails from latitude
685 miles. Find the latitude arrived at. Ans. DifT. in latitude = 569.6 miles
Required latitude
A
5.
ship sails S.W.
by
S.
=
30'
+9
569.6'
29.6'
=
= 56
a distance of 1225 miles.
between the
difference in latitude
47
=
first
and
9
29.6'.
59.6'.
Find the
last positions of the ship
and the departure made.
A
6.
ship sails from latitude 10 24' N. and after 30 hrs. reaches 26' N. Its course was N.N.E. Find the average speed
latitude 15
Ans. 10.9 miles per hr.
of the ship.
58' N. on a course between S. and 7. A ship sails from latitude 35 E. a distance of 359 miles to a point whose latitude is 32 16' N. Ans. S. 51 48' E. Find the course of the ship.
A vessel
8.
sails
from latitude 5 21' S. on a course N.E. by N. a Find the new latitude and the departure. Ans. 8 ii'N., 542.3 milts.
distance of 976 miles.
9. A steamer bearing W. by N. with a speed of 12 knots has a * broadside across her track which after 5 hours current setting port to located 108 miles from her starting point. an island her brings
Find the true course
Ans. N.N.W.
of the ship.
is 19 miles due N. Two two ports at the same time, one from B sailing due E. at the rate of 9 knots an hour, the other from A. The vesDetermine the speed and the course sels meet 5 hours out of port.
10.
One port A
of a second port B.
vessels leave the
Ans. 9.77 knots,
of the second vessel. 11.
67 07' E.
From a "crow's nest" no
depression of a rock *
S.
The
ft. above the water, the angle of above the water was found to measure just
left-hand side of the ship as one faces ahead,
opposed to starboard.
RIGHT TRIANGLES AND APPLICATIONS
SO]
15
103
Find the distance from the rock to the foot
36'.
of the
Ans. 394
mast.
A
12.
ferry,
whose speed in
still
water
is
ft.
4 miles per hour, crosses
a channel whose current is 3! miles per hour. How much will she " " in order to make the run straight across, and bear up have to how long will it take her to cross, the channel being 7 miles wide?
An
13.
flash of a
observer on board ship notices that the time between the gun from a fort located N.W. by W. and the report is 5
After sailing N.E. by N. the gun was heard again, and this time the interval between the flash and the report was 10 seconds. seconds.
Find the distance
sailed
and the bearing
of the fort
Ans. 9440
position of the ship.
(Assume the velocity
of
sound to be 1090
ft.
from the second
ft.,
S.
63 45'
W.
per second.)
A ship sailing due N. observes two lighthouses in a line due two hours later the bearings of the lighthouses are found to and W., be S. by W. and S.W. by W. respectively. The distance between the Find the rate at which the lighthouses is known to be 10 miles. 14.
ship
is
moving.
A
man-of-war sailing due N.E. at a uniform speed of 20 knots 15. observes at 9.30 A.M. a fort bearing N.N.W. Twenty-four minutes Find the distance and bearing of the fort later the fort is due N.W. Ans. 20.54 miles, N. 64 55' W. from the ship at 10.15 A.M.
50.
Geographical and Astronomical Problems.
EXERCISE 28 1. The shadow of a vertical pole 35 ft. high the sun's altitude (angle of elevation)?
2.
In Fig. 45,
circle center
the circle center
let
M the moon.
The
angle
E
is
51
ft.
long.
Ans. 34
What
represent the earth and the
PME,
EM
and a formed by the line of centers line drawn from tangent to the earth, is known as the moon's equatorial horizontal parallax and measures 57' 02". EP,
M
the earth's
Determine
mean
EM,
from the earth,
radius,
is
3959 miles.
the distance of the
Ans.
By
is
27.6'.
Rlg
moon
use of S and
T
*
45
'
table, 238,650 miles.
PLANE TRIGONOMETRY
104
In Fig. 45, the angle
3.
and a
line
drawn from
REM,
[CHAP,
formed by the
line of centers
v
ME
E tangent to the moon, is known as the moon's
angular semidiameter and measures 15' 34". Use the result of the last problem and determine RM, the moon's radius.
Ans.
By
use of S and
T
table, 1080.6 miles.
The sun's equatorial horizontal parallax (see Problem 2) is 8.8". radius of the earth is 3959 miles. Find the distance of the
4.
The
sun from the earth.
Also the sun's diameter, the angular semi-
Ans. 92,798,000 miles, 865,620 miles.
diameter being 16' 02".
The
largest angle between Venus and the sun as seen from the 47 30'. Using the sun's distance as given in Problem 4, find the distance of Venus from the sun, the orbit of Venus being assumed 5.
earth
is
circular.
In Fig. 46,
6. (
~
3959
miles).
circle, latitude
66
EO
represents the earth's radius the radius of the arctic y
Find
AP
Ans.
32'.
1577 miles.
7. Prove that lengths of two parallels of latitude are to each other as the cosines of the latitudes.
One
degree of longitude on the equator is approximately 69.1 Determine the length of a degree of longitude at Seattle, Ans. 46.5 miles. 47 40' N. latitude. 8.
miles.
9.
Prove that the lengths
of the degrees of longitudes at different
latitudes are to each other as the cosines of the latitudes. 10.
If
one minute of arc of longitude in latitude 60
1012.7 yards, how long to be a sphere ?
A
11. ship sails due difference in longitude
ship?
is
measures
the radius of the earth, assuming the earth
W. 540 miles in latitude 36 N. What is the between the initial and final positions of the Ans. 9
40'.
12. How high above the Pole would an observer have to be to have the Arctic Circle for his horizon? (Use the data of Problem 6.)
Ans. 357 miles. 13. Beginning at latitude 40 N., two consecutive section lines run How far apart are they directly north for a distance of 100 miles.
at their northern end?
Ans.
5166
ft.
RIGHT TRIANGLES AND APPLICATIONS
So]
The
14. is
21
IOS
shortest shadow cast by a vertical rod 25 ft. long at noon the longest shadow cast by the same rod at noon is 56 ft. Find the approximate latitude of the place.
ft.,
The
Explanation.
shortest shadow, 05, Fig.
on June 21, the longest shadow, OL, on December 21. Halfway between these dates the sun will be on the equator; its elevation above the horizon at noon will then be the
47,
.
is
cast
Fig. 47.
latitude of the place, that
the figure, where 15.
EP
bisects the angle
A wall runs east and west;
to the wall,
is
10
wide.
ft.
its
The
LPS.
angle
is,
OPE
Ans. 52
in
59'.
shadow, measured at right angles
altitude of the sun
is
25
30', its
azimuth (angular distance west of the south point) is 27 45'. Determine the height of the wall.
Let
(Suggestion. wall, FL the
WE
width
of
represent the the shadow
measured at right angles to the
MP the direction of the sun. FMP MFL
is
its
the
wall,
Then angle
and angle The problem in-
sun's altitude
azimuth.
MFL
and PFM, in which FL, volves two right triangles, namely, PF is to be found.) are and and given angle angle
PMF
MFL
two observers notices a bright meteor. To the first the meteor appeared directly south and at an elevation observer, A, To the second observer, B, stationed 40 miles west of A, the of 54. 16.
Each
of
meteor appeared 56 times of
observation
been observed.
Compute
PF
and
the height at which the meteor was seen. L to be A's position, M, B's
In Fig. 48 consider
(Suggestion. position,
east of the South point. On comparing the it was ascertained that the same meteor had
P
the position in which the meteor was observed.
represents the height of the meteor.) 17. Find the greatest distance at sea at which a mountain 14,500 ft. high can be seen, the earth being considered a sphere, radius 3960 miles, and
the distance sought being the chord joining the point at sea to the foot of the mountain.
Ans.
By
S and
T
functions, 137.4 miles.
PLANE TRIGONOMETRY
106
The distance sought is SF =
(Suggestion.
cos
OS SOF = -
OM
,
OS and
but as
[CHAP,
SO
2
sin
OM are nearly equal,
v
\ (SOF) and
it is
better to
use the formula in Article 21 for the determination of angle SOF.) 51. Geometrical Applications. Many geometrical problems can be solved by properly dividing the given figures into right triangles and solving these. Thus, an isosceles triangle is divided into two equal right triangles by drawing a perpendicular from the vertex to the base; any rhombus is divided into four right triangles by its two diagonals; any oblique triangle is equal to the sum or difference of two right triangles, formed by drawing the perpendicular from
to the opposite side;
any vertex as
many
any
regular ploygon
equal isosceles triangles as the figure
has
is
sides,
divided into
by the
lines
joining the vertices of the polygon to its center; etc.
EXAMPLE
Two
i.
length of one side
rhombus meet at an angle. Find the lengths of the diagonals.
sides of a
is a.
By
Solution.
The
definition the sides of a
rhombus are equal and from geometry
A^?^-
1.
^^>>c
^a^^l^-^"^^ D
rhombus.
,
Flg s '
AB = AD =
it
known
that the diagonals intersect at right angles and bisect the angles of the is
'
Let
BAD =
e be the given angle,
and
a, the given side.
In the right triangle
ABO, two
parts, namely, the hypotenuse a
a
and the acute angle - are known, hence we may
find
AO
and OB.
2
AO =
a cos ^0,
and
BO =
AC = 24O=20cos0,
and
BD = 2 BO =
asin|0,
hence
In particular,
if
the given side
AC =
BD =
30
EXAMPLE
sin 19
2.
The
=
is
15
and the given angle 38, we have
30 cos 19
9.768,
20sin0.
=
by the use
radius of a circle
28.365, of natural functions. is r.
To
find the perimeter
of a regular inscribed polygon of n sides. Solution. By definition the sides of a regular polygon are equal.
and area
RIGHT TRIANGLES AND APPLICATIONS
Si]
Let O represent the center of the circle inscribed polygon.
O
The angular magnitude about there are
n
AOB
Triangle
from
O
angle
AOB
angle
sides,
is isosceles,
perpendicular to
one side of the
and, since
360
.
so that
AB,
AB
n if
OC
is
drawn
will divide the tri-
it
two equal right
into
is
^360
AOB =
and
triangles.
In the right triangle AOC, the hypotenuse equals ^ 1CJ = I nP hence and the acute angle AOC equals 3 ,
,
107
~1 1
.
I
}
n
AC
and
OC
can be found.
OC =
ylC=rsin^-, w
=
^41?
vlC
2
ft
=
2 r sin
1 80 >
w and the perimeter
= n AB = -
2
w sin
Also the area of the triangle AOB
and the area
of the entire
In particular,
if r
=
polygon
=
=
1 80
\
n ^1
5 OC = r2 sin ^-^- cos ^-w
n times the area
95 and the figure
is
of triangle
a heptagon, n
=
7,
AOB
and we
have
O O
Perimeter A Area
log 14 log 95
= =
=
=
7
95
14
sin
2-180 sin Z
1-97772
i 80
cos
lg
1.14613
=9- 6 3737~ I Q
log sin
95
2 log
7
95
=
log sin
0.84510 3-95544 9- 6 3737
o O
=
2.76122
log cos
=
9.95471
7
perimeter
-
10
-
10
7
7
log perimeter
= =
=
577.06.
log area
area
= =
4-39262 24696.
>
w
'
PLANE TRIGONOMETRY
108
EXAMPLE
[CHAP,
v
A
right pyramid has for its base a square whose angle formed by a face and the base of the pyramid Determine the altitude, slant height and lateral edge of the is 6. pyramid, also the angle made by a lateral edge and the plane of the base, and the angle between the lateral edge and the edge of the base. side
3.
The
is 2 a.
V-ABCD
Let
Solution.
and VM its slant and angle VMO =
= = = a =
height. 6.
represent the pyramid, and 0. Then
Join
VO
altitude
its
MA = MO =
a,
Further, let
h
altitude
5
slant height
MV,
I
lateral
A F,
angle
M
FO,
edge
VAO, made by a lateral
edge and the plane of the base,
=
VA M, made by
angle
a lateral edge and an edge of the
base.
In the right triangle found.
VMO,
a and
are given, hence h
=
a tan
s
6,
=
a sec
In the right triangle VMA, a is given and hence / and may be found. Solving, ft
=
tan" 1 -
a
,
I
=
may be
a sec
jS,
s
6.
has just been found,
or
from the right triangle VAO,
As a numerical example take the = 60. Then a = 5, the angle and
s
Solving,
h
Finally,
and
side of the base equal to 10,
= 5 tan 60 = 5 N/3, slant height, s = 5 sec 60 = 10, = V$* + io = 5 \/S, lateral edge, a = sin- V ^ = 50 46' 08", = tan- 2 = 63 26' 06". altitude,
h
2
/
/
1
1
and
RIGHT TRIANGLES AND APPLICATIONS
Si]
109
EXERCISE 29
The base
1.
vertex
is
The chord
2.
an
of
isosceles triangle is 12
Find the altitude of the
48.
subtended by
of a circle
it is
42
10'.
is
20
and the angle at the Ans.
triangle.
Find the radius of the
circle.
Ans.
The
angle between two lines
13.48.
long and the angle at the center
ft.
27.80.
50 21' 24" and a circle whose radius is 2380 ft. is tangent to both of them. Find the distance from the intersection of the two lines to the point of tangency, (a) When the circle lies in the smaller angle, 3.
(&)
lines
When
the circle
lies
is
in the larger angle
formed by the two
Ans.
-
4.
The
5.
A
5062.8, 1118.8.
radius of the inscribed circle of an equilateral triangle Find the radius of the circumscribed circle.
chord of a
In the same
circle,
subtends at the center an angle of 80 24'. large is the angle subtended by a chord half Ans. 37 39.4'.
circle
how
as long? (Suggestion.
is r.
Call the length of the chord a.)
6.
The
side of a regular octagon
7.
The
radius of a circle
Find the area.
is 7.
Ans. 236.59.
n
regular polygon of
One side
is r;
show that a O
sides is 2 r tan
o ^L
side of a circumscribed
m
n
of a right triangle
is 27.5 and the adjacent acute angle the length of the perpendicular from the vertex of the right angle to the hypotenuse, and the segments into which the hypotenuse is divided.
8.
is
54
9.
38'.
Compute
Solve the preceding problem, using a for the given side and
B
for the given angle.
= a cos J5, n = a sin B tan B, where p is Ans. p = a sin B, the perpendicular, and n the segments of the hypotenuse, being the segment adjacent to B.
m
m
m
In an oblique triangle two sides and the included angle are = 25.37, * = 3 6 12 C = 35 27'. Find the remaingiven, namely a Ans. A = 43 35.9', B = 100 57.1', c = 21.34. ing parts. 10.
-
(Suggestion.
,
Divide the triangle into two right triangles by drawing
a perpendicular from one of the
vertices,
A
or B, to the opposite side.)
HO
PLANE TRIGONOMETRY
11.
given,
[CHAP, v
In an oblique triangle one side and two adjacent angles are c = 10, A = 60, B = 75. Find the remaining parts. Ans. C = 45> a = 5 ^6, * 5
namely
7
Divide the triangle into two right B triangles by drawing a perpendicular from B to the (Suggestion.
,
side opposite.)
A
regular parallelepiped has for its base a = 8, A rectangle whose dimensions are 10, 12.
D=
AB
and
altitude
its
the diagonal
A A' =
AC
Find the angles which 15. makes with AD, with AC and Fl'g- S3-
with AB.
13. A right pyramid has 2 a for an edge of the regular hexagon which forms its base and an altitude equal to a. Find the angles which a lateral edge makes with an edge of the base, with the plane
of the base,
and the angle which a
makes with the plaae
lateral face
of the base.
Ans.
tan~ l
=
2
63
26',
tan~
l
\
=
26
34',
tan~
!
\
V$ =
30.
14. Verify trigonometrically the following practical rule for in-
scribing a regular pentagon in a circle:
Let radii.
and
O
be the center of the
Bisect
CR
OA
in
as a radius,
P.
M.
circle,
Take
and
draw an arc cutting the
PC
will
be the side
(Suggestion.
From From From Fig. 54.
OA
OC
MR equal to MC.
two perpendicular
With C
circle in
P.
as center
C and
Join
of the pentagon.
Assume the
radius of length
triangle
OCM find CM =
triangle
COR
triangle
COS (SC=|PC)
find
CR =
r.
RM.
CP. find angle
SOC, which should be 36.)
Oblique Triangles Solved by Right Triangles. Every may be solved by decomposing it into right triangles. This is done by drawing a perpendicular from one of the vertices of the triangle to the opposite side. In three of the four cases the perpendicular can be so chosen that two of the given parts become parts of one of the right triangles. This triangle having been solved, two parts of the other right triangle become known. The second triangle may now be solved, and with this all the parts of the original The fourth case (given the three sides) triangle become known. 62.
oblique triangle
RIGHT TRIANGLES AND APPLICATIONS
52]
requires a
somewhat
different
method.
We
will
III
take up each case
separately.
Case
Given one side and two adjacent angles, as
I.
c
b,
C, A.
c
ABC represent the triangle. From A or C, say C, draw a perCD to the opposite side, or opposite side produced. Let CD = p, AD = m, DB = n, angle ACD = 6, angle BCD =
Let
pendicular
.
Three
different figures
may
arise,
the left-hand figure, when the middle figure, when A
the right-hand figure,
In the right triangle and 6 may be found.
A CD,
b
A and B is
are both acute,
acute and
when A
and angle
is
B
obtuse,
obtuse.
CAD
are
known, hence
From C and 0, may be found. and p, we know two parts of the Having found BCD, hence a and n and the angle CBD may be found. Knowing m and n, c may be found.
p,
m,
<
To
check the answers, we repeat the solution, drawing the per-
pendicular from
Case as a,
right triangle
by
II.
A
instead of from C.
Given two sides and the angle opposite one of these
sides,
A.
In this case draw the perpendicular from that vertex which
between the two given sides a and If a < #, two different triangles as in Fig. 56,
left, if
a
=z &,
lies
b.
exist
which have the given parts,
only one triangle
exists.
PLANE TRIGONOMETRY
112
In the right triangle CAD, b and (= angle ACD) may be found.
A
[CHAP, v
are given, hence p,
Next consider the right triangle BCD. n and (= angle BCD) may be found.
m
and
9
p and a are known, hence
Finally,
AB = c = m + n, C = 0+<, B = The second given
solution,
if
there
is
-
180
(A
+ C).
one, as in the figure to the
left, is
by
AB' = c'=m-n, ACB' =C" =0-0, AB'C=B'=iSo-(A+C). From
Check.
From and
since
triangle
ACD,
triangle CDJ3,
m+w=
c,
Gwew
III.
Draw
n =
b cos A,
a cos
J5,
we must have cos
Case
m=
A
+ a cos 5 =
c.
two sides and an included angle, as
the perpendicular from
In the right triangle and w may be found.
ACD,
b
B
Here, as in Case
B
acute and
A
check
and
I,
A
.
or C, say C.
and angle
CAD
are known, hence
Knowing m and c, n may be found. Having found p and n, we know two parts of the BCD, hence a and angle C8Z> may be found. = 180 - (A + angle ABC). Finally, angle ACB
A A A
b, c,
p
right triangle
three different figures are possible, according as
are both acute (left-hand figure),
B obtuse (middle figure), obtuse (right-hand figure), but the above analysis applies to each figure
alike.
obtained by repeating the solution with the perpendicular drawn from B. is
RIGHT TRIANGLES AND APPLICATIONS
52]
Case IV. In the
Given the three
first
113
c.
sides, a, b,
we were
able so to choose the vertex, from drawn, that one of the right triangles
three cases
which the perpendicular was contained two of the given parts.
In the present case this
The apparent
overcome as
possible.
difficulty is
easily
is
not
follows,
Fig. 58.
In either of the figures 58 we have
from the triangle CAD, from the triangle CDB, hence
b2
from which
b
m = a2
p a2
n2
= m2
nz
2
2
p
2 2
= =
-w a2 - n2
2
b2
,
,
,
=
(m
+ n) (m
n).
(i)
We now have, in the right figure,
in the left figure,
m+n and from
m
c,
n=
c,
(i),
m+n m from which,
n
=
m
m may be
We
n
c
since a, 6, c are given,
n
m
or
found.
then have in either case,
(m
We now know CDB, and from
+ n) + (m
two parts these the
n)
_
in each of the right triangles
angles A
and
B may be
CAD
computed.
and
PLANE TRIGONOMETRY
114
[CHAP, v
As a check the solution may be repeated with the perpendicular drawn from one of the other vertices.
EXAMPLE.
Given a
=
=
45.652, b
62.735, c
~
S1
to find the
-^;
angles.
Solution. C
+ = b - a = c = a
b
62.735
+ 45-652
62.735
45-652
-
= =
108.387,
log (b
= = colog c log(w- n) = m n= log (&-<*)
i7-83>
51.238.
m=
(m + n)
+ (m-n) =
51.238
(m
2
+ n)-(m-n) ^
51.238
the right triangle
log
w= = ^4 = ^ =
colog b log cos
From
CAD,
1.64036
c
+a=
c
a
= 6 =
10
9.84285
45 Si' 46",
51.338
45-652
(^
= =
log (c
96.890, 5-5 86
+ n)+(m-n) =
62.735
(m
(c
1.98628
a)
+ 8.6272 ^ 3S-68lj
+ ri)-(m-n) ^ 62.735 ~
8.6272
=
2
* If this result were greater than
and we should have taken
-
find
2
2
Fig. 58,
+ a) =
we
= 0.74710 = 8.20249 b colog (m n) = 0.93587 m w = 8.6272.
log
>
2
=
CDB,
= 0.87795 = 8.34054 = 10 log cos 9.21849 B = 80 28' 49", = 5339'24". )
62.735.
w=
'
colog a
8.20249
+ 45.652
*
''^
Interchanging b and c in the above formula, 51.238
1.55796
36.138
n
log
n
=
1.23257
8.29041
the right triangle
log
C=i8o-U + Check.
36.138
2
2
From
-
2.03498
^
+ 36.138 =
2
=
+ a) =
m
c,
-f-
--
we would have n
=
-
-
the right-hand figure in .
RIGHT TRIANGLES AND APPLICATIONS
52]
log
m=
= colog c = log cos A
8.34055
C=
9.77278
C=
53 39' 24",
log
8.29041
colog a
,4=45
B=
= =
n
1.55243
log cos
9.84284
180
5i' Si",
-
(A
+ C) =
115
i.43 22 3
80 28' 45" *
EXERCISE 30 Only a few triangles are given here for solution by the method of right triangles, for soon we shall study a better method, by means of which the computation can in most cases be shortened. 1.
Given a
=
342.56, b
Am. A = Given
2.
b
=
=
C=
125.72,
37 42' 24".
B= A = 127
124 44' 28",
134.5, c
=
235.2,
Ans. a 3.
c
33' 08",
17
=254.97.
36.3'.
=
334.7,
B=
18
33.9'.
A =
= 50 50' 50", c = 278.98. 25 25' 25", B Ans. C = 103 43' 45", a = 123.29, b = 222.70. C = 127 36.5', A = 283i. 3 b = 312.9. Ans. c = 612.55, a = 369.22, 5 = 23 52.2'. a = 630.50, b = 527.39, A = 6537' 12". 4^. B = 49 3/ 38", C = 64 45' 10", = 626.13. /
4.
5.
6.
,
b
=
1825, c
=
1563,
Ans. 7.
a
8.
a
*
5=
C=
14
22
13.7'.
54.8',
4 =
Find the remaining parts. 142
51.5',
a
=
2913.
= 3.537, & = 6.667, c S-ooi. Find the remaining parts. = 4,6=5,c = 6. Find the remaining parts. Ans. A = 41 24.6', B = 55 46.3', C = 82 49.1'.
When
checking five-place distances and angles expressed to seconds obtained tables, the results will generally be found to agree only to four places for distances and to the nearest tenth of a minute (6") for angles. This is because, as we have already observed, the fifth place of a number and the sec-
from five-place
onds of an angle obtained from a five-place table are not necessarily accurate. When cut down to the nearest tenth of a minute, the results of the two computations in the above example agree, each giving
A = The
solution
may
45 Si.8',
therefore be
B =
80
28.8',
assumed to be
C -
correct.
53 39-4'.
PLANE TRIGONOMETRY
Il6 9.
A =
Find the
50,
B=
ratio
60,
between the
C=
10. In a quadrilateral following parts are known
AB =
673,
= angle B
BC =
sides of a triangle
70. Ans.
589,
[CHAP,
a
ABCD,
:
b
:
c
=
0.7660
:
whose angles 0.8660
:
:
CD - 223, C = 127 38'.
105 06', angle
It is required to find the length of
nearest unit.
AD to
Ans.
the
1017.
are,
0.9397.
Fig. 59, the
Fig. 59
.
v
CHAPTER
VI
FUNCTIONS OF AN OBTUSE ANGLE Reason for a New Definition. In a right triangle no angle can exceed 90, but when the triangle is oblique one of its angles may be obtuse, that is, one of its angles may have any value between 90 and 1 80. In order to solve oblique triangles in the simplest way possible, we must define the trigonometric functions for obtuse This is best done by means of the conception of rectangular angles. coordinates.
63.
Let
Rectangular Coordinates.
X'X and Y'Y
be two
lines,
The two lines indefinite in length, intersecting at right angles at O. divide their plane into four parts, known as the II \y
*
m
IV
r
,
p.lg * 6o
*
first,
second, third
and fourth
quadrants respectively, as indicated by the numerals I, II, III, IV, in Fig. 60. Let P be any point in the plane of the lines X'X and Y'Y, and let PF be the perpendicular drawn from P to X'X.
O and
Join
from O, is always positive and OF, generally represented by
#,
P.
P
It is considered positive left of Y'Y.
if
is
P.
OP, the distance
of
P
designated by r. is called the abscissa of the point is to the right, negative if to the
FP, generally represented by y, is called the ordinate of the It is considered positive if P is above, negative if below point P. X'X. Considered together,
OF
or x
and
FP
or y are
known
as the rec-
tangular coordinates of the point P. X'X and Y'Y are called the coordinate axes, or axes of reference; X'X is the tf-axis, or axis of abscissas Y'Y is the y-axis, or axis of ;
The second.
O
called the origin. abscissa of a point is always written first
ordinates]
is
Thus, by the point
(a,
6),
117
and the ordinate
we understand
the point for
n8
PLANE TRIGONOMETRY
which x
=
ordinate
is b.
=
y
a,
that
6,
From what has been In the
[CHAP, vi
the point whose abscissa
is,
said
a and whose
is
plain that:
it is
quadrant, x is positive, y is positive, r is positive; in the second quadrant, x is negative, y is positive, r is positive; first
in the third quadrant,
x
is
in the fourth quadrant,
negative, r
is
is
negative, y
x
is
positive,
negative, r
y
is
positive;
is
positive.
For every point on the jc-axis, y = o; for every point on the ;y-axis, x = o; for the origin, x and y are each o. Thus, in Fig. 61, if
for
=
JPi,*
then for for
P2 P
= 4, = 3; = - 4, y = * = 4, :V=-3; # = 4, = o;
,
x
3,
x
forP 4
,
for ^4,
=
,
;y
then for 5,
#
for A',
x * x
3;
for 5', for
3;
but each of the distances OPi, (XP2 25 54.
=
,
O,
OP8 OP*
= o, y = 3; = - 4, y = = o, y = = o, = o;
3;
3;
equals
,
o;
V^ + 2
2 4 =
5.
Definition of the Trigonometric Functions of
Any Angle
Less than 180. Let angle XOB = 6 represent any angle less than 180. Take O an origin, OX for the positive #-axis, and draw OY perpendicular
for
to
OX.
Then
OB
will
be in the
quadrant according as 6
Now x
take any point
P
on
first
or second
acute or obtuse.
is
OB
r the distance of this point
and denote by from the origin,
and by x and y the rectangular coordinates of this point with reference to OX and OY Fig. 62.
The
as axes.
trigonometric
functions of 6
are then defined as follows,
sine cos 8
= r
csc e
=-= r
'
,
abscissa
-
sec 6
cosB'
distance
= x
=
sin 6
distance
cot 6
i
tanO'
FUNCTIONS OF AN OBTUSE ANGLE
56]
be observed that when the angle
It will
agree with those given in Art.
7
;
119
6 is acute, these definitions
for x, y
and
r are the base, altitude
and hypotenuse respectively of the right triangle which we there used in defining the trigonometric functions of an acute angle.
The Signs of
the Functions of an Obtuse Angle. In the first as well as r, are positive. x and Hence all the functions y quadrant, are positive for an angle in the first quadrant. In the second quad55.
y
rant,
x
is
negative, y
negative, while
and
r are positive,
^ remains
hence the ratios - and * are T X
positive, that
the cosine and tangent
is,
their reciprocals are negative, while the sine and its reciprocal are positive. The functions of an obtuse angle are all negative, except the sine and
and
which are
its reciprocal,
56.
positive.
Fundamental Relations. Of sin 6
esc 6
=
i,
cos8.sec8= tan8 cot 8 =
the six fundamental relations,
+ cos 3 8 =
58 sin 8
i,
i,
tan*8+i =sec 3 8,
i,
cot*6
+i=
csc*8,
first three rest upon the definitions of the cosecant, secant and cotangent, and hold, therefore, whether 6 is acute or obtuse, and the 2 = 2 r2 which is true last three depend upon the relation x y
the
+
,
These six fundamental relations positive or negative. as well as for the hold, therefore, for the functions of obtuse angles functions of acute angles. whether x
is
Also, y.
tan 6
= % =
-
x
x
=
^-^, whether x
is
cos0
positive or negative, hence
the two relations,
cos 8 also hold true
when
6 is
,
an obtuse angle.
sin
PLANE TRIGONOMETRY
I2O
[CHAP, vi
Functions of Supplementary Angles. be any angle less than 180, and draw OP' so as to make angle P'OX' 57.
Let
XOP = 6
XOP.
equal to angle = 180 - e. If
Then
angle
XOP'
OP
and OP' are taken equal, the two and OA'P' will be geometrically and we have
triangles
equal,
OAP
sin(i8o-e) =
AP ^
A'P'
OP
OP'
OA -^= OP
csc(i8o
-cose,
-6)=sin(i8o
sec(i8o-6) = cot(i8o-
6)
csc
6,
sec
8,
6)
= cos(i8o-0)
-cos0
tan(i8o-0)
-tan0
=
= -cote.
comparing our results we observe that the signs on the right are those of the functions in the second quadrant, hence it appears
By
that, Q
Any function
of (i8o
ff)
equals plus or
minus
the
same function
of 6, the sign being that of the function in the second quadrant.
The rule just given enables us to express the functions of an obtuse angle in terms of the functions of an acute angle, thus: sin
11
6=
sin (180
cos
116=
cos (180
tan
116=
tan (180
csc
116=
csc (180
sec
116=
sec (180
cot
116=
cot (180
= - 64) = - 64) = - 64) = - 64) = - 64) = 64)
sin
64,
- cos 64, - tan 64, csc
64,
- sec 64, - cot 64.
FUNCTIONS OF AN OBTUSE ANGLE
S9l
68.
121
Functions of (90 + 9) Let angle be any acute angle and draw OP' .
XOP =
Then angle XOP' = If OP' is taken equal to OP, the 90 +0. triangles A OP and A' OP' are geometrically equal, and we have perpendicular to OP.
Fig. 64.
cos
OP
OP'
-?=
=
+ 6) =
csc (90
+ 9) = sin
(90
+ 0)
+ 9) = cos
(90
+ 0)
sin
+ 0)
-cot0
sec (90 cot (90
-
OA>
\
= -1= -
tan (90
sec
6,
csc
6,
cos
i
+ 9)
-sin 9,
OA
A'P'
tan (90
.
9,
-
i
tan6.
Again the signs on the right are the signs of the functions in the second quadrant, hence,
+
0) is equal to plus or minus the corresponding of (90 the unction sign being that of the function in the second quadrant. of 8, cof This rule, like that of the preceding article, enables us to express
Any function
the functions of an obtuse angle in terms of the functions of an angle less
than 90.
Thus, sin 116
cos 116
tan 116
= = =
sin (90 cos (90
tan (90
Functions of 180. If in must coincide with OX', the
59.
OP
+ 26) = cos 26, + 26) = - sin 26, + 26) = - cot 26, etc. is taken equal to 180, Fig. 62, abscissa of will be r and its
P
ordinate zero, hence sin 180
=
cos 180
=
tan 180
=
-
=
-
csc 180
o,
=
o
-
= = r
oo,
o,
coti8o
=
i,
oo.
o
PLANE TRIGONOMETRY
122
The
results in the last line
[CHAP, vi
need some explanation. The tangent and is negative, hence their linfiting
cotangent of every obtuse angle
180, is negative. Numerically these limiting values are o and oo, the minus signs merely indicate that these values have been approached through a succession of negative
values, as the angle approaches
magnitudes. 60. cosine, tive,
Since the Angles Corresponding to a Given Function. obtuse and of secant, tangent angle is negacotangent every
we can
tell whether the angle corresponding to one of these obtuse or acute by noting the algebraic sign of the But the sine and cosecant are positive whether the angle
functions
is
function.
In fact, since the sines and cosecants of suppleare equal in every respect, there will always be two mentary angles one and the other obtuse, which will correspond equally acute angles, This is expressed by saying that well to a given sine or cosecant.
is
acute or obtuse.
the angle corresponding to a given sine or cosecant
EXAMPLES.
If cos
and if cos
and
If
tan
if
tan
6
If sin 6
= = = = =
i,
ambiguous.
must equal 60, equals the supplement of 60, or 120.
J, i,
is
must equal 45, i,0 equals the supplement of
^,0may be
either
30 or
its
or 135.
45,
supplement, 150.
EXERCISE 31 1.
=
y
Locate the points whose coordinates are x = 8, y = 6; x = 8, = = 6; # = 8, y = 8, y 6; and in each case compute 6; #
the distance of the point from the origin. 2.
=
x
3. i
;
Locate the points # o,
y
= -
The
=
5,
y
=
o;
#
o,
y
=
5;
#
==
i,
y
==
i;
i.
distance of a point from the origin and locate the point.
is 2,
and
its
abscissa
is
find its ordinate
Ans.
is
=
Two
solutions,
4. JThe distance of a point from the origin ^50; find the abscissa and locate the point. 5.
Make
out a table containing the
each of the angles 120, 135, 150.
is
y
10 and
sine, cosine
= its
V^. ordinate
and tangent
of
FUNCTIONS OF AN OBTUSE ANGLE
6i]
6.
Construct the angles, having given the following functions: = = $, tan B 3, sin 6
A =
cos
123
.
7. Express the following functions in terms of functions of the supplementary angles: sin 115, tan 165, cos 125, cot 100, sec 170, sin 145, cos 136, tan 95.
8.
Express in terms of an angle sin
95,
less
than 45 the following:
cos 120, tan 100, sec 114, sin 125.
Express in two ways in terms of an acute angle, first by means of the rule of Art. 57, second by means of the rule of Art. 58, each 9.
of the following:
Which 10.
two methods
of the
Use the
tions: sin 112
tan 126
being any obtuse angle, prove the following relations: cos
(6
tan
(0
-
90) 90) 90)
= = = -
cos0,
esc (6
sin 0,
sec (6
cot0,
cot (0
90) 90) 90)
= = = -
sec0, csc0,
tan0.
Review.
What
(a)
base a ?
(b)
What
(e)
Given log 4
is
is
meant by the logarithm
Show
(c)
that
the logarithm of
=
=
log ab i
?
(d)
log a
of a
number
+ log b
y
log a
0.60206; find log 16, log
Vio, Vio, Vioo.
,
,
(c)
to a given b
=
b log a.
What is meant by a cologarithm ? 2,
log \.
2. (a) What is meant by the common logarithm Give from memory the common logarithms of 10, (6)
io r
25.3".
?
table of natural functions to find the following func30', cos 156 25', tan 162 50', sin 105, cos 175 10',
sin (6
1.
the easier
14'.
11.
61.
is
n'
54' 36", tan 140
35' 13", cos 157
sin 127
What is meant by
of a
number?
100, o.i, o.oi,
the characteristic
logarithm? What by the mantissa? (d) Give the of the logarithms of the following numbers: 15, (e) Give the rule for the characteristic of a 153, 6.23, 0.05, 0.0105. number greater than i; of a decimal fraction less than i. of a
common
characteristics
3.
how a table of common logarithms might be conextracting square roots only. Find the number whose f , without consulting the table.
Explain
structed
by
logarithm
is
PLANE TRIGONOMETRY
124
4.
Prove that logJV
5.
What
is
=
^6 ;
meant by the
also
[CHAP.VI
show that
log&0
loga&
=
principle of proportional parts?
i.
Read
again Art. 32. 6. 7.
Work Problems 13 and 14, Exercise 18. What is meant by -an exponential equation ?
Solve Problem 18,
Exercise 18. 8. (a) Read again Art. 44. (b) What accuracy is called for in the angle of a triangle, when the sides are given correct to three places? To four places? (c) How accurately can a number be determined with the aid of a five-place table of logarithms ? An angle ? (d ) When
When
should a six-place table be used? 9.
pass?
by N.?
Between N.E. by E. and E.N.E.?
opposite to the direction S.E. 10.
Explain
by means 11. (b)
a seven-place table?
How many degrees constitute a point of a mariner's (b) How many degrees in the angle between N. and
(a)
(a)
The
how
(c)
What
comN.E.
direction
is
S. ?
by
to solve each of the four cases of oblique triangles
of right triangles.
What
is
meant by the rectangular coordinates
of a point? a point P is 5 units in length and with the positive direction of the #-axis.
line joining the origin to
makes an angle
of
30
P
What
are the rectangular coordinates of the point ? (c) What are the coordinates of the point P, if OP makes an angle of 150 with
OX? 12. (b)
and tangent of an obtuse angle. cos A, A, cos (180 - A) Complete the equations
(a) Define the sine, cosine
Prove that
tan (90
sin (180
+ -4) = ~
tan (180
-
Find
A)
-
cot A.
=
,
A) (c)
sin (90
=
sin
+ A) =
,
cos (90
+ A)
.
123, cos 136, tan 105 30'. (b) Find x in 13. (a) each of the equations: sin x = 0.3423, cos x = 0.9061, tan# 0.0913, x being in each case the angle of a plane triangle. sin
CHAPTER
VII
PROPERTIES OF TRIANGLES IN this chapter we shall develop certain properties of triangles which will enable us to compute, from a sufficient number of given parts, the remaining sides related to the triangle.
and
The
developed in this chapter will 62.
The Law
angles, the area
and other magnitudes
principal applications of the results
be treated
in a separate chapter.
of Sines.
Let ABC be any plane triangle. Draw the perh of the vertices C of the triangle to the oppofrom one pendicular site side AB (Fig. 65), or AB produced (Fig. 66). (a) First proof.
Fig. 66.
Fig. 65.
In Fig. 66,
In Fig. 65, in the right triangle
h
and
=
b sin
ACD
A
h
h
t
in the right triangle
=
in the right triangle
BCD
and h
a sin B.
=
=
b sin
A
ACD
,
in the right triangle
a sin (180
- B) =
BCD
a sin B.
Hence, whether the triangle is acute or obtuse, we have h = b sin A = a sin B, or
_
a sin
A
b sin
B
from Similarly, by drawing a perpendicular or the opposite side produced, we obtain
_
a sin
A
c
sin 125
C
'
B
to the opposite side,
PLANE TRIGONOMETRY
126 so that
we may
write
a sin
Equation
(i)
[CHAP.VH
= A
ft
sin
may be otherwise a b c = sin A :
:
B
c
=
(x)
.
sin
C
written thus, :
sin
B
:
sin C.
(2)
Equation (i) or (2) embodies what is known as the Law of Sines, which states that, In any triangle the sides are proportional to the sines of the opposite angles.
The Law of Sines may be proven in another (b) Second proof. way, which at the same time brings out the meaning of the ratios in equation (i).
Circumscribe a
about the triangle
circle
ABC
D
the diameter BA', drawn through one of the vertices, as B. Join A' and C. A'BC is
and denote by
a right triangle (Why?), and therefore Fig. 07.
D But angle A'
=
angle
'
sin
A
A'
(angles inscribed in the
same arc are
equal),
hence
_
a
D
sin .4'
and
similarly
a sin
A
9
~'
~'
sinC
from which
sine
sinJB that
is,
77^
ra&*0 0/
awy
numerically equal
side of a triangle to the sine of the angle opposite is diameter of the circumscribed circle.
to the
63. Projection Theorem.
In Fig. 66,
In Fig. 65,
AD =
A DB = acos. b cos
AD= b cos A,
y
BD =
acos (180
- B)= -
acosS.
Moreover, c
= AB = AD + DB.
c
= AB =
AD- BD.
PROPERTIES OF TRIANGLES
64]
Substituting for
=
c
AD, DB, and
A
b cos
BD
their values,
+ a cos B.
Hence, whether the triangle
c
is
= =
b cos b cos
acute or obtuse,
127
we have
A 4
a cos 5)
(
+
a cos B.
we have
= a COS B + bcosA.} a = 6 cos C + c cos .B, | 6 = c cos A + a cos C. J c
Similarly,*
(4)
We may consider the line DB (Fig. 66) the negative of BD, = - BP. In that case is, DB = AD + DB, just as in Fig. c = AB = AD-BD=AD-(- DB) A D = b cos A is called the projection oi AC on AB, DB = a cos J5 is called the projection of CB on 4, so that the relations (4)
may be
The Law
65.
stated thus,
In any triangle, each side is equal to the algebraic two sides upon it. jections of the other 64.
that
sum
of the pro-
of Cosines.
(a) First proof.
In Fig. 66,
In Fig. 65,
c
Substituting in the
second and third, &2
first
equation the values of h and
.
from the
DB =
BD +
_ Elf.
AD from
the
we have
=a2 -]^2 +c2 -- 2c DB+DB* = a2 + ^2 _ 2C J)^.
Now
-
i2
=a2 ~^D2 +c2 + 2c BD+1H? = a2 + C + 2C 5ZX -
2
figure
a cos 5.
BD =
a cos (180
-
B)
= -
a cos J5.
*
c
The second formula may be obtained from the first by replacing a by 6, b by c, by a, and 4 by B, 5 by C, and C (should it occur) by A. In the same manner
may be obtained from the second, and the first from the third. of these formulas is given, the other two can be supplied by one any
the third formula
That
is, if
cyclic substitution.
PLANE TRIGONOMETRY
128
[CHAP,
vn
we obtain
Substituting these values in the equation just preceding, in either case
b2 c
Similarly,
2
= c 2 + a2 = a2 + 62 *
2 ca cos 2
JB]
ab cos C, [
(5)
2bc cos -4 J
These formulas embody the so-called Law of Cosines: In any triangle, the square on any side is equal to the sum of the squares on the other two sides diminished by twice the product of those two sides times the cosine of the included angle. (b)
easily
Second proof. The law of cosines may be proved even more than above by the aid of the projection formulas. We need
only to multiply the first of the formulas second and to subtract b times the third.
=
b (c cos
A + a cos C)
2 ac cos
By
by c, to add a The result is
(4)
times the
from which
65. Arithmetic Solution of Triangles. The law of sines and the law of cosines are sufficient to solve each of the four cases of
oblique triangles.
one side a and two angles are given, the third angle is found B C = 180. The remaining immediately from the relation A I.
If
+
sides
may
+
then be found from the law of ^
_
a
B AT
__
sin :
sin
sines, thus,
c
9
a sin C T :
sin
A
sides b and c and the angle opposite one of them, say B, third side a may be found by the law of cosines; for, the given, of equations (5), considering a as the unknown first the solving II.
If
two
is
quantity,
we
find
This gives two values for a, as has in general two solutions.
it
should, for
we know
that this case
PROPERTIES OF TRIANGLES
6s
Having found the third found from the law
7 b
III.
two
If
third side a
equation
sides b
may
sin B,
C may now
be
C=
sin
7 sin B. b
and
and the included angle
c
A
are given, the
be found from the law of cosines, for the third
(5) gives
a
The
and
of sines, thus:
A =
sin
A
the angles
side,
129
sides
= V&
2
+c
2
2 be cos
A
and one angle being known, the law
.
of sines will give the
other angles.
IV. If the three sides are given, the three angles
from the law
may be
we have from
Thus, to find A,
of cosines.
found
the third
of the equations (5),
cos
"r c
A =
ft .
.
2 be
While the law of
sines
and the law
of cosines are theoretically all
necessary to solve triangles, the law of cosines, which would have to be used in two out of the four cases, is not adapted to
that
is
The numerical work
logarithmetic computation.
necessary to solve
triangles will
be greatly shortened by the use of other formulas
which we
develop in the following
will
articles.
EXERCISE 32
and reduce the
of sines to a right triangle their to simplest form. sulting equations 1.
Apply the law
2.
Apply the law
when the
angle
of cosines,
A = o, 90, Ans.
a2
= W+
c
2
2 be cos
re-
A, to the cases
180.
A = o, a = b2 + c2 - 2 be = (b - c) 2 A = 90, a2 = 62 + c2 A = 180, a2 = 62 + c2 + 2 be = (4 + c)*. theorem to the case A = 90; to the case 2
.
.
3.
Apply the projection
A=B. Solve the following problems without the aid of logarithms, 4.
Given
A =
35,
B=
75, a
=
7; find b
and
c.
Ans. b
=
n.8, c
=
11.5.
PLANE TRIGONOMETRY
130 Given
5.
A =
b
65,
=
10,
a
=
B
15; find
and
B=
Ans.
Given
6.
A =
b
16,
=
15,0
=
6; find the
B = 43 B' = 136
Ans. or 7.
Given a
8.
Given
=
=
150, b
200,
33.5',
27
c.
By means
9.
=
10', c
37
C=
120 27
^o'; find
= c' =
26.5', c
16.2.
33.5',
18.76,
10.07.
c.
Ans.
a=2,6 = 3,c = Ans. A = 28
vn
remaining parts.
= 26.5', C
C=
[CHAP,
=
c
96.3.
4; find the angles.
B=
57.3',
=
46 34.1', c
104
28.6'.
of the law of sines prove that the bisector of an
angle of any triangle divides the opposite side into segments proportional to the adjacent sides.
Derive the law of sines from the law of cosines.
10.
""
/Suggestion.
Form
the ratio
\ it is
equal to
/ Vi -
CO ** A ;
=
^4B and show
that
C
as a
sin
cos2 J5
)
/
The Law
66.
of Tangents.
First proof.
(a)
and
center
Z>,
Let
ABC
be any
With a vertex
triangle.
the shorter of the sides
adjacent to C, as a radius, draw a circle in P and BC produced in Q. and AQ. Triangles ACP and ACQ are isosceles and QAP is a right angle (Why?). Denote the whole angle at A by w, and the three parts by x, y, z,
cutting
Draw
BC
AP
as indicated in the figure, then angle
(Why?). x
thus,
x and angle
AQC =
z
Also
+ y = A,
xy =
Solving these equations for
Now
APC =
#
B,
x, y, z
+2=
90,
x
and w, we obtain
apply the law of sines to each of the triangles
AB = BP
x) __ sin x
sin (180 sin
y
sin y
,
APB
AB =
sin z
BQ
sin
w
and AQB,
PROPERTIES OF TRIANGLES
66]
131
or
(6)
Dividing the
first
of each pair of equations
a+b ~ tangU + JB) a- b
-
tan | (A
B)'
e+a c- a" Formulas
(7)
embody
b
by the second gives
+ c ~~ tanjCB+CQ -
b
tan%(C
tan 5 (B
c
-
C)' (7)
+ A)
tang(C- A)
the
triangle, the sum of two sides is to their of tangents: as the the sum of the angles opposite is to the tangent of half difference
Law
In any
tangent of half their difference.
The formulas (6), which we shall have occasion we shall refer to by the name of Double Formulas.*
to use hereafter,
Second Proof. The law of tangents can be proven more double formulas. In Fig. 68 easily without the intervention of the draw PR parallel to QA, then angle APR = angle QAP = 90. (b}
From
the similar triangles
BQA
and BPR, we have AQ_
A == AQ = RP RP' AP = = BP a + b, BQ
BQ BP
but an(l
42 =
tan
A PQ =
AP
tan x
=
tan?
=
a
-
b,
tan } (A
+ J5),
* Also called Mollweide's formulas, after the German astronomer (1774-1825) who introduced their use. The cosine form of these formulas appears in Newton's Arithmetica universails (1707)-
PLANE TRIGONOMETRY
132
[CHAP.VH
hence
a+b a- b The law of tangents and double formulas are adapted to the logarithmic solution of the third case of oblique triangles, that is, when two sides and the included angle are given. Suppose the given parts are a, b lows: 1.
2.
ist. 3.
and C.
\ (A
+ B)
| (A
E)
is is
The
different steps in the solution are as fol-
-
180
C.
found from the law of tangents, formula
(7),
equation.
Adding and subtracting the - B) = A B) + | (A \(A
+
c is
4.
c
c
J5,
\(A
and
i
we have
2
+ B) - } (A - B) = B.
(6).
could have been determined from the
_
__ = asinC sin
this
results of ,
found from the double formula
Having found A and law of sines, thus,
but
+B=
A
found from the relation
.
or
A
c
= 6sinC ~-> ;
sin />
would require us to look up three new logarithms, namely,
those of a, sin
C, sin
A
or
,
6,
sin C, sin B.
new
while the double formula requires but two
+ B), sin| (A - 5), and
these
may be
or
cos
(4
(a)
+ 5), cos
(A
-
5),
taken out at the same time and the same opening B) and tan \ (A B).
of the table with the logarithms of tan | (A
67.
logarithms, those of
+
Formulas for the Area of a Triangle. In terms
of the base c
and
the altitude h.
base of the triangle, Figs. 65, 66, h
its altitude
If c represents the
and
T
its area,
then
by elementary geometry
T=\ch. (6)
In terms of two
sides b
and
c
and
(i)
the included angle
the right triangle, Fig. 65 or 66, h substituted in (i) gives
we have h
T=
\1>C8in A.
=
b sin
A.
A.
From
This value of
(2)
PROPERTIES OF TRIANGLES
6;]
In terms
(c)
and
of one side c
the angles
133
A, B, C.
By
the law of
sines,
sinC Substituting this value of b in (2)
T=
we
In terms
w
-
of the three sides a,
b, c.
(*)
C
sin
2 (d)
obtain
It is
shown
in plane
geometry
that
T = Vs(s-a)(s-b)(*-e)* where
s
=
(a
+ b + c),
that
is,
s equals half the
(4)
sum
of the three
sides.
(e)
In terms of
s
and
the radius k of the inscribed circle.
be the center of the inscribed
Let
c
to the vertices of Join the triangle, and draw the radii of the inscribed circle to the points of tangency. circle, Fig. 69.
These
k
radii will
be perpendicular to the
respective sides.
Area
BOC = J ka, \ kb, triangle COA = \ kc. triangle AOB
of triangle
Area of
Area
of
Adding,
T=
or
where
ks,
s
=
\(a
+ b + c).
(5)
If the three sides are known separately, (5) enables us to find the radius of the inscribed circle of a triangle, for we have I".
'-?-V on substituting the value *
Hero
This formula
is
known
of
l
T
from
(4).
as Hero's formula for the area of a triangle, after who, so far as we know, was the first to
of Alexandria (ist century B.C.),
prove and apply this remarkable formula.
PLANE TRIGONOMETRY
134
Functions of Half the Angles in
68.
[CHAP,
Terms
vn
of the Sides.
In Fig. 69,
BD = BF DC = EC AF + (BD + DC) = BF + (AE + EC),
Adding,
and
sum
since the
of the six
+b+c=
segments equals a
2 s,
we have
BF +
or
from which
=
b
s,
BF = s-b.
Also, since the lines.4O
and
BO bisect the angles;!
we have .
tan
A =
Ar
2
Substituting for
k
tan
,,
t4--*-
and
B -=
their values
from
B respectively,
k ;.
BF
2
AF and BF
(i)
(i),
we have
f
tan*= =
tan
similarly J
2
Again
,
C
ft
+ AF = k*+(s- a) - a)(s - b)(s - c) + s (s - a) _ ^^j 2
2 AO* = OF
2
2
(j
.
sin
A = OF _ 7: ^4O
2
Substituting for ylO
its
ft
AO -77:
cos
>
A
(3)
AF = 5-
value from (3) and reducing the result gives,
sin 2 T>
Similarly
(4)
sin2 .
sm
C
J(.-q)(5-6)
2
y
/jo
)
PROPERTIES OF TRIANGLES
68]
In each of the radicals the positive root of the angles
ARC 222 ,
,
135
to be taken, since each
is
1
,
necessarily less than
is
90.
The formulas (2) and (4) are adapted for the logarithmetic computation of the angles of a triangle when the sides are given. For the quantities s, s a, s 6, s c, can easily be found, and with these known, the right member of each formula involves only multiplications, divisions
and the extraction
In general, the angles
of square roots.
may be found from
either the sine, cosine or
tangent formulas, but since an angle near 90 cannot be accurately found from its sine, nor a very small angle from its cosine, the sine formulas are to be avoided when the angle is greater than 45 and the cosine formulas when the angle is less than 45. When all the angles are required, the tangent formulas (2) should be used, since they require but four logarithms to be taken from the tables, that is, the s b and s logarithms of s, s
EXERCISE 33 1.
Express in words each of the rules for the area of a triangle in
Art. 67. 2.
From each
two others by a 3.
of the formulas (2)
cyclic
By comparing sin
advance of
and
(3) in Art. 67, write
down
letters.
the formulas (2) and (4) in Art. 67 show that
~ \/ s (s
A
a)(s
b)(s
c).
be
Why
could not the angles be found by this formula as well as
or (4), Art. 4.
By comparing the A
expressions for sin
expressions for sin
A
22 ,
cos
sin
By
Problem
A =
2 sin
forming the ratio of sin 3,
A
show that sin
A
A
in (4), Art. 68,
A -
cos
(2)
:
sin
:
in
Problem 3 with the
show that
A .
2
2
5.
by
68?
sin B, using the values given in
B=
a
:
b.
This constitutes another proof of the law of
sines.
PLANE TRIGONOMETRY
136
From
6.
we have
(3), Art. 62,
of the circumscribed circle. (2), Art. 67
sin
A =
[CHAP.VII
~> where
D is the
diameter
substituting this value for sin
By
A
in
show that
T = ^c
2D
7.
By
From
we have
(5), Art. 67,
for the radius of the inscribed circle
using the relation (Fig. 70)
- SCO show that
Fig. 70.
*= 8-V a
*6 =
;i
where
fc
a , ^6,
,
-^T
are the radii of
c
b
9
fc
,
T = -i-, $
the escribed
c
circles,
touching the
sides 0, by c respectively, externally. 8.
Prove that
L=JL + i+JL kn
A;
Prove that
9.
Let
10.
sides
lcn
ABCD
and Q
+
Jcb
kc
k=
+&
2 J>.
be an inscribed quadrilateral,
(Fig. 71)
Show
its area.
kb
0,
ft,
c,
d
its
that (a)
Also
by comparing
two expressions
the
for
the
diagonal
show that Fig. 71.
2 (ad
and by substituting
this value in sin 2
-f-
be)
A =
i
cos 2 A,
be
where
$
By substituting (c) in
(a),
Q = V($
=
| (0
+ 6 + c + d).
show that a)(s
i)(^
-
show that
PROPERTIES OF TRIANGLES
68]
ii.
ABC
9
137
It was shown in Art. 65, II, that if two sides 6, c, of a triangle and the angle B opposite one of these sides, are given, the third side a may be found from the A relation
a
=
c cos
B
Interpret the result geometrically. Ans. The two terms on the right 72-
and
C
are respectively the distances from B drawn from A to EC or EC
to the foot of the perpendiculars
produced.
CHAPTER VIH SOLUTION OF OBLIQUE TRIANGLES Solution of Oblique Triangles. In the present chapter we computation of each case of oblique triangles by a numerical example. Since five-place tables are used in the compu69.
shall illustrate the
tation, the results are given to only five significant figures
and the
angles to the nearest second. In every case a check has been applied Such a check is to be looked upon as an to the results obtained. essential step in the solution, since no computation, no matter by whom, can be relied upon if it has not been checked. Instead of
the analytic checks here given, graphic checks are often resorted to in practice, but such checks, while more easily applied, are of course errors of construction and are therefore unsatisfactory, except as checks against gross errors. The solutions given below are arranged in a form which may serve as a model to beginners. It is customary for computers to make out
open to
a complete schedule of work (as
is
illustrated in the first case below)
before referring to the tables, so that
when the
tables are once
opened no writing remains to be done except that of numbers taken from the tables. 70. (1)
(2) (3)
Case
To To To
I.
Given two angles and one
+
+
B find C, apply the relation A find a and 6, apply the law of sines.
side, as
C =
filling in
A & y
the
and
c.
180.
check, use the double formula.
EXAMPLE
i.
Given
a
=
24I l8 > '
6=165.68,
.
Flg 73 '
C=
'
68 12' 15"'
Schedule of Work. (1) (2)
To To
C=
find C.
find a
and
b.
By
180 - (A B) the law of sines,
+
138
=
68
12' 15".
SOLUTION OF OBLIQUE TRIANGLES
]
a -
=
A or C = c log sin yl = sin C = = log a sin --
--
a
=
-
csinyl --
sin
c
log
colog
sin
b -
.
C
sinE
=
__
colog
Check.*
By
the double formula
-
sn
_
- 5) = log c
--B) colog sin |(^l
+ J5)
log (a
b)
(6), left,
*
19' 37",
15
= = = =
_ ;
a _
sin
c
~= csmB sin C
= B= sin C = = log & b = log c
log sin
a= (3)
,
or b
sm C
c
139
W + B) = 55
a
b
S3' 5*"-
=
b must agree with above. (2) Having completed a schedule as above, we now turn to the tables and complete the solution by filling in the missing numbers as If the
computation
the value of a
is
correct this value of a
b obtained
from
follows:
Solution.
= 2.37388 = 9.97626 log sin A = 0.03221 colog sin C = 2.38235 log a a = 241.18. logc
_ 10
log
colog
= = sin C = = log 6 b = logc
2.37388
sin J3
9.81318
_ 10
0.03221
2.21927 165.68.
Check.
logc= log sin I (A
colog sin J
(A
log (a
B)
+ B) -
6)
= = =
2.37388
9.42214
10
0.08195
a
1.87797,
-6=
75-5-
Beginners will do well to follow the above form. Expert comc in the puters save the repetition of recurring numbers such as log above example by employing a more compact arrangement. For instance, the above solution and check can be put in the following *
Some authors check by
check
is
unreliable, for
it fails
the law of sines, a to detect
an error
:
b
=
sin
A
:
sin B,
in either c or log sin C.
but this
PLANE TRIGONOMETRY
140
EXAMPLE
tCHAP. VIII
Find the area of the triangle given in Example i. we use formula
2.
Since one side and two angles are given,
Solution. (3), Art. 67,
logc log c
A
c2 sin
sin
2
= =
2.37388
2 log c log sin
4.74776
A = 9.97626 = 9.81318 log sin B = 0.03221 colog sinC = 9-69897^ colog 2 = 4^6838] log T
B
2sinC
10 10
ip
T=
18552.
EXERCISE 34
The student must check his results when no answer
= = A =
Given A Find a Given Find b Given
A
Given A Find a 5.
6. 7.
= = Given 5 = Given A = Find a = Given A Find b
46
=54i8',
36',
354-4,
79 59', 56.8,
54 34',
69 30.2', 592-7,
29 41.2 40.68,
78
64
,
given.
b = 396.1, a = 79.5. B = 44 41', c = 66.4, C=552o'. c = 67.9. Find a, b,C. B - 43 56', B = 66 39. 4', c = 438.36 = 581.0, C = 435o.4'. a = 32.84. 5=37 C
C= 45-6', = 56' 18", B
895.14,
is
c = 479C=79o6'.
b
61.27,
C=
= 63 32. 9', = c 47 29' n", C 728.40, a
112
28.4'.
8.875.
Find
913.45-
67
34' 3i".
b, c,
A.
SOLUTION OF OBLIQUE TRIANGLES
7i]
Given B =48 24' 15", Find b = 1337.2,
8.
Find the radius
9.
C= a =
31
= A =
13' oo", c
1758.9,
141
926.74.
100 22' 45".
of the circumscribed circle in 8.
Ans. Ans.
R= =
894.06.
Area
10.
Find the area
11.
Find the area and the radius of the circumscribed circle in Check by using the relation in Problem 6, Exercise 33.
Case
71.
To To To To
(1) (2) (3) (4)
&,
find
301,360. 2.
Given two sides and the angle opposite one
II.
them, as a,
of the triangle in 7.
of
A.
J5,
apply the law of
sines.
+B+C=
find C, apply the relation A find c, apply the law of sines.
180.
check, apply the double formula or the law of tangents.
determined from its sine, as the angle B above, it admits of two values which are supplements of each other. Whether one or the other or both of these values are to be used
When an
angle
is
depends on the conditions imposed by the problem.
construct-
By
may arise ing the triangle graphically, A b and the between given parts a, depending on the relations the to Construct angle BAG equal given angle A, making CA it is
seen that various cases
.
equal to
b,
and from
C
as a center, with a radius equal to
a,
draw a
circle.
than the perpendicular distance from C to AB, the the line AB. In this case it is impossible to concut not circle will the given data are a struct triangle having the given parts, that is, If
a
is less
inconsistent. If
a
is
equal to
distance from
the perpendicular
be tangent to the line AB at have a right angle at B. will triangle tance from C to A B is b sin A hence in
circle will
,
If less
a
is
than
C
b,
and there are two angle AB'C.
AB
in
AB,
the
.
from greater than the perpendicular distance the circle will cut
to
B (Fig. 74), and the resulting Now the perpendicular disthis case a = b sin A to
AB
but
B and B (Fig. 75), triangle ABC and the tri-
two points
solutions, namely, the
C
1
AB
in a cut the line equal or greater than b, the circle will of intersection falling on second the B point 76), (Fig. single point or to the left of A. In this case there will be but one solution. If
a
is
PLANE TRIGONOMETRY
142 So
far
obtuse, a
[CHAP,
we have assumed the given angle A to be acute. If A must be greater than b (since A is greater than B, and
=
6 sin
A.
b sin
A <
<
a
b.
Fig. 75-
Fig. 74-
is
in
In this
triangle the greater angle is opposite the greater side). case there can be but one solution. (Fig. 77.)
any
a
vm
Fig. 76.
we
disregard the case in which the triangle is right-angled (Fig. 74) as not properly constituting a case of oblique triangles, we have the following simple test for the number of possible solutions, If
EXAMPLE
a
~
a
<
b,
one solution,
6,
two
solutions.
i.
Given a
=
Required
=
34S-46,
43
'=136
&=
56' oo",
04' oo",
C=io9
i6'28", C' = 17 08' 28",
531-75, Fig. 78.
c
-
7 2 3-45-
c'= 225.88. Solution, (i)
To
<
a
find
B
ft,
hence there are two solutions.
and a
B
1
A
sin
the law of sines, .
or
log 6 log sin A
9.65394
colog a
7.46160
10
9.84125
10
log sin JB
2.72571
=
C and (7. C=i8o- (A+B) - 109 (2)
By
.
To
-
sin
6 sin
B
A
10
find
16' 28",
C'
*
180-
S'
28
SOLUTION OF OBLIQUE TRIANGLES
:]
(3) 07
To
and
find c
By J
c'.
the law of sines
,
sin
colog
By
__
or
,
a sin C' sin
= 2.53840 sin C = 9.97495 sin A = 0.34606 = 2.85941 logs c = 723.45-
Check.
(4)
^4 sin
A
A
= 2.53840 = sin C 9-46942 log = 0.34606 colog sin A = 2.35388 logc' c = 225.88. log a
log a
log
=
a
143
10
10
'
the double formula,
a
o
sin
or
c'sin J
(' - 4) =
a
b
= A) =
logc log sin
%(B
=
~ (ft
a) sin
186.29.
10
9.17327 2.03268
= (5'- ^4) = log c
log sin J
'
2.35388
9.91143
-
log sin i
-
a)
2.27019
9.76249
10
2.03268
= = (&' + ^)
log (b 1
-
= = log sin %(B + A) log (b
2.85941
a)
2.27019
9-995*3
IO
2.26532
2.26531
Compact Arrangement Check.
Solution. b
log 2.72571
=531.75
A = a = B= C= c =
26 47' 32"
345.46
43 56' oo" 109 16' 28"
-A)=
&-a=
=
colog 0.34606
colog 7.46160
log 2.53850
log 9.84125 log 9.97496
log 2.85941
723.45
i(B i (B 4. A)
log 9.65394
8 34' 46"
35
186.29
21'
46"
log 2.85941 log 9-17327
colog 0.23751 colog 7.7298 1
0.00000
PLANE TRIGONOMETRY
144 JB'
C' c'
= = =
} (ff
i3 1
ICHAP. vra
604'oo" 08'
7
28"
log 9.46942
lg 2.35388
225.88
-
A) =
log 2.35388
log 9.91 143
5438'i4
colog 0.00487
a
=
colog 7.72981
186.29
9-99999
EXERCISE 35 Solve the following triangles: 1.
=
a
Ans. 2.
=
a
=
b
840,
B= =
6
41.4,
52-8,
A = C= A =
124
C'=i5i/',
485, 12
14',
21
31'.
146
40
3-
4.
a
i==
3-25,
Ans. 5.
a
=
91.97,
Ans. 6.
=
b
= A = b = A = ft
242,
978.7, ^4w5.
7.
6
8.
a
678.5,
48.134,
Ans. 9.
ft
=
216.45,
Ans.
10. 11.
72.
= a =
ft
24',
2 -57,
767,
10
55',
93.99,
57 23.7',
^ = B=
32
54'.
36
53'.
C=
132
1 2',
120
35'.
#= C=
= 871.6, B = 44 01. 5', = = * B= c
= B=
14.332, 342.6,
Case HI.
c
c ft
= =
135 58.5', 423.1, 35- 826
>
26 12' 38", 177-01,
45
2 3'
13-617, 745-9>
28",
=
(1) (2)
find
To To
01. 3',
c
=
947-
c
=
3.85.
%(A
find i (4
97 44-
a'
= =
c
=
72.022.
=
300.29.
3',
C= C= 4 =
C= 4"
117
23' 22",
35 36' 20".
99*00'
I39S-. 142.
12^,
45 23' 54". 43 35- 6 '^15. Impossible.
Given two sides and the included angle, as
A and 5, we first find
2
'=5 47-3', '=53 234'. A = 36 24' oo".
a, &, C.
To
c'
c
5'=
= =
= 63.6. = 16.9.
c
5'=
1272.
19'.
'
= =
=
c
15',
find % (A
+ B)
and
\(A-
+ B), apply the relation A + -
)>
+
B).
C=
180.
Art. 66. apply the law of tangents,
SOLUTION OF OBLIQUE TRIANGLES
7*1
(3) (4)
(5)
-
A = \(A
+ B) +
To To
apply the law of sines.
find
c,
(4
B),
B=
).
i.
Given
= 6 = C=
Ji
a
.
+ 5) - | (4 -
check, apply one of the double formulas.
EXAMPLE
c
J (4
145
J^
12.346,
5-7213,
*
65 3 o'io".
^
Required
^ =
V \*
c
B= c =
86 55' 57", 27 33' 53", 11.250.
Fl'g- 79-
.
Solution. (1) (2)
a
b
To To
=
}(A+ B). | (,1 + B) = J (180 - C) = find | (/I - B). By the law of tangents find
+*=
a
6.6247, log (a
=
b)
18.0673.
0.82117
+ )= 8.74310 + B) = 0.19162
10
colog (a log tan | (^4 log
(3)
To
find
yl
To
find
c.
$ (A
+ B)
From c
A
is
J5)
= =
-
10 9.75589 02". 29 41'
a
-
i (A
-B) = - B) =
sin ~~.
C A7
'
an angle near 90
or
c
it is
86 55' 57", 27 33' 53".
we have
the law of sines
sin
but since
-
-*
and B.
B(4)
tan| (X
57 14' 55".
=
,
b
sin
C
sin
B
preferable to find c from the
second expression.
= = log sin C = colog sin B logc = c = log 6
either
0.75749 9.95903
0.33464 1.05116 11.250.
10
PLANE TRIGONOMETRY
146 (5)
Check.
the double formula (Art. 66, (6)),
By
a-b logs log sin J
=
s
log (a
1.05116
)=
(A
[CHAP.VIII
9.69480
+
10 log sin ? (4
b)
=
J5)
=
0.82117 9.92481
10
0.74598
0.74596
Compact Arrangement Check.
Solution.
a
=
*= = a+ b = C= \(A + B) = - B) = \ (A A = 5= c = &
12.346 log 0.75749
5-72I3
Iogo.82ii7
Iogo.82ii7
6.6247
colog 8.74310
18.0673
65 30' 10"
log sin 9.959 3
55" log tan 0.19162
log sin 9.92481
29-4i' 02" log tan 9.75589 86 55' 57"
colog sin 0.30520
14'
57
27
col
33' 53"
_
g sin 0-33464
1.05118
1.05116
11.250
EXERCISE 36 1.
Given a
=
C = 5* 3 6 A = 83 15', B - 45 09', c = 383.5. = 364, & = 640, C = 53 14'. Find A = 34 38', B = 92 08', c = 513. = 875, b = 567, C = 34 S*'Find A, B, c. = = 233.4, 6 = 557.2, C 18 23.0'. Find A = 12 22.0', B = 149 is.o ,*; = 343.7.
=
486, b
347,
'-
Find
2.
Given a
3.
Given a
4.
Given a
7
5.
6.
7.
8.
= 39.90, /I = 92 11.3'. Find B = 72 40.7', C == 15 08.0', a = 152.7. = 35 37.9'. Find C, A, b. Given c = 453.9, a = 478.1, Given a = 51.269, 6 = 14.687, C = 62 09' 24". Find A = 101 32' 32", B = 16 18' 04", c =* 46.269. Given b = 467.92, c = 612.34, ^ = 45 29' 16". - 438-36Find 5 = 49 34' 05", c = 84 56' 39", Given
b
=
145.9,
c
<*
SOLUTION OF OBLIQUE TRIANGLES
731
9.
10.
Given
c
Given a
n. Show
= =
345-67, a
44745, &
when
that
147
= 654.32, B = 67 45' 45". Find C, A, b. = 216.45, C = 116 30' 20". Find the area. Ans. T = 43336.
the included angle
is
a right angle, the law of
tangents gives tan
%(A
-
=
B)
^
?
.
Case IV. Given three sides, a, &, c. Each of the angles A, B and C is found by applying one 73.
formulas
of the
(2) or (4), Art. 68, for the tangent, sine or cosine of half
the
respective angle, but for the reasons stated in Art. 68 the tangent formulas are generally to be preferred. To check, apply the relation A C = 180.
+B+
EXAMPLE
i.
c
S\.
Given
= b = c =
a
!2.6 53
,
17.213,
A
//
Required
A = 32 36' 22", B = 47 08' 42", C = 100 14' 56".
\\
c
23.106. Fig. 80.
Solution.
By
formula
(2), Art. 68,
tanl=-*-,
ti
tan*=-*-,
a
S
2
2
S
b
--*-, 2
S
C
where
=
a
+b+c
t_J(s-a)(s-b)(s-c) V 5 = 8.57698 10 s = 26.486 cologs = 1.14092 s a = 13.833 a) log (s = 0.96722 b = 9.273 s b) log (5 = = s c c) 3.380 0.52892 log (5 2 = 1.21404 log& logk = 0.60702 = 0.60702 = = log k log k 0.60702 log i 0.60702 = = = 1.14092 log (s 0.96722 log (5 a) log (s c) 0.52892 = 9.46610 log tan 1 5 = 9.63980 log tan | C= 0.07810 log tan $A \A =16 18' n" K=5o o 7 28" =23 34' 21" ^ = 32 36' 22", 08' C=ioo 14' 56". 42", 5=47 Check. ^+J5 + C=3236 22"+47o8 42 '+ioo i4 56 =180. s
f
2
ft)
'
/
/
/
/
//
[CHAP, vnt
PLANE TRIGONOMETRY
148
Compact Arrangement Check.
Solution.
A =
B= C=
36' 22"
32
47 08' 42" 100 14' 56"
i8ooo'oo"
EXAMPLE inscribed
Find the area of the triangle and the
2.
and
escribed circles for the triangle in
The
=
a
Solution.
12.653,
=
b
area of a triangle in terms of the sides
Example
c
=
is
given
17.213,
radii of the r.
23.106.
by
(4),
Art. 67.
T = Vs(s-a)(s-b)(s-c). The formulas
for the radii of the inscribed
given in Problem
7,
T s
L
log
(s-
log (s log
(s
a) b)
-
c)
= = = =
T2 = = log T
log
s
i,
b
we have
1.42302 1.14092
0.96722 0.52892
4.06008 2.03004
T= k =
= 0.60702 log k = 0.88912 log&a
ka
1.06282
kb
= log k b =
T
_
a' a
Using the results of Example log 5
circles are
Exercise 33.
ka s
and escribed
1.50112
= =
107.16, Fig. 81.
4.046, 7-747,
n.55 6
>
SOLUTION OF OBLIQUE TRIANGLES
731
A
Check.
Problem
8,
convenient check
1
k
I
g
obtained by using the relation in
is
Exercise 33,
T
log
-
=
colog k
R
L= ka
=
log
=
-
x
7 ka
4- JL-4- JL L L Kb KC
= kc
=
9.II088
10
colog k b
=
8.93718
10
=
8.49888
=
0.24716
=
0.12909
~=
0.08653
-
10
9.39298
colog k a
colog k c
"
'
'
^ka
Kb
Kb
log
149
-
~=
10
0.03154
KC
f+
#a
6
+ ^ = 0.24716 =fK #c
EXERCISE 37 1.
2.
3. 4.
5.
6. 7.
8.
9.
10.
11.
= 463. = 43 "', C = 99 15'= Find^ 37 34', B = 4.02. Givton a = 3.21, ^ = 3.61, c Find A = 49 24', B = 58 38'. C = 71 $*' = 74.6, 6 = 81.9, c = 90.0. Find the angles. Given Given a -= 3544, 6 = 2 77-9 * = 4<>i-3Find .4 = 59 39-5', B = 42 35-3', C = 7745-2'. Given a - 1.961, 6 = 2.641, c = 1.354Find .4 = 46 03.7', iJ = 104 07.6', C = 29 48.8'. Given a = 87.06, 6 = 9.16, c = 79.02. Find A, B, C. Given a = 33594, 6 = 4216.3, c = 4098.7. Find ,4 = 47 38' oo", B = 68 01' 06", C = 64 20' 54". Given a = 33.112, b = 44.224, c = 55-336= 53 03' 08", C = 90 n' 38". Find A = 30 45' M", Given a = 14.493, * = 5543 6 ^ = 66.913. Find the angles. Given a = 46.78, b =* 35.90, c = 77.00. Find the area. Ans. T = 573-Qi-
Given a
=
== 286, 6 321, c
<*
>
Find the area
in
Problem
2.
Check.
PLANE TRIGONOMETRY
ISO
[CHAP.VHI
12. Find the radii of the inscribed circle, of the escribed circles and the circumcircle of the triangle in problem 10: Check by using the relation k a + kc k = 2 D = 4 R, Problem 9, Exercise 33.
h+
=
a
13.
4, b
=
5, c
cosines directly, that
=
Find the angles by applying the law of
6.
is,
cos
^= P + ct-a*
B=
etc
2 be
The
14.
sides of
an inscribed quadrilateral taken in order are
a
=
56, b
=
33, c
=
16,
d
=
63.
Q of the quadrilateral and the angles A, B, C, D, the notation being that employed in Fig. 71, Problem 10, Exer-
Required the area cise 33.
Q=
Ans.
1428,
A =
44
Practical Applications.
74.
numerous applications
finds
46',
among
BThe
90,
C=
solution of
in the various arts
H',
135
#=
90
oblique triangles sciences. Chief
and
these are surveying, engineering, physics, astronomy and The simpler applications which involve the solution of
navigation.
single triangle need no explanation, since they may be immediately referred to some one of the four cases treated in the preceding sec-
a
of the practical applications, however, involve several triangles which must be successively solved in whole or in part before the required distance or angle can be ascertained. Sometions.
Many
times the intermediate triangles to be solved are not apparent from by some auxiliary geometrical con-
the figure, but must be sought
Again, it may happen that no single triangle exists conthe requisite number of parts; in such cases the solution is taining effected by solving the equations which arise by applying the forstruction.
mulas
of
illustrate (a)
Chapter VII so as to involve the unknown parts. each of these cases by an example.
We
EXAMPLE
c
System of
triangles.
i.
Given one side
quadilateral ABC'C (Fig. 82) and two angles A and this side, also the angles a, ft which the diagonals d\, d*,
and
B
respectively,
make with
x opposite the given side Hansen's problem.
c.
B
shall
of
a
adjacent to
drawn from A
the given side, to determine the side
This problem
is
sometimes known as
SOLUTION OF OBLIQUE TRIANGLES
74]
Let
Analysis.
angle angle angle
CAC' = 4CC' = AC'C =
AC =
b,
= P, angle CC' = 0', 0, angle = 0'. 0, angle BC'C ABC, one
two adjacent angles A d2 may be found. In the triangle
(2)
BC =
CBC'
a',
In the triangle
(1)
a,
151
are known, hence b
ABC,
may
side c
and
are known, hence
,
one side
c
and two adjacent angles
a,
5
be found.
Then, in the triangle CBC, two sides J2 b and the included = 5 /3 * are known, hence # may be found. angle ft' (3)
(4)
,
Compute x
Check.
again, using the triangle
previously found a and di from the triangles
ABC
CAC' having ,
and ABC'
re-
spectively.
In order to determine the length CC (Fig. 82) of a be built across the end of a lagoon, a distance AB, 500 ft. long, was measured off, and the following angles were measured with a transit: Illustration.
trestle to
CAB =
f
35
Required the distance Solution,
(i)
-
Triangle
ABC.
sin
,
dz
= A, C BA = = a, CBA = 17' CC = x.
105 30'
C'AB =
c
By
A
and
ACB = 180 logc = 2.69897
(4
47 32'
=
B, p.
a
=
+ 0) =
26
58'.
= 2.69897 = 9.86786 log sin/3 = Q.34345 loS a = 2.91028 a = 813.36.
= 9.9839^ log sinyl = 0.34345 colog sin ACB = 3- 026 33 log dz = d2 1062.5. *
5' =
the law of sines ,
,
95
A 5, C, C' are in the same pfane, otherwise be given in addition to the data of the problem. ro's are omitted In this and the following problems the 10. t 9*98391 when this may be done without danger of confusion. This is a common practice It
is
the angles
assumed that the points
CAC and CBC' must
among computers.
,
PLANE TRIGONOMETRY
52 (2)
Triangle
ABC. By
ACB =
colog
Triangle
-
log tan | (<'
&i
=
= =
2.81971 660.25.
CAC.
the law of tangents
= 48
18'.
0-^1=153.11,
2.83196
0.02020
O = 46
2.69897
B = 9.99775 = 0.12299
Triangle
=
ft)
0')
53'.
log^i
+ b) = 6.83988 colog = 0.34836 tan + tf) \ ($' log
log (a
a+Ji= 1473.61,
di)
=
2.18501
+ = 6.83162 = 0.15302 log tan J (^ + 0) colog (a
log tan | (0
i
19' 55".
B)
=
(*-*)-
9.16965 24' 25"-
the double formula involving the sines,
sm
= log (c/2 + 6') = log sin \ = 0') colog sin i (<' = logo; # = ft)
('
(ft)
)
=
(dz
it
+ a = 48
log sin
6=679.15, ^+6=1445.85, log (dz
By
(B
CBC.
-j8) fa
180
3 8 3-35-
By
vm
the law of sines
= 2.69897 logc = 9.76164 sin a log sin^C^B = 0.12299 = 2.58360 log b *
(3)
[CHAP,
<>-
sin
2.83196
= 2.18501 = sin \ + 6) 9.91278 = 0.83505 sin 6) (0 logs = 2.93284 x = 856.7 (check). log (a
9.96022
log
0.14065
colog
2.93283 856.7,
di)
(c/>
EXAMPLE 2. In Example i may be found without leaving
Auxiliary geometrical constructions.
was shown how the distance
the line
AB.
CC
Another important problem
is
to determine one's
position from the angles which the sides of a known triangle subtend from that position. This is known among surveyors as the three-
SOLUTION OF OBLIQUE TRIANGLES
74J
153
point problem, sometimes as PothenoVs problem* It may be stated thus: Given three points A, B C whose mutual distances are known, 9
P
to find the distance of a fourth point from either of the points A, B, C, having given the angles, which the sides AB, BC, CA subtend at P.
C
Analysis. Let A, B, mutual distances are,
be the three points whose
AB = c,BC = and
BC
CA =
a,
b,
P
represent the fourth point at which subtend the angles a and /? respectively. let
the distances
P from A B
d\, d%, d$ of
Circumscribe a
,
and
C
about the triangle
circle
AC It
Fi
and
-
8 3-
required to find
is
respectively.
ABP, and
let
C
be the
PC produced. Draw AC and BC. Then angle BAC = angle BPC = 0, being inscribed angles subtended by the same arc, and likewise angle ABC' = angle A PC = a.
point at which this circle cuts
PC
or
f
ABC
In
the triangle one side c and the two adjacent angles ', 1 are hence can be found. a, known, (2) In the triangle ABC, the sides are known, hence A, the angle opposite the side a, can be found. (1)
AC
AC
In the triangle ACC', two sides b and = A are known, hence angle angle (3)
CAC
and the included
ACC =
71 can be
found. (4)
In the triangle
hence the sides (5)
d\,
A PC, one
and
Similarly, d%
side b
and two angles
a, 71 are
known,
can be found.
can be found from the triangle BPC,
having previously computed the triangle BCC. (6) Check. Compare the value of d$ found in
(4)
with the value
of fa in (5). Illustration.
A,
C
B,
(Fig.
83)
CA =
3 miles.
From
are three hostile
forts
whose
4 miles, B'C = 2 miles, a battery planted at P, AC subtends an angle
mutual distances are known to be 34 30' and BC an angle of 23 from each of the forts.
45'.
AB =
Find the distance of the battery
* Pothenot's problem and Hansen's problem are named after the men who were supposed to have first formulated and solved these problems. It is now known that both of these problems were previously solved by Snellius, the first
Hence, if any one's name ought to be that of Snellius.
in 1617, the other in 1627.
these problems hereafter
it
is
to be associated with
PLANE TRIGONOMETRY
154
With
Solution.
[CHAP, vni
the notation indicated in the figure,
given
a =2,
6
=
3,
c
=
4,
a
=34
30',
=
ft
23
we have 45';
to find d\j dz y d$.
ABC. By
Triangle
(1)
the law of sines
sing
BC=c
-55L
sin (a
= = c log = sin a log = sin + ft) (a colog
+
log^C'
=
30' +23
34 0.60206
0.07040 0.42559
logC" =
A
/I
By
-a = +c
2
cos
!
B=
c
2
+4 2-3-4
A (3)
By
28
= a gyj
=
1.8945.
2
ACC.
2
4
+
ca
-
2
3
=
6 g 75
2-4-2
B =
57'.
Triangle
0.27749
!
2 2
9.60503
0.07040
+a -b
2 6c
3
0.60206
the law of cosines
ft*
-
ft
BC =
2.6644.
ABC.
ft)
15'.
log sin
9.75313
2
cos
58
= = =
AC = (2) Triangle
45'
=
+
46
34'.
Triangle
the law of tangents
tan^^T,
ton^g-'ft tan
- a=
22
22
12
04'.
. .^"-.^"'gyQ^, gggta_.'8o"-V.g3 5g
ft
ft
(ft
>
~ 4C'=
0.3356,
a
- BC =
+ AC=
5.6644.
a
+ 5C' = 3.8945.
- AC) = = colog (b + AC) log
>
9-52582 9-24685
log (a
- BC) =
9-02325
+ BC) = 9.40954 "^ ^ ^ = log tan 0.97596 colog (a
r2
1.34285
0.1055,
SOLUTION OF OBLIQUE TRIANGLES
74]
log tan
-
AC'C
log tan
0.11552.
I5S
BCC
72
=
2
9-40875
3*'
AC'C +
7i
ACC-yi
yi
2
2
=
=
34
69
52'.
APC.
Triangle
(4)
Triangle
By
the law of sines
BPC. sin 72
"
sin sin
,
CAP _
sin
ft
log 6 log sin 71
a
'
%in/3 ^
sin
sin (a
(|8
+ 72)
.
.
a
+ 7i = 34 + 72 =23
= =
35'-
sin
+ 34 45' + 69
30'
.
sm p
a S^'
=
35'
69
22',
93 20
.
= = sin 72 log = colog sin/3 = logdi log a
0.47712
9-757*4
cologsina =0.24687 0.48113
0.30103 9.97182
0.39497 0.66782
3.0277. log
ft
= logsin(a + 7i) colog sin
a
log dz
^3 (c)
= = =
+ 72)
colog sin
0.24687
0.69520
0.30103 9.99925
0.39497
0.69526 4-957 (check).
4-957-
From a point O AP, PQ, QB
3.
84), the segments
straight line
PQ =
(Fig.
of a
subtend the angles a, 7,
respectively.
and
log sin 08
= = = =
Solution by solving a system of simultaneous equations.
EXAMPLE
/3
log a
0.47712 9-97 121
The
distances
AB =
d,
known; required distances AP = x and QB = y. Fig 84 In this case no single triSolution. number of known parts to angle contains a sufficient law of sines to the various solution, but on applying the
we
c
are
obtain the following equations,
afford
a
triangles,
PLANE TRIGONOMETRY
156
[CHAP, vra
(2)
(4)
bin x>
7;
Dividing
* .
#
+
(i)
by
sin (a sin
<;
and
(2),
(3)
+ 7) ^ OP a
Multiplying
by ,
?
by
we
y
From
sin
+ c)(y + c)
sin (a
the conditions of the problem
x
The simultaneous x and
y,
which
^
+y
=
sin yp -r j)
+c
=
/
t^f
OP
sin ft
6x
a slight reduction,
(6) gives, after
xy (x
obtain
y
v
*
OQ
(5)
(4),
we d
a sin ft
+ 7) sin also
-
W ,
"
(ft
+ 7)
<.
have
c.
(8)
equations (7) and (8) contain but two unknowns, be found from them by the familiar methods of
may
algebra.
Second
Call the
solution.
The area
(Fig. 84).
of
common altitude of the triangles h may be expressed in two ways,
each triangle
as one-half the product of the base
first
by the
altitude, second as
one-half the product of two sides into the sine of the angle included by these sides (Art. 67, (2)).
Accordingly we have 2
2 2 2
APO = xh = AO area QPO = yh = QO area PQO = ch = PO area ABO = dh = AO area
PO
sin
a
BO
sin
ft
QO BO
sin
7
(9)
(10)
(n)
sin (a
+7+
(12)
ft)
Dividing the product of (9) and (10) by the product of (n) and (12), and canceling the factors which appear in both the numerator and the denominator, gives
2
._.
cd
sin
a
sin
ft
+
ft
sin 7 sin (a
_
sin
,
line
An
between two
island
cities
A
PQ
sin
(Fig. 84)
(8),
is
/
ft
sin7sin (a +ft
+'7)'
Equation (13), together with equation determination of x and y. Illustration.
a
x
+7)'
sufficient for the
one mile wide
lies
in
a
direct
and B on opposite shores of a river. From
CHAPTER IX THE GENERAL ANGLE AND General Definition of an Angle.
81.
ITS
MEASURES
In elementary geometry an
defined as the difference in direction, or the amount of openwhich meet or tend to meet in a point. Obviing, between two lines in direction, the opening between them ously, two lines differ most
angle
is
is
greatest,
when they extend
in opposite directions.
this definition, no angle can be greater than a fact every angle considered in plane geometry
According to
straight angle, is less
and
in
than two right
angles.
For the purposes of trigonometry and higher mathematics in genconvenient to think of angles as formed by revolving a
eral it is
a plane about a fixed point. The line in its first or initial it is called the position is called the initial line, in its final position The amount of vertex. terminal line, the fixed point is called the
line in
rotation which brings the line
from
its initial
position is called the angle between the
With
this definition, it is plain that
two
position
to its
terminal
lines.
an angle
may have any
magni-
tude whatever, for there is no limit to the amount of rotation which a line may undergo. The angle described by the big hand of a clock when it has run 15 increases so long as the clock continues to run, minutes the angle described is a right angle; when it has run 30 mindescribed is a straight angle; after an hour the big utes the angle
to its initial position, yet the angle which it has not zero but four right angles. As the hand continues
hand has returned described
is
to move, the angle between the
hand and
its
initial position still
After two complete revolutions this angle to eight right angles; after three complete revolutions, twelve right angles, and so on.
increases.
Before this
we can
wider sense,
assign a magnitude to
an angle
we must know more about
it
is
equal to
in
than
Fig. 118. the angle AOB merely the position of its sides. By indicated as of an angle by the short 45 (Fig. 118) may be meant
indicated arrow, or an angle of 405, as 177
by the long arrow, or
PLANE TRIGONOMETRY
l?8
[CHAP, ix
an angle of 45 increased by any number times 360, depending on how many complete revolutions OB has described in reaching its final position. The angles 45, 405, and all the other angles which AOB might represent, are said to be coterminal. In elementary geometry, and generally when nothing to the contrary is said or implied, in referring to an angle, the numerically smallest of the coterminal angles
all
is
understood.
This value
is
known
as the
principal value of the angle.
and Negative Angles. Sometimes it is convenient between the two directions in which rotation of a line about a point in a plane may take place. This
82. Positive to distinguish
done by prefixing a plus sign to the resulting angle if it has been described by rotation in a is
Vc
*a
^
a clock
move
minus sign sign
120.
first\
120
is
read
if
When the (clockwise), Fig. 120. considered, the initial line is always read
posite direction
x ig.
which the hands of (counterclockwise), Fig. 119, and a the rotation has been in the op-
direction contrary to that in
Fig. 119.
" angle
is
thus, each of the angles in figures 119
and
AOB"
83. Complement and Supplement. If the algebraic sum of two angles is equal to a right angle, each angle is said to be the complement of the other ; if their algebraic sum equals two right angles each That is to say, is said to be the supplement of the other.
A If A If
A
+ B = 90, A and B are complementary angles, + B = 80, A and B are supplementary angles. 1
B may have
or
95 4" (~~
any magnitude,
5) = 9>
95 an d
positive or negative. 5
Thus, since
are complementary angles,
and since
iioH84.
290= 1 80,
110 and 290 are supplementary angles.
Angles in the Four Quadrants.
Let the vertex of the angle
be taken as the origin of a system of rectangular coordinate axes,
and the #-axis.
initial side of
the angle for the positive direction of the 90, as angle
It follows that every angle less than
THE GENERAL ANGLE AND
8s]
Fig.
ITS
MEASURES
121, has its terminal side in the first quadrant, it is said to be an angle of the first quadrant.
reason
and
179 for this
Every angle
which is greater than a right angle but less than a straight angle, as angle AOP^ is said to lie in, or to be an angle of, the second quadrant. Similarly, the angle AOPs is an angle of the third quadrant, angle AOPt an angle of the fourth quadrant, but an angle greater than four
right angles
and
less
than
,_
five right angles falls
again in the first quadrant, and so on. Negative angles also lie in the quadrants determined by their terminal sides. first
AOB
Thus, while angle
quadrant, the angle
AOB
(Fig. 120)
lies in
(Fig. 119) lies in the the fourth quadrant.
Sexagesimal Measure of Angles. It is familiar to all that measures may be employed in the measurement of a given magnitude. Thus, any distance may be expressed in inches, feet, yards, rods and miles of the English system, or in millimeters, centimeters, meters and kilometers of the metric system. Values may be expressed in dollars and cents, in pounds and shillings, in francs and centimes, etc. Just so, angular magnitudes may be 85.
different units of
expressed in terms of different units of measure. In order to express a measure given in one set of units in terms of the units of another system, it is only necessary to know the relation between
the units of the two systems.
The system
of angular
measure which we have used thus
called the sexagesimal system.* *
From
The
far is
total angular
magnitude about
+ decimus,
one- tenth) , meaning
the Latin word sexagesimus (sexa, six
one-sixtieth.
The sexagesimal of a unit of time,
The sixtieth part scale was once applied to many measures. and length and weight was called its primate or prime; each
prime was divided into sixty seconds, each second into sixty thirds. The divisor 60 has now disappeared among Western nations, except in measures of angles and of time. It is probable that the division of the angular space about a point into 360 degrees originated with the Babylonians, whose year consisted
The Latin word for degree is gradus, the Greek, bathmos, each of which means step; that is, a degree originally meant the daily step of the sun eastward among the stars. The Chinese knew many centuries ago that the year consisted more nearly of 365! days, so they divided the total angular space about a point into 365 i degrees. The sexagesimal system is unscientific and is doomed of 360 days.
to be replaced sooner or later
by a
better system.
PLANE TRIGONOMETRY
l8o
[CHAP,
u
a point is divided into four equal parts and each part is callec a right angle. The ninetieth part of a right angle is called degree (), the sixtieth part of a degree is called a minute ('), anc the sixtieth part of a minute is called a second ("). The sexa gesimal system of angular measures is the one most frequently usec in
life.
every-day 86.
Decimal Division of Degrees.
minutes and seconds, the degree into tenths,
is
Instead of subdividing intc
sometimes divided decimally
The decimal
hundred ths and thousandths.
division
o:
the degree has been used more or less ever since the invention o: decimal fractions in the sixteenth century. Tables based on th( decimal division of the degree have been published at various times.*
Centesimal Measure of Angles.
Another system of angulai as the centesimal, is obtained by dividing the righl angle into 100 equal parts, each of which is called a grade (g), eacl: ), and each minute into 100 seconds grade into 100 minutes 87.
measure, known
O
An
angle of 25 grades 16 minutes and 78 seconds would be writter s
25" i6 78", or
more simply,
25.1678^.
The
centesimal system of angular measures was introduced as i of the metric system by the French reformers! at the time 01 part the great revolution. It possesses many advantages over the sexa-
gesimal
system, but owing to the fact that nearly all reference tables, all records of observation, the graduation of al
books and
astronomical instruments and of most engineering instruments,
a$
well as the scales of geographical and nautical maps and charts, an based on the sexagesimal system, the progress of the introductior
has been slow. the system
system
is
is
But
in spite of the
making steady gains
used.
many
obstacles to be overcome
in the countries
The centesimal system
is
where the metric
now used
exclusively the field surveys in France, Belgium, Hessia and Baden, and it It is regularly taught legally recognized in several other states. many European high schools and technical schools. *
The decimal system
is
now taught
Harvard University and a number t
The
alongside the sexagesimal system
ir is
ir
ir
of other Eastern institutions of learning.
invention of this system and the first attempt to introduce it dates back is due to a German by the name of Johann Karl Schultze.
to 1783 and
THE GENERAL ANGLE AND
88]
88.
The Circular
ITS
MEASURES
or Natural System of Angular Measures.
181 In
practical investigations and in nearly all theoretical work, it is convenient to employ what is known as the circular or natural
many
system of angular measure. It is shown in geometry that in concentric a' (Fig.
tional to the radii r
and
angle
a
a
-
by any
r
and
circles
the arcs a and
at the center are proporf r of the circles, that is,
122) intercepted
again,
a> ->
r
M
(i)
that in the same circle two central
angles a and
cepted arcs a
ft
and
are to each other as their interb,
that
is,
-!-*/; From
that the ratio of the length of the arc to the independent of the length of the radius, ratio is constant so long as the angle is this that other in words, and from (2) that this ratio varies as the angle, and constant may therefore be used as the measure of the angle. This ratio is (i) it follows
radius of the circle
is
as the circular measure of the angle.
known
measure of an angle is the ratio of the length of its intercircle whose center is at the vertex of the angle, to the a in cepted arc, radius of the circle. The unit of circular measure or natural unit is obtained by making
The
circular
the angle such that the intercepted r, that is, by taking arc equals the radius. This unit is called a radian. A radian is an angle which, when placed with its vertex at the center arc equal in length to the of a circle, intercepts an radius of the circle.
a equal to
Thus,
if
the arc
to the radius
AB
OA, the
(Fig. 123) is equal in length
angle
AOB
measures one
radian.
One peculiarity of the circular system is that it has no subsidiary or derived units; that is, no All other units which are multiples or sub-multiples of a radian. the this of in terms single unit, angles large or small are expressed
radian.
PLANE TRIGONOMETRY
182
[CHAP. IX
of Sexagesimal and Circular Measure. To between the two kinds of units, degrees and radians, best to compare the two measures for the entire angular space
89.
Comparison
find the relation it is
about a point, that is,, of four right angles. Expressed in circular measure, the measure of this angular space is equal to the circumference of a circle divided by the radius. Now the circumference of a
circle is equal to the diameter, or twice the radius, multiplied by 3.141594-. Denoting this number by the Greek letter TT,* we have for the circular measure of four right angles
circumference
=
2
IT.
radius
Measured in degrees the same angular space obtain the fundamental relation 2
or
TT
radians
=
360 degrees,
IT
radians
=
180 degrees.
is
360, hence we
(i)
This relation enables us to reduce radians to degrees, and vice Dividing both sides of (i) by the number TT, we obtain
versa.
i
radian
=
^-^
degrees,
7T
hence, *
The number TT is a most marvelous number. Tt is incommensurable, that is, cannot be exactly expressed by a fraction whose terms arc whole numbers. Nor can it be found by taking the square root, cube root or higher root of some commensurable number. Neither is it the root of any algebraic equation. For it
this reason it is called
a transcendental number.
However, the value of w may be computed to any desired degree of accuracy. Archimedes showed that it is less than 3^ and greater than 3} J. The former value
commonly employed in rough approximations. The Hindus, as early as the sixth century, computed
is still
the value of
TT
from the
= 3.1416; perimeter of a regular inscribed polygon of 384 sides and found TT the value now generally used where 3^ = 3.1428 is not sufficiently accurate. In recent times, the value of ir has been computed to 707 places of decimals. first 32 places were computed by Ludolph van Ceulen, a Dutchman, who devoted a good portion of his life to this task. For this reason TT is frequently Its first 30 places are referred to as Ludolph's number.
The
TT
=
3.141,592,653,589,793,238,462,643,383,279.
THE GENERAL ANGLE AND
8g]
To reduce radians
to degrees,
MEASURES
ITS
183
multiply the number of radians by 7T
(equals 57.3 nearly*).
Again, dividing both sides of (i) by 180 and writing the second first, we obtain
member
i
degree
=
-^-radians; 1 80
hence,
To
reduce degrees to radians, multiply the number of degrees by -^1 80
(equals 0.0175 nearly).
EXAMPLE
i.
How many
Since
Solution.
degrees in
^radians? 2
w radians
=
1
80, -radians
=
2
and
radians
=
90,
270.
2
EXAMPLE
2.
Solution.
How many
Since 180
=
radians in an angle of TT
radians, 60
It is best to retain the letter
TT
=
6o?
-radians. o
in the answers unless there
is
some
reason to the contrary. For instance, putting for IT its value 3.1416 and dividing by 3, the answer to the second example might have
been written 60 wise the
first
=
1.0472 radians, but x/3 radians is preferable. Likeexample might have been stated thus: How many de-
grees in 4.7144 radians? but the statement as
first
given
is
preferable.
When
a number represents the measure of an angle, and no unit is expressed, the natural unit is understood. Thus, when we speak of the angle T/2, we mean not 7r/2 degrees but TT/ 2 radians. The angle TT means TT radians or 180, the angle 2 means not 2 but 2 radiWe may look upon TT, ir, 2?r, 2mr, when referans, or 114.5916. ring to angles, as abbreviations of
90, 180, 360 and n times 360
respectively. *
More
The value
accurately, i
radian
i
degree
i
minute
i
second
= = = =
57-295,779,513 degrees,
0.017,453,290 radians, 0.000,290,888 radians, 0.000,004,848 radians.
of the radian in degrees has
been calculated to 43 places of decimals.
1
PLANE TRIGONOMETRY
84
[CHAP, ix
EXERCISE 44 (Use 3} for
TT
and
57.3 for the value of
n represents any
radian,
i
positive integer.)
Express decimally the following angles
1.
:
45 48' 36," 185 59' 15", 35 30' 30", 375 oo' 47"Ans. 45.81, 185.9875, 35.5083, 375.01305.
Express in degrees, minutes and seconds, 67.003, 7 of a right angle. Ans. i62i 67 oo' 10.8", 153 09' 21.6", 2.
I
16.35,
53* I 56,
/
64
,
Express in radians,
3.
o
o
315
75
i
90, 45, 22^, 60, 15, 30, 135, 270.
Q
.
TT
A
Ans.-,
360
>
17' 08.57".
IT
TT
-,
-,
TT
4 Express in sexagesimal units the following angles
4.
7T
7T
7T >
,
10
18
2 7T
x
-, 23
2
2 W7T,
7T,
340, f
$
right angles,
TT,
angles:
105, -105,
2
725,
State in which quadrant a line
6.
:
-
TT .
angles:
etc.
-J,
3
Ans. 10, 18, Give a geometrical representation
5.
2,
.
,
12
I
-,
,
7T,
5
IT
-, 3
8
2
is
.
114.6, 28.65, 18.23.
.
of
TT,
each of the following
w TT,
2
(2
+
n
TT.
i)
after describing the following
2
^,
-fir,
4
-
~,
+375, 2W7r-|7r.
2,
3
-4s.
2d, 3d,
...
ist, 4th.
7. Name three angles coterminal with an angle 30. 8. Name two positive and two negative angles coterminal with Ans 345, 705, -375) -735-iS-
9. What is With -465.
10. ,
Name
the smallest positive angle coterminal with
Ans. the
smallest
With -625? 11. The principal value
negative
of
an angle
If 6 represents
coterminal
is
write in one formula
j
TT;
Ans.
any angle
less
+
quadrant can be expressed by 2 nir in each of the other three quadrants. Ans. 13.
II.
What
complement?
(2^+1)71is
0.
III.
with
angle
coterminal angles. 12.
465?
105, 255.
than 90, 6.
all
2
mr
(2n+i)T + B. TT
all
+ 6?
What 6,
first
angles
IV. 2Uir
Ans. |?r
all
J v.
angles in the
Express similarly
the supplement of the angle J
+
735?
e.
is
the 0.
THE GENERAL ANGLE AND
90]
What
14.
is
the complement of J
IT
+
MEASURES
ITS
Of -
?
185
6?
o
Ans. $
ir
-0,1 + 6. 6
Express both in radians and in degrees:
15.
(a) (b)
Each of the angles of an isosceles right triangle. Each angle of a triangle whose angles are to each other
as
1:2:3. Find the number of radians
1 6.
in
each of the angles of a regular Ans.
pentagon, octagon, w-gon. 17.
187'
2Z,
2/ 2J[, v^~"
54
7r .
n
Express decimally the following angles: if 18* 19"", 25" 65* 75", v Ans. 17.1819", 25.6575", etc. 95", i' oo 15".
5*
Express 35 30' 30" and 35* 30^30" each as a fraction of a right Ans. 0.394537,0.35303. angle. 18.
19.
Express 35 30' 30" in centesimal units.
20.
Prove the following
diminish the number 21.
Make
Ans. 39" 45* 37^.
To
convert grades into degrees, of grades by one-tenth of itself. rule:
a corresponding rule for converting degrees into grades.
22. Find two regular polygons such that the number of degrees in an angle of one is to the number of degrees in an angle of the other
number
as the
of sides of the first
is
to the
number
of sides of the
Ans. Triangle and hexagon.
second.
23. Find three pairs of regular polygons such that the number of degrees in an angle of one is equal to the number of grades in an angle of the other.
d and r represent respectively the number in any angle, prove that radians and degrees 24.
If g y
n (g
d)
=
20
of grades,
r.
90. Relation Between Angle, Arc and Radius. If an angle 6 is subtended by an arc 20 ft. long and the radius of the circumference is of which the arc forms a part is 10 ft., the number of radians in an subtended is 20 ft. -r- 10 ft. or 2, and generally, if an angle by arc s units long and the radius measures r of the radian measure of the angle is
e
=
i.
same
units, the
d)
PLANE TRIGONOMETRY
186
Solving this equation for s and r respectively, *
Equation
-
*0,
we
get
r = |. u
(2)
(i) enables
[CHAP, ix
(3)
when we know
us to find the angle
the length
and the radius, (2) gives us the length of the arc in terms of the radius and the angle, and (3) gives the radius when the length of the arc is known as well as the angle which it subtends. These three equations enable us to solve a great variety of inter-
of the arc
esting problems.
EXAMPLE i. The diameter of the earth subtends an angle of 17.5" at the center of the sun. Assuming the diameter of the earth to be 7917.6 miles, what is the distance of the earth from the suri? Solution. may consider the diameter of the earth approxi-
We
mately equal to an arc which subtends an angle of 17.5" to radians
we
=
v
if'S
60X60 hence, applying equation (i) miles
EXAMPLE
T
we have
-
for the required distance in
3500
^^
*7'5
of logarithms, putting
A
2.
T
=
feet radius.
N
is
it
3.1416.)
entered the curve.
,
In what direc-
leaves the curve, assuming that rate of 50 miles per hour. uniform at a going it
train
when It
N
leaves the curve at B.
it
is
required to find the angle
Angle equation M angle
it
be the direction of the Let A on entering the curve and TB its direction
Solution.
-
93)322>ooo miles
railroad train going due north strikes a curve of The curve turns to the left and the train leaves
the curve just 40 seconds after tion does the train move after
Fig. 124.
Reducing
180'
ooxo =
"
(By the use
17. 5".
find
AOB
NTB =
angle
AOB
(Why?), and by
(i) v '
(in radians)
=
NTB or 0.
~^^-
THE GENERAL ANGLE AND
goa]
OA
AB
it
move over
is given and arc takes the train to
OA = 3&
ITS
MEASURES
187
can be found, since we know the time which it at a given speed.
AB =
arc
miles,
X4 X 60 miles,
5
60
5280
hence ^
=
S
_X 4o^aS oo =
60X60
_
88
v X
The
_
180
88
7T
105
_88_ radi
5280
105
X
180 ~~
105
X
X 7 _ ,o 4
22
train leaves the curve in the direction
The
Area of a Circular Sector.
90a.
N. 48 W.
area of a circular sector
is
equal to the area of a triangle with a base equal in length to the arc of the sector and an altitude equal to the radius, hence the area
A
is
rs
AA =
>
2
which by
(2), Article 90,
becomes .i
where 8
is
=
ir*e,
expressed in radians.
EXAMPLE
(i)
Find the area of a circular sector bounded by an arc
i.
10 feet in length, the radius of the circle being 50 feet. Solution.
and
=s
By r
=
(i), Article
50,
90,
hence by
the radian measure of the angle
(i)
5
A =
I
2
2
5o
12
=
250
sq.
ft.
50
EXERCISE 45 (To obtain the answers as given, use 1.
Express in radians
arc 50
ft.
long on a circle
tr
=
3}).
and in degrees the angle subtended by an whose radius is 200 ft. Ans. J radian
2.
What
is
the radius of a circle
tends an angle of
i
?
on which an arc 100 Ans.
=
ft.
14.3.
long subz
ft.
of 75, on a 3. How long is the arc subtended by a central angle Ans. 130.95 ft. circumference whose radius is 100 ft.?
PLANE TRIGONOMETRY
1 88
4.
Find the length
of
an arc
[CHAP, ix
= 3960 (r Ans. 69} miles.
on the earth's surface
of i
miles). 5.
i
How
long must a line be to subtend i" at the distance of
mile ?
Ans. 0.3 inch (approximately).
A
fly-wheel 10 ft. in diameter is revolving 200 revolutions per minute. Find the speed per second of a point on the rim. 6.
AnSm 12oi
ft .
3 7.
The diameter
and the distance
of the
sun subtends an angle of 32' at the earth is about 93,000,000 miles. Required the
of the sun
diameter of the sun.
Ans. 866,000 miles.
The earth moves around the sun once each year. Assuming its to be a circle, find the velocity of the earth per second. path (Use the distance of the sun given in Problem 7.) 8.
Ans.
The
18.5 miles.
earth revolves on
its axis once in 24 hours. Find its 9. angular velocity per second, that is, the angle through which the earth turns per second, and hence the velocity in miles of a point on the
Use r = 3960 miles. Ans. Angular velocity 15" per
equator.
sec., linear
velocity 0.288 mi. per sec.
The
latitude of the city of Seattle is 47 40'. Find the shortest distance from Seattle to the north pole. Ans. 2926 miles. 10.
(Distances on the earth are of course measured along the arc of a great
circle.
11.
Use
=
r
The diameter
3960 mi.)
of a
graduated circle is 12 inches and the graduFind the approximate distance between
ations on its rim are 15'.
two consecutive
divisions
on the rim.
Ans.
of
an inch.
40 12.
Two
meet at
railroads
are connected
They The inner curve
is
right angles at O.
by a quadrant of a 2000 ft. long. What
distance of either point
A
or
B
In Problem
how much curve?
the
from
Ans.
13.
O? OA =
circle. is
4222
ft.
Fig. 525.
the rails are 4 ft. apart, be the outer rail than the inner will longer 12,
if
rail
Ans.
2
of the IT it.
THE GENERAL ANGLE AND
9i]
A
14.
A
sq. in.
The straight sides of the lot are 100 ft. each and them is 60. The lot was sold at $100 per foot
what was the
Ans. $87,120.
price per acre ?
(a)
State and prove the law of sines, (c)
Show
that
R=
2
.
the radius of the circumscribed
State and prove the
(b)
a sin
is
26^
Ans. 30.
city lot has the shape of a circular sector, the curve border-
law of cosines,
R
189
Review.
91. 1.
of
of the sector.
ing on the street. the angle between frontage,
an area
circular sector, radius 10 inches, has
Find the angle 15.
MEASURES
ITS
= A
circle,
sin (d)
=
B
-rsin
C
where
Prove the projection
theorem by means of the law of cosines. 2.
= -
(a)
In any triangle show that sin (A
%(A+B)=*
cos C, sin Prove the double formula of
+ B) = sin C, cos (A + E) + B) = cot C. (b)
C, tan % (A
cos
-|
and the law of tangents, (c) Apply the law to two the legs a and b and the opposite acute angles of tangents
a right triangle and obtain tan ^ (A 3.
B)
""
=
-
a
+b
Express the area of a triangle: (a) In terms of two sides and the (b) In terms of the three sides, (c) Interpret geometri-
included angle,
cally the quantities
4.
Show how
s- a,s - b, s - c, and k =
to solve a triangle
when
the three sides are given,
Without logarithms, (b) With logarithms, (c) Write down the formulas needed in (a) and (b). (d) Prove the half angle (a)
formulas in terms of the sides. Discuss the solution of each of the four cases of oblique triangles giving in each case the formulas necessary for the logarithmic 5.
solution 6.
for checking the answers.
Review the method
triangles 7.
and a formula
by
division into right triangles.
Given the three a
=
Compute one tities:
of solving each of the four cases of oblique
sides of a triangle as follows:
301.9,
or
more
b
=
673.1,
c
=
422.8,
of each of the following sets of related quan-
PLANE TRIGONOMETRY
190 (a)
Angles,
(b)
Altitudes,
A = i8i2. 4 A = 294.5,
(c)
Medians,
(d) (e)
C=
50.8',
25$^
^=210.3.
0^=541.4,
w&= 147.3,
*,.= 476.9.
Angle bisectors, 64=512.8, Area and radii of
65=132.4,
bc=4o6.2.
and
cumscribed
cir-
T =44,458,
^=483. i.
^=1723, by the graphic method.
rc =i6i.o.
Radii of escribed
Check
.
= 112.0,
ra
results
r
=63.61
circles,
circles, (g)
B =135
^=132.1,
inscribed
(/)
',
[CHAP.IX
(a) Give a general definition of an angle of any magnitude. Define a negative angle, (c) Give a general definition of the
8.
(b)
complement and supplement
of
an
angle,
(d)
What
is
meant by
the principal value of a set of coterminal angles, (e) Give three positive and three negative angles coterminal with 30. 9.
(a)
Explain each system of angular measures and define the (b) State some of the advantages of each system.
unit in each, (c)
State the relation between radians and degrees,
the following relations in radian measure, sin (180 = cot 6, cos 180 = tan (90 i. e)
-
(d) 0)
Express
=
sin 0,
-
10. Prove the following relations, where a denotes the side of any regular polygon, p the perimeter, n the number of sides, r the radius of the inscribed circle, R the radius of the circumscribed circle, and the area of the polygon.
A
(a)
a
(b)
p
(c)
= 2 R sin- =
2 r tan
n
= 2 nR
sin
-
=
2
n
n
nr tan -
n
A= w^sin^cos^ n
n
!1
2
=
nr * tan I.
n
CHAPTER X FUNCTIONS OF ANY ANGLE IN this chapter a right angle will be denoted by R, rather than by 90 or ir/2. The advantage of this notation is that it frees our results from any particular system of measure, that is, our results hold equally well whether the angles are expressed in degrees, grades or radians. Whenever R is expressed in a given unit it is understood that the other angles are expressed in the same unit. Thus, if we write is expressed in degrees, but if sin (90 6) it is understood that we write sin (IT/ 2 0), 6 is to be expressed in radians, while the expression sin (R measured.
0)
does not specify the unit in which the angles are
92. Definition of the Trigonometric Functions of Any Angle. functions of any angle, positive or negative, are defined in
The
same way as were the functions of an angle less than Let 6 be any angle. Take the vertex O of the angle for an
exactly the 1
80.
Fig. 126.
Fig. 127.
Fig. 128.
Fig. 129.
and the initial side of the angle for the positive direction of Let P represent any point (not the origin) on the terminal side of the angle. Let x and y denote the rectangular coordinates of P, and r its distance from the origin. Then whether 8 is positive or negative, and whether P falls in the first, second, third or fourth quadrant, we have in every case, origin
the #-axis.
csc6
sine
PLANE TRIGONOMETRY abscissa
cose= 2
[CHAP,
--
sec e
r
x
cos 6
cote-
-
tan 8
a?
Signs of the Functions in Each of the Quadrants.
93.
Since the sine
(a)
the ratio of the ordinate of
is
P
to its distance
from 0, and the distance is always positive, the sign of the sine is in the first and the same as the sign of the ordinate, which is in the third and fourth quadrants. second,
+
(b)
Since the cosine
the ratio of the abscissa of
is
P
to its distance
from 0, and the distance is always positive, the sign of the cosine is in the first and the same as the sign of the abscissa, which is
+
in the second
fourth, (c)
and third quadrants.
Since the tangent
is
P
the ratio of the ordinate of
to the
+
when the ordinate and abscissa of P, the sign of the tangent is abscissa have like signs, that is, in the first and third quadrants, and
when they have
unlike signs, that
is,
in the second
and fourth
quadrants. (d)
Any number and
of the cosecant, secant
have like signs, hence the signs and cotangent are the same as the signs of
its reciprocal
the sine, cosine and tangent respectively. The student must make himself perfectly familiar with the signs The following figure will of the functions in the various quadrants.
prove an aid to his memory. tan
If the ter94. Periodicity of the Trigonometric Functions. minal side of an angle is revolved in the plane of the angle through 4 R, or any number of times 4 R, it will return to the position from
which
it
started,
and
this is true
whether the revolution
is
in the
It foldirection. positive (counterclockwise) or negative (clockwise) remain of lows that the trigonometric functions unchanged any angle
FUNCTIONS OF ANY ANGLE
95)
when
the angle
of times 4 R.
is
increased or diminished
That
4^) =
is
4 nR) = sin
6,
cos (9
4 nS) =
cos
6,
tan (6
4 nM) =
tan
0, etc.,
sin
sin
375
=
sin
=
sin
=
sn
-
(-
15
2
~ sin
TT
360) = siri
]=
1
6
W
+2
-
15,
sin
345
,
4
/
|
\
sin
360)
+ 2
(
\4
4
(r)
integer.
sin (375
(- 15) = sin
similarly for
In general,
sin (6
any positive or negative
Thus,
and
sin0,
similarly for each of the other functions.
where n
by any number
or
is,
sin(0
and
R
by 4
193
)=
TT
sin
^, 6
/
any other f unction.
be observed that by means of formula (i) the function of any negative angle can be replaced by the same function of some It should
positive angle.
Since the trigonometric functions remain unchanged when the is increased or diminished by 4^, they are called periodic
angle
It will be shown presently that the tangent and cotangent have the smaller period 2 R. There are other periodic functions besides the trigonometric functions.
functions with the period 4 R.
95.
Changes Changes from
in the Values of the Functions while the Angle to 4 R. Let a point P start from a position A and
move in the positive direction along the circumference of a circle whose radius equals unity.
OA
Join
P
to the center
of the circle
y represent the coordinates of
_ B
'
positions with reference to
and
let
x and
in its various
as origin
and
OA
as
the positive #-axis. Let us consider the changes in the values of the various functions of the
angle A OP circumference of the circle. Fig. 131.
O
P
=
0,
as the point
P
moves along the
PLANE TRIGONOMETRY
194
By
=
definition sin
taken equal to
by the
^-> cos 6
hence sin 6
i,
and tan
abscissa #,
by
P
While
First quadrant.
is
= -~>
tan 6
[CHAP,
=
2
But
-
OP
was
represented by the ordinate y, cos 6 the ratio of the two.
moves from
A
to
J3,
that
is,
while 6 changes from o to R y is positive and changes from o to 6 is positive and changes from o to f
hence
sin
x hence
hence
is
positive
cos 6
is
positive
y/x
is
positive
is
positive
tan
Second quadrant.
While
P
while 6
and changes from and changes from
i,
i
;
i
to o,
i
to o;
and changes from o to , and changes from o to oo
moves from
changes from
B to A that R to 2 R, f
is
,
and changes from and changes from
i
.
to o,
y
is
hence
sin 6
is
positive
x
is
negative and changes from
o to
i,
hence
cos 6
is
negative and changes from
o to
i
y/x
is
oo to
o,
tan B
is
negative and changes from negative and changes from
to
o.
hence
positive
While
Third quadrant.
P
moves from A'
while B changes from v
hence
hence
hence
is
2
to
R
B
1
',
to 3
i
x
is is
y/x
is
tan B
is
Fourth quadrant.
oo
that
;
is,
negative and changes from
o to
i,
and changes from negative and changes from negative and changes from
o to
i
P moves from B
while B changes from 3 is
r
to
o,
i
to
o;
to A, that
R
;
i
positive and changes from o to positive and changes from o to
While y
to o;
7^,
sin B is negative
cos B
oo
,
oo
.
is,
to 4 R,
negative and changes from
and changes from and changes from o and changes from o
i
to
o,
i
to
o;
hence
sin B is negative
x
is
positive
hence
cos B
is
positive
y/x
is
negative and changes from
ooto
o,
tan B
is
negative and changes from
oo to
o.
hence
x
to
i,
to
i
;
FUNCTIONS OF ANY ANGLE
97]
and
secant
Cosecant,
Since
cotangent.
195
these
functions
the
are
reciprocals of the sine, cosine and tangent respectively, their variations can be immediately written down from the variations of the
Remember,
latter.
That the reciprocal of a number i, and vice versa.
(a)
than
less
i is
some number
greater than (b)
That the
(c)
That
reciprocal of o
reciprocals
have
is oo
,
and
vice versa.
like signs.
Changes in the Value of the Tangent of an Angle as the to 4 R. Some students find it difficult to Angle Changes From 96.
follow the changes in the tangent from
The
y/x when x and y both change. cussion
is
free
from
the ratio
following dis-
this difficulty.
P
Let (Fig. 132) move as in Fig. 131, but instead let us consider the of the coordinates of the point coordinates of the point T or T' in which pro-
P
OP
duced meets one of the tangents to the A and A'. While
First quadrant.
o to
GO
,
therefore tan
Second quadrant.
from
+ 00
to
o.
from to
+
o to
to
-
1
.
OA
While
6
i
changes from
-
A'T' -
,
therefore tan 6
2
R
to 2 R,
A'T' -~
R
to 3 R, f
f
\JA
A'T' changes
changes from
= A T7- = A T f
oo
==
While 6 changes from
A'T
oo
changes
f
changes from
+o
I
oo.
Fourth quadrant.
from
-
at
changes from o to R, AT changes from A T = changes from o to oo
A T*
+ o, therefore tan 6 =
to
Third quadrant.
-
=
6
circle
-
oo to
-
While
o, therefore
changes from 3 tan 6
=
A T*
=
\JA
R A
to 4 R,
AT
changes
T
1
changes from
oo
I
o.
97.
Summary
of Results.
The
results
of
the two preceding
articles are brought together in the following table:
PLANE TRIGONOMETRY
196
The student should (6)
than
observe,
Every sine and cosine has some value between Every secant and cosecant has some value
(a)
+
i,
(c)
A
(d)
The
x
[CHAP,
or less than
+
i
and
either
i.
greater
i.
tangent or cotangent may have any value whatever. functions change only at the points between the quadrants, that is, when the angle has one of the values R,2R,3R,4 R, etc, and then only when the value of the function is either o or oo .
To
a given value of a function correspond in general two different angles between o and 4 R. To a positive sine correspond two (e)
angles, one in the first the other in the second quadrant; to a negative sine correspond two angles, one in the third the other in the fourth quadrant. To a negative tangent correspond two angles, one in the second the other in the fourth quadrant, etc.
98. Fundamental Relations. All the fundamental relations between the functions of an acute angle (Article 12) hold true when the angle is unrestricted in magnitude. The argument is an exact It follows that all trigonometric repetition of that used in Article 56. identities which have been proven for the case when the angle does not exceed a right angle, hold universally, that is, whatever be the magnitude of the angle provided that radical expressions such as
Vi
cos 2
0,
Vi +
tan2
be given the proper which 6 lies.
0, etc.,
depending on the quadrant
in
99. Representation of Trigonometric
sign,
Functions
+
or
,
by Lines. by lines
Until recent times the trigonometric functions were denned connected with a circle as follows:
Let a
be any angle, AP the arc which this angle intercepts on drawn with O as a center and any length OA as a radius.
A OP
circle
FUNCTIONS OF ANY ANGLE
991
Draw
IQ7
OA' perpendicular to OA, the initial side of the tangents to the circle at A and A', and produce OP to
the radius
Draw
angle.
perpendiculars
PF
T
T and From P draw the respectively. and PF' to OA and OA' (produced if necessary)
intersect these tangents in
respectively. A'
Fig. 133.
The former
FP = = AT = A'T' = OT = OT = FA = F'A' = F'P
r
Fig. 134.
definitions
sine of arc sine of
were then as follows:
A P.
complementary arc A'P
tangent of arc AP. tangent of complementary arc secant of arc
=
cosine of arc
A'P =
A P.
cotangent of arc
A P.
AP.
secant of complementary arc A P.
A'P =
cosecant of arc
AP.
versine of arc
versine of complementary arc
A'P =
co versine of arc
A P.
The definitions just given apply to any arc, provided the conventions regarding the algebraic signs of the various lines be carefully observed. These conventions are, as already stated in Article 53, with the additional one that the distances OT, OT' are positive
if
they pass through the extremity of the arc in question, that is, if the '; and negative if they do not, that point P lies between O and T or
T
is, if
the point O lies between P
and
T or T.
Thus
for
an arc
AP
in
the second quadrant, Fig. 135,
FP, the OF, the
sine, is positive;
cosine, is negative;
the tangent, is negative; A'T', the cotangent, is negative;
AT,
OT, the
secant,
is
OT', the cosecant,
negative;
Fig. 135.
is positive.
According to the old definitions, the length of each of the lines which defines the function depends on the length of the arc; and
PLANE TRIGONOMETRY
1 98
on the
since the length of the arc depends
[CHAP,
radius, it
x
was necessary to
If, however, all lengths specify the length of the radius employed. are expressed in terms of the radius as unit, the old definitions agree
with the modern
definitions.
tan arc
=
Now
OA
AT- =
and
OA
so that, line
if
AT
we
Thus
AP = AT
in Fig. 133, (old definition).
the measure of the angle which the arc subtends,
0,
the measure of
AT using OA
as the unit of measure,
AP
substitute for the actual lengths of the arc and the measures in terms of the radius, the old definition
their
becomes tan
= AT OA
Similarly, the old
which
is
the
and the new
modern
definitions of each of the other
may be shown
definitions, are
still
useful in
many
ways.
By
ation of the functions in the various quadrants cos2 9 = and the fundamental relations sin2
+
cot 2 6
definition.
to agree. definitions of the functions as lines, while
functions
The
,
+
i
=
esc2 6
plain the origin of
no longer used as means the varimost readily traced,
their is i,
tan2 6
become manifest at sight. Above the names of the functions.*
+ = i
all,
sec2
6,
they ex-
EXERCISE 46
Make
1.
out a table giving each of the functions of 45, 135,
225, 315.
Make
2.
out a table giving each of the functions of 30, 150, 210,
330. *
In the light of the historic definitions the origin of the terms tangent and The origin of the terms cosine, cotangent and cosecant has is obvious.
secant
already been explained. The origin of the term sine is probably as follows. The " " Latin word from which the word sine is derived is sintis, meaning bay or " ''bosom." The Arabic word was dschiba, meaning half the chord of double an
Owing to the practice of the Arabs to omit the vowels in writing, dschiba " " " was confused with dschaib meaning bay or bosom," and it was this word Roman translators which the dschaib properly rendered sinus. The word arc " a bow." The versed sine was formerly comes from the Latin arcus, meaning called sagitta, an arrow, because it occupied the position of the arrow in a bow. The modern conception of the functions as ratios dates from the second half
arc."
of the seventeenth century.
The
radius of the circle were used
by many
old definitions modified
by
using unity for the
writers less than twenty-five years ago,
FUNCTIONS OF ANY ANGLE
loi]
Given
3.
=
sin
be the values
What
f find the values of 6 less than 4 R. ,
than 4
less
199
R
which
6
may have
if
=
cos 6
f
will
?
Ans. 30, 150; 60, 300.
Find the values
4.
tan 405, sin
(- 45),
of the following functions: sin
cos
390, cos 765,
(- 30), tan (- 135). Ans.
Trace the changes of the cosine through each
5.
rants from the changes of the sine
cos0
by means
= Vi
those of the tangent
quad-
sin2 0.
Trace the changes of the secant in the
6.
of the four
of the relation
by means of the relation sec
6
first
quadrant from
= Vi
+
tan2
6.
7. Construct the lines representing the various functions of an arc in the third quadrant; of an arc in the fourth quadrant.
What sign must be attached to the
8.
when
6
is
What
9.
sign
must be attached to the sec0
when
is
radical in sin 6
an angle in the second quadrant
an angle
= Vi
in the third
+
?
cos2 0, ?
radical in
tan2
quadrant
= Vi
In the third quadrant
?
0,
In the fourth quadrant
?
Trigonometric Functions to the First Quadrant. In Articles 57 and 58 it was shown how to express any function of an angle in the second quadrant in terms of functions It remains to be shown how functions of of an angle less than R. fourth third and in the quadrants may be expressed in terms angles R. than less When this has been done, the of of functions angles value of any function of any angle can be found from the tables 100. Reduction
of
which contain the functions of angles from o to 90.
in the first quadrant, that
is,
101. Reductions from the Third Quadrant. Any angle 3 in and 3 R, hence every such angle lies between 2
R
the third quadrant
may
be expressed by either 3
where 9 anc}
is
= 2 R + 0,
each
less th?tn
or
R,
3
= 3R-
,
PLANE TRIGONOMETRY
200 (a)
=
3
2
R+
Let
0.
3
=
angle
AOPS
quadrant, and let 6 = angle Produce PsO to P, making OP = OP3 = the
third
=
6. angle AOP coordinates of
Let
P
;y3 )
any angle
represent
in
A'OP* then
r,
denote the
Draw PF
respectively.
PF
and 1
OP/* yz
P&
and
(#3 ,
(x, y),
x
[CHAP,
3 3 perpendicular to A A', then the triangles and OP3 F3 are geometrically equal and
= ~
y, x*
=
sin 9 3 = sin
)=^r =
04)
we
^
=~
r
^
=
sinO,
r
= -- = -
= -=
cos 63=
From
Fig. 136.
Hence we have,
#.
obtain
csc%
=
esc (2
+
sec03
=
sec (2
^+
0)
=
fl)
- csc0,
.
_
= COS0
COS
tan
Observing that in both
(2#
+ 0)
and
(.4)
Sec 0,
tan0
(A') the signs
the signs of the functions in the third quadrant,
on the right are
we have
the simple
rule:
Any function of
0,
of (2
R+
0) is
equal
to
plus or minus the same function
the sign being that of the function in the third quadrant.
EXAMPLE,
sin 204
cos 204
tan 204
Put R Hence
<
0,
= = =
<|>)
R
<|>)
83= cos
tan 6 3 =
(3
cos (180
tan (180
then sin
sin 9 3 = sin (3 JR
cos
sin (180
tan(3JR-
<)>)
=
+ 24) = + 24) = + 24) =
cos 0, cos
R + 0) = cos (2 R + 0) = tan (2^ + 0) =
sin (2
=
sin 24
cos 24
tan 24
= =
= cos0 =
tan0=
0.9135,
==
sin <, tan
sin0
0.4067,
0.4453.
=
cot
cos
<|>,
sin
4>,
cot<|>,
(E)
FUNCTIONS OF ANY ANGLE
IO2J
201
and from (B) sin
3
=
cot 03
=
sec
sec
3A
-
sec
cot
(
)
-CSC0,
cosU/c
(3^
<
W
tan.
tan
(3^
<)
cot
In (J5) and (') the signs on the right are again the signs in the third quadrant of the functions on the left, hence the second rule: <) is equal to plus or minus the correspondAny function of (3 R in the third ing cofunction of <, the sign being that of the functions quadrant.
EXAMPLE,
sin 204
cos 204
tan 204
= 66) = 66) =
= sin (270 = cos (270 = tan (270
cos 66
66)
sin 66
cot 66
= = =
0.4067, 0.9135, 0.4453.
from the Fourth Quadrant. Any angle 0* in the fourth quadrant lies between 3^ and 4^, hence every such 102. Reductions
angle can be expressed Ot
where 6 and (a) 04
=
either
= 4R-0,
or
04
= 3^+0,
are angles less than R.
<
4
by
-R
quadrant, and
-
Let 4 = angle AOPi be any angle in the fourth us put angle P^OA = 0. Draw
0.
let
=
OP = OP*
== AOP 0. From P r, making angle Then the and P\ draw perpendiculars to OA. are and OPF OPtF geometrically equal, triangles
and and
if
P
yO denote the coordinates of P =$ = y, %* respectively, we have y 4
(#, y), (#4, 4
Fig. 137-
Consequently sin 8 4 =
cos
=
sin(4-B-
84= cos (4* -e)=
tan 8 4 = tan(4JB~8) =
~^=
-sin 8,
r 4
=
cos 6,
= -2=
-tan8.
=
X
PLANE TRIGONOMETRY
202
From
[CHAP,
x
(A) follows
CSC04
=
CSC (4.K
0)
sec0 4
=
sec
(4^
0)
(4^
- 0)
=
C0t0 4
COt
In (4) and
=
=
-r
7-^cos (4 /c
0)
R
0)
tan (4
=
=
-^
sec0,
cos
\
(A')
tan a
the signs on the right are the signs of the functions
(-4')
in the fourth quadrant, hence 0) is equal to plus or minus the same function Any function of (4 R 0/0, the sign being that of the function in the fourth quadrant.
EXAMPLE,
2I
sin
=
sin
f]
cos
(2*
6
=
sinf o
6/
\
=
cos
2*
[
6
-
=
6/
cosfo
o/
o
-}
\
=
,
tan 6 (b)
Put
\
4
=
-R
0,
=
then sin
cos
0,
cos
<
=
sin 0, cot
=
tan
0,
hence
= cos 8 4 = tan 6 4 = sin 6 4
and from csc04
+ )= sin (4^-0) =- sin0 = ~ cos<|), cos0 = sin, cos (3 R + <|>)= cos (4^-0)= tan (3 R + $)= tan (4^-0) =- tan = - cot sin (3
-
<|>,
(-B)
=
(B)
esc (3
R + <)=
.
f
sin (3
,. Rp + 9) ,
=
-
L
cos
7
9
=~
sec^,
(-#') the signs on the right are the signs in the fourth functions on the left, hence the of quadrant 0) is equal to plus or minus the correspondAny function of (3
In (B) and
R+
ing cofunction of
quadrant,
,
the sign being that of the function in the fourth
FUNCTIONS OF ANY ANGLE
103]
EXAMPLE,
=
sin
sin
6
&+*
\2
=-
3/
cos 3
203
--
i,
cos 2
tan
HZ ==
tan /3JT
6
V 2
3
+ = _ cot ? = - i x/J. !T\
3/
3
103. Functions of Negative Angles. 6 = angle -40P' be any negative angle. Let Construct the angle AOP = 0. Take OP = OP' = r, and let (#, y), (#', y') denote the coordinates of ]A f the points P and P respectively, then #' = x, r
y Fig. 138.
=
sin
y,
and we have
(-e)--z-r
z
sine,
r
cos
(A)
8,
and from (A) esc
The
(-
=
0)
sec(-0)
=
cot
=
signs
(-
on the
6)
(A')
right are the signs of the functions in the fourth
quadrant, hence 0) is equal to plus or minus the same function of of ( sign being that of the function in the fourth quadrant.
Any function 0, the
EXAMPLE,
sin
(
13
25')
cos
(-
13
25')
tan
(
13
25')
= =
sin 13
cos 13
tan 13
= = 25'
25'
25'
* -
0.2320,
0.9727, 0.2385,
PLANE TRIGONOMETRY
204 104. Table
[CHAP,
and General
Principal Reduction Formulas
of
x
Rules.
The principal results of the last three articles, together with the corresponding results for the first and second quadrants (Articles I0 > S7> 5&)> are brought together in the following table for purposes of comparison
and
reference.
Quadrant I
~
sin(JR
<)
=
cos<,
cos (R -<(>)
-
tan (R
<)
'
COt .
Quadrant II
cos (2
R
=
0)
-
tan
sin0,
sin
cos
cos
0,
+ (1? +
(R
cos<,
<
-
sin<,
tan0.
Quadrant III
= =
R + 6) cos (2 R + 0) sin (2
tan
(2^ + 0)=
sin 0,
sin (3 -R
cos
cos (3 J?
0,
tan (3
tan0.
= <) =
cos
$)
<
sin
.#$)=
cot
Quadrant IV sin (4^ cos (4 tan (4 R
0) 0) 0)
= = =
sin (- 0) = = cos ( 0) tan (- 0) = -
sin0,
sn
cos 0,
cos (3
^ + ^) = tan (3 R + $) = -
tan0.
sin
cot
We observe that each equation on the left involves a pair of named
same-
R
are even numbers, 2 or 4. On the right each equation involves a pair of conamed functions and are the odd numbers i and 3. In either case the coefficients of
functions and the coefficients of
R
the signs are the signs of the functions in the respective quadrants. By increasing the angles on the left by multiples of 4 R (which will
not change the value of the functions), we obtain formulas for
the functions of
6-0,
6
+
0,
&R-0, 85 + 0,
and by increasing the angles on the obtain
.,
right
.,
by
.,
2nR6,
multiples of 4 R,
we
FUNCTIONS OF ANY ANGLE
104]
205 two form-
All the foregoing results are therefore included in the ulas,
Any function (2 nR 0) = Any function (2n+ iR
same function
=
cf>)
6,
cofunction <,
the sign being the sign of the function on the left in the quadrant in which the angle falls when 6 or are acute angles.* <
EXERCISE 47 Express in terms of same-named functions of angles
i.
sin
146, cos 235, tan 317,
-
Ans. sin 34,
cos 55,
-
tan 43,
-
,
sin-,
Express in terms of cof unctions of angles
2.
tan 95, sin 272, cos 115
10',
cot
cos
5,
sin 25
2,
than R,
sec
,
tan i^-
- cos, -
-, cot $%R.
sin
5
tan
cos-,
esc-,
10',
tan--
than R,
less
o
Ans.
less
484 404 662
sin^,
cos 2
Express in terms of functions of positive angles less than 45 15', cos 143 15', tan 243 10' 15", sec 284 30', cot 127.
3.
sin 143
Ans. sin 36
-
45',
cos 36 45', cot 26 49' 45", esc 14 30',
-
tan 37
4. Use natural functions table to find
sin
in
30', cos 253
i2/ tan 134, sin 317
Ans. 009304,
Find
5.
cos
sin
(- 150),
(- 100).
6.
If sin 6
7.
If tan
value of
0.2890,
_
(- 5445)>
_
If cos 6
= 0.5831, what values less = 4.3897, and sin 8 is known
9.
If sin 6
= =
sin
147, show that one value
cos 5
0,
2
(
7~\
I?
6.
8.
m
0.1320.
_ i v^ _ Q I?36 than 4 R may 6 have ?
i Q 8ogo> ?
cos 9735'-
0.6788,
1.0355,
cos 3564, tan
Ans
15',
to be positive, find the
Ans.
=
of 6 is
303.
50'.
15, and another
show that one value
of 6
0.4561; find tan <.
Ans. tan
is
102
7S10. *
Given
sin
<
=
<
=
0.5125.
These rules hold not only for the sine, cosine and tangent, but for the coseThe latter have been omitted from the sumcant, secant and cotangent as well. mary on account of their lesser importance.
PLANE TRIGONOMETRY
206
[CHAP,
x
105. Generalization of the Preceding Reduction Formulas. In the proof for the formulas for the functions of a negative angle was not restricted in magnitude. These formulas (Article 103), therefore hold true for any angle, but in the formulas for the functions
+
of 2 tf 102), 6
0,
-
(Article 101), and of 4 R 6, 3 R (Article were assumed to be angles between o and R. This not necessary and will now be removed. In other
sR-
and
<
restriction is
+
words, we will now show that the formulas of Articles 101 and 102 hold for any value of and <. The complete list of reduction formulas includes
R
we
the formulas for the functions of
shall
0, of the angles.
show that these formulas
2
R
and
also hold for
of
any value
For the sake of brevity the proofs will be confined to the first one of the formulas in each set. The proofs of the other formulas are left as exercises for the student.
Functions of
(a)
+
(2
The
0).
angles (2
R+
ff)
and
6 differ
R no
matter how large 6 is and whether is positive or negative. Consequently the points P$ and P (Fig. 136) must always lie on a straight line through the origin. The coordinates of P3 and P will
by
2
therefore be numerically equal but opposite in sign. 0, positive or negative,
Hence
for
any
value of
sin(2* This establishes the value of (b)
+ 0) = 2- = .=^ = 3
-sin0.
(i)
of the relations (A), Article 101, for every
first
6.
Functions of (2
R
0).
If in (i)
we put
for
0,
0,
R-8) = - sin(-0). = sin for every sin ( 0)
we obtain
sin (2
But by (A)
y
Article 103,
value of
0,
therefore
sin(2# This establishes the
first of
0)
=
sin0.
(2)
the relations in Article 57 for every value
of0.
The functions of an angle are not Functions of (4^ 0). is increased or diminished by 4 R (Article 94), the if changed angle (c)
hence sin
(
4R
0)
=
sin
(
0), for
every value of
0.
FUNCTIONS OF ANY ANGLE
xosl
But
sin sin (4
therefore
This establishes the
= =
0)
(
R
0)
first of
sin0, sin
by
207
Article 103,
0.
(3)
the relations (A), Article 102, for every
value of the angle.
Let be any angle and let 0' be the (d) Functions of (R ). smallest positive angle coterminal with 0. Then 0' can be written in one of the forms 0, 2
R
#
0, 2
according as
'
His
0,
jR
4
0,
an angle
where
in the
is first,
positive
and
less
than R,
second, third or fourth
quadrant.
= 2 R - 0, then sin (R - 0) = - cos0 = cos0'. sin CR - 0') = sin (- R + 0) = + 0,then If0' = 2 sin (R - 00 = sin (- R - 0) = - sin (R + 0) = - cos0 = cos0'. If 0' = 4#-0, then sin(tf- 00 = sin(- 3# + 0) = -sin (3^-0) = cos0 = costf/.
If 0'
We
see then that, whether
quadrant,
'
is
in the
sinCR- 00 =
and hence sin is
(R
=
0)
first,
second, third or fourth
cos0', cos
(4)
established for every value of 0. (e)
Functions of (#
generality
+ 0),
= R
(3
R-
0), (3
+ 0).
Then by applying has been already established, we find
angle and put
sin
(R
6.
+ 0) =
R 0) sin (3 R + 0) sin (3
= =
sin
(2^
0)=
R + 0) sin (4 R 0) sin (2
= =
sin
sin sin
Let
be any
the formulas whose
= = =
cos
(5)
cos 0,
(6)
cos 0.
(7)
EXERCISE 48 Repeat the argument of the preceding and respectively,
values of 1.
tan
(2^
2.
tan
(R
3.
tan
(3
R
= 0) = = 0)
tan0.
0)
=F
cot0.
cot0.
article to
show that
for all
PLANE TRIGONOMETRY
208
Show
4.
sin (6
cos (0
tan
(0
5.
-
~ = =-
(b) (c)
2
sin0, cos
2
R) R)
2
)
cot0, tan
(0
(0
When When When
is
= =
sin 6, sin (6
3 R)
cos0, cos
3 R)
=
tan0, tan
is is
$)
=
sin
(0
(0
3 #)
= = =
cos0. sin0.
cot0.
,
an angle in the second quadrant. an angle in the third quadrant. an angle in the fourth quadrant.
General proofs of the reduction formulas for the angles
(2/2-0),
may be the
0,
cos0, sin (6
Prove geometrically that cos (R (a)
6.
R) R) #)
that for every value of
[CHAP.X
(2#
+ 0)
obtained from considerations of symmetry.
same
ff-axis,
(4#-0),
origin
and taking the
then no matter
how
large
initial
line
Referred to
to coincide with the
and whether positive or negative*
0,
R
(a)
The terminal
(b)
situated with respect to the y-axis. The terminal sides of and (4^ 0) are symmetrically situated with respect to the #-axis.
(c)
The
sides of
terminal sides of
and
(2
and (2^
0)
+ 0)
are symmetrically
are symmetrically
situated with respect to the origin. It follows that the ically equal (a) (b) (c)
and that
same functions
of
each pair of angles are numer-
in
The sines have like and the cosines opposite signs. The sines have opposite and the cosines like signs. The sines and cosines each have opposite signs.
In each case the sign of the tangent
may be
relation
tan0
=
^. COS0
determined from the
CHAPTER XI TWO OR MORE ANGLES
FUNCTIONS OF 106. Addition
Theorems
for the Sine
and Cosine.
denote the sides of any triangle and A, B, C the angles opposite these sides, we have from the law of sines First Proof.
If a, b, c
(Article 62, (3)),
= D sin A, = D sin B, c = D sin C,
a b
and by the projection theorem
A
(Article 63,
Fig I39 '
'
(4)),
=
c
a cos
B+
b cos
A
.
and
Substituting in this equation the above values for a, b dividing out the constant factor D, we get sin
Now C =
1 80
cos2 (A
= = = = =
i
B+
A sin J5. C = sin (^
cos
2
i
(i
A
2
cos
cos
B (cos A cos 5
2
5
A)
cos2 ^4 cos2
Taking the square root cos (A
it
cos
+ B) = - sin (A + B) (sin A cos jB + cos A sin J5) sin
i
A
and
+ 5), whence + J5), therefore sin = A ^4 ^ sin cos ^. sin cos + (i) (A + B)
2
i
sin
(A
sin
Again
C=
c,
+
2 sin 2
cos
A
2
U
sin J5 cos
A
cos B
2
2(...)
(i
2 sin ^4 sin -B cos ^4 cos J5
sin
of
A
sin
both
B) =
cos
5)
cos2 sin
+ sin A 2
A
4)
sin2
sin2
#
B
sin 2 .#
2 .
sides,
A cos B
sin ^4 sin B.
(2)
Since the last expression was obtained by extracting a square root may seem that the double sign, , should have been put before
the right-hand
member
of (2),
but on putting 209
B=
o,
the minus
PLANE TRIGONOMETRY
210
[CHAP, xi
would yield the result cos A = cos A, which shows that the minus sign cannot be used. Formulas (i) and (2) embody the so-called addition theorems for the sine and cosine respectively. In words, The sine of the sum of two angles is equal to the sine of the first angle times the cosine of the second plus the cosine of the
first
angle times the
sine of the second.
The cosine of the sum of two angles is equal to the product of the cosines of the separate angles diminished by the product of their sines. It is plain that by means of these theorems the sine and cosine of the sum of two angles may be found of the separate angles are known.
EXAMPLE sine
Given the functions of 45 75.
i.
and cosine
Solution,
and
of
30,
to find the
of
=
sin
(45+ 30) = sin 45
cos
75=
cos
(45+ 30) = cos 45
cos 30
sin 7 5
cos
107.
the sines and cosines of each
if
30+ cos 45
A and B
30
sin45sin3o
Generalization of the Addition Theorems.
going demonstration,
sin
In the
are angles of a triangle, their
fore-
sum
is
therefore necessarily less than 2 R. This restriction may be removed, in other words, the addition theorems hold for angles of any mag-
nitude and whether positive or negative.
To prove their
sum
this let
is less
sin^i sin (Ai
cos 04 1
+ B)
+ B)
= =
= = = = = =
A and B
* If the student finds
two cases Ai
= A
be two angles each
less
than R, so that
A = A /?, then = cos A, cos A T sin A, (Art. 104, and R + B) = cos (A + B) sin (A sin A sin B) (cos A cos B cos A cos B =F sin A sin B sin A cos B + cos A sin B. R + B) = T sin (A + B) cos (A T (sin A cos B + cos A sin B) T sin A cos B ^ cos A sin B * sin A sin B cos A cos B than
+ R,
2
R, and
let
i
i
i
\
i
it difficult
A^
i
to follow the double signs, let
= A -
R, separately.
Ex. 48, 4)
(i)
(2)
him consider the
FUNCTIONS OF TWO OR MORE ANGLES
io8]
Equations
(i)
and
(2)
211
show that the addition theorems continue
the angle A is increased or diminished by R, and the same reasoning applies to the angle B. By a repetition of the process just employed it is clear that the theorems will continue to hold true to hold
if
replaced by A*, where A 2 erally that A may be replaced by
if
Ai
is
and B
by n and
But
108. Addition
A
An = A
nR,
Bm - B
mR,
theorems for
this proves the
some integer and
R=
Ai
AI
2
R, an d gen-
m being two arbitrary integers. values of the angles, for any
all
be put in the form some angle less than R.
positive or negative angle is
=
may
A
nR, where n
Theorems. Second Proof. The addition theorems
be proved without making use of the law of sines and the pro-
may
jection theorem.
XOM =
In Fig. 140, let angle B, then XON
On
PT
ON
and
MON =
angle A, and
=
angle (A+B). take any point P and from
P
draw
and
OM
f
PQ
perpendicular
to
OX
respectively.
From Q draw QR perpendicular to OX and F g J40 parallel to OX. The triangles QOR and QPS are similar (Why?), hence angle j
QS
QPS =
angled.
Now
OP
OPOP
OP
?Q QQ. + SL 01 OQ OP QP OP sin A cos B + cos A sin B. . '
. *
QQ _SQ QP OQ 'OP QP' OP = cos A cos B sin A sin B.
^OR
PLANE TRIGONOMETRY
212
In the figure we have taken
[CHAP, xi
+ B less than a right angle, but the
.4
proof just given will hold for any angles, provided proper attention be given to the algebraic signs of the lines which enter the figure.
Theorems
109. Subtraction
for the Sine
the addition theorems have been
as for positive angles,
and
replace
B
and Cosine.
Since
to hold for negative as well
The equations
B.
by
(i)
then become
(2), Article 106,
sin
we may
shown
B)
(A
cos 04
B)
= =
sin
cos
A cos ( A cos (
+ cos A sin
B)
sin
B)
A
sin
(
B)
(
J5),
from which
(A
B) =
cos (A
B) =
sin
A cos B -cos A sin B, cos A cos B + sin A sin B.
sin
(i) (2)
These formulas enable us to compute the sine and cosine of the difference of two angles if the sines and cosines of the separate angles are known.
EXAMPLE. Given the functions and cosine of 15.
and 30; to
of 45
find the sine
Solution. sin 15
cos 15
=
=
-
sin (45
cos (45
30)
=
30)
=
cos 30
sin 45
cos 45 cos 30
-
cos 45 sin 30
+ sin 45
sin
30
EXERCISE 49 1. 2.
sin (x 3.
Find the sine and cosine
=
Given
sin x
+ y)
and cos
Find
(x
$,
from the
of 15
cos x
=
f
,
sin
y
=
relation, -jfV,
cos
i5=6o y
=
45. find
\%\
+ y).
sin
90 and cos 90 from the relation 90
=
sin
=
30
o and cos o from the relation o
4.
Find
5.
Apply the addition and subtraction theorems
60
+ 30. 30.
to find the fol-
lowing: sin (90
sin (360
x),
-
cos (90
y), cos (45
+ x\ sin (180 + x), cos (270 sin (30 + y)> cos (60
- y),
x), y).
FUNCTIONS OF TWO OR MORE ANGLES Show
6.
213
that
+ B) + sin (A - E) = 2 sin4 cos, - sin (A - B) = 2 cos.4 sin, sin (A + B) = 2 cos cos, cos (4 + 5) + cos (A = ~ 2 sin 4 sin 5. cos (A cos (4 + sin
(A
,4
)
)
)
7.
By putting 5 = A in Problem 6 show that sin24 = 2 sin A cos 4, cos 2^ = 2 cos ,4 2
i
*
i
-
Prove the subtraction theorems geo= metrically by means of Fig. 141. XOQ
2sinM.
8.
angle A,
(Ato
B).
MON =
angle angle B, PT are perpendiculars QR, QP,
OX respectively.
OX, OM,
A
angle 9.
Angle
o
that the answers to Problem 17,
Exercise 38, may be put in the forms M __ a sin ft cos a sin (a - 0)
Show
,
a sin a sin
sin(a-j8)'
__
may be
__
a sin a sin a
Vsin
Show
=
M
Fig. 141.
that the answer to Problem 19, Exercise 38,
written
11.
SQP =
(Why?).
Show
10.
0V*
XON =
(a
- a) sin (a + a')
that cos ^4 cos B
and hence that tanff
tan 4 12.
Show
that
and hence that similarly
tan
_
J5
and
Show
sin
(4
+ -
)
B)
+ i) = sin nS cos + cos wB sin = sin w cos + cos w sin i, sin n+ = sin n sin cos w cos cos w+ 0,
sin (n
i
i
i
i
Hence, if the sine and cosine of may be readily computed. 13.
sin (A
i
are known, those of 2,
+ T) + sin (A - }TT) -i sin (A + i T) + cos U
that cos (A
TT)
o,_
V
2 (sin
3,
etc.,
PLANE TRIGONOMETRY
214
[CHAP, xi
Show that
14.
cos A sin (B
sin4
sin
Two
15.
cross at section,
+ cos B sin (C A) + cos C sin 04 B) - vl) + sinCsin (4 - B) C) + sin .B sin (C
C)
-
(B
straight roads
CM
and (XB
(Fig.
o, o.
142)
an angle a. From 0, their point of intera straight road is to be laid out to a
point P, which is p miles from the first road and g miles from the second. Required the angle 6
OP
which
will
1=
make with OA.
I42
whence
.
sin (a
q
tanfl
0)
/>
cos
+q
16. The area T of a triangle was computed from two sides, b and c, Afterwards it was found that an error and the included angle A had been made in measuring the angle A. Show that the corrected .
area
is
given
by the formula
T= 17.
Two
T (cos e +
sin
6
cot A).
p and q act on levers of lengths a ahd b respectively, which are inclined at an angle a at
parallel forces
the
common
forces
may be
What angle 6 must the the lever a in order that there
fulcrum O.
make with
equilibrium?
(Suggestion.
Equating moments about
we
have -
143-
ap
from which
sin 6
tan 6
=
bq sin (a
Show
=
19.
sin 2
(A
x, y, z
bq cos
.)
a
that
+ B) sin (A- B) cos (A + B) cos (4 - B) sin
0),
^ sina a/>
1 8.
-f-
2
cos
.4
.4
- sin J5 = - sin B = 2
cos 2
2
2
cos
B - cosM, 5 - sin 4. 2
being any angles, show that
sin (x
+ y + 2) =
cos (#
+ y + z) = cos # cos y cos z
sin
+
# cos y cos z sin y cos z cos # sin x sin ;y sin z. cos # sin y sin
2?
+ sin z cos # cos y cos y sin s sin x
FUNCTIONS OF TWO OR MORE ANGLES
no]
215
Show that
20.
+y
sin (x
x + y + z) *= + y + z) + 4 sin x sin y sin x + y + z) = y + z) + cos cos (# + y + 2). 4 cos x cos y cos z
+ sin (x
z)
y
+ z) + sin
(
sin (x
cos (x
+y
z)
By
21.
+ cos (x
z.
(
6, c
eliminating a,
from the equations
(Art. 63),
= acosB
+ 6 cos A, b cos C + c cos a 6 = c cos + cos C, A + cos J5 + cos C + 2 cos A cos c
,
.4
show that
2
2
cos2
cos
C=
i.
Solve this equation for cos C, and obtain
C= C=
cos
that
Remembering lower sign which
is
cos (A
cos A cos
180
(A
5
+
inadmissible (Why?),
+ B)
=
cos
A
cos
A
sin
sin 5.
B), and disregarding the
we
B
find
sin
A
sin B.
This constitutes another proof of the addition theorem
for
the
cosine. ~
110.
we
Tangent
of the
divide the sine of the
Sum sum
the tangent, thus (A 4- B}
= sm
(^
cos (A
On
and Difference of
of
Two
Angles.
If
two angles by the cosine we obtain
+ ^) = + B)
sin
A
cos
cos
-4
cos
B+ 5
cos sin
A A
sin
B
sin
dividing both the numerator and denominator of the rightcos A cos B, we have
hand member by
sin i.
f A
i
T>\
tan (A-rJt>)~
that
A cos B cos B
cos
^4
cos
^4
cos .#
cos
^4
cos
B
,
_
cos
A
sin
^
sin ^1
cos
^1
cos
#
cos
sin
A
sjin
cos ^4 cos
.
A
sin
B
cos J5
ff
sin
A
sin .g
B
cos
-4
cos .5
is
(i)
To obtain the tangent of the difference of two angles, B, thus only put in (i) for B, tan A ten (~ B tan (A B) = \, i- tan 4 tan (- E)
+
we need
PLANE TRIGONOMETRY
2l6 that
is,
[CHAP, xi
is,
Of course we might have deduced B) by cos (A by dividing sin (A
(2) just
as
we deduced
Double an Angle. If B and .tangent of the sum of two
111. Functions of for the sine, cosine
(i)
that
B).
=
A, the formulas become
angles
+ A) = sin A cos A + cos A sin A, cos (A + A) = cos A cos 4 sin A sin .4, tan A + tan ^ tanU + X)tan tan sin
(A
^4
i
that
-4
is,
= -4 =
sin 2 ^t
2 sin
cos 2
cos 2
By means easily
A cos ^4, A sin2 ^i =
(i)
2 sin2
i
A==
2 cos 2 -4
of these formulas the functions of twice
i,
(2)
an angle are
computed, provided the functions of the single angle are
known.
an Angle. It is frequently necessary to of half an angle in terms of the functions of functions the express the whole angle. This is most easily accomplished by means of (2), 112. Functions of Half
Article
we may
in.
Since these formulas hold for any value of the angle, replace A by J A, thus cos (2
%A) =
i
2 sin
A =
i
2 sin
cos
or
If
we
solve the
first of
2
2
1
A =
\
A
2 cos
2
\A
these equations for sin \ A,
i,
we obtain
2
and the second solved
for cos \
A
gives (3)
_
FUNCTIONS OF TWO OR MORE ANGLES
2]
Dividing
by
(2)
(3) gives
217
EXERCISE 50 1.
75
Given the tangents and 15.
Ans. tan 2.
Given tan
3.
Show
A =J
f
and 30, compute the tangents
of 45
75=
=f
tan
3
+
^3 -V 3 3
+
J5)
I
tan (4S^
,
f
Show
5.
tan (#
-f"v 4- z)
tan x
=
Express sin 4
6.
A
,
+ tan y + ten z
tan x tan y
'
C
iACOtB+I
.
cos 4
yl
,
tan x tan y tan
tan y tan z
tan 4
tan
g
tan x
in terms of the functions
^4
A.
Given the functions 60.
7.
of
30;
find the sine, cosine
_
_
Given the functions of 45; find the
8.
22^.
Ans.
sine, cosine
shi22j=\/2-\/2, cos22|=J\/2+\/2,
_
Given the functions
9.
of
E).
that
i
of
^3.
+ ^3
and tan (4
- ^) -
COt(A-B)=
of
-
Show that
4.
of 2
3
that
+ A) = i- tan
tan (45
15=
3
find tan (4
,
tan
,
of
15.
sinis= \/ 2 -\/3,
Ans. 10.
Express sin 3
A
of
30;
cos
_
15=^
in terms of sin
V + \/3i= 2
11.
tan
and tangent
22|=\/ 2- 1.
find the sine, cosine
and tangent
i5=2-
\/3-
A.
Ans. sin 3 (Suggestion.
tan
and tangent
A =
3 sin
A
8 4 sin A.
$A=2A+A.)
Express cos 3
A
in terms of cos
A.
Ans. cos 3
A
=4 cos A 3
3 cos A.
[CHAP, xi
PLANE TRIGONOMETRY
2l8 Express tan 3
12.
A
in terms of tan A.
A
3 tan
A
Ans. tan 3
tan3
2 3 tan
1
Find the sine and cosine of 18.
13.
=
Ans. sin 18
^
+
cos 18
->
4
=
(Suggestion. Let x cos 3 x. sin 2 x
= Now express
and
A
A
Show
sin 2
15.
sin 2
16.
2 -4
Use
sin %
cos 2
Fig. 144 to
A =
=
2 sin
cos
2
#
+ 3 z = 90,
x
2
=
90
-
3 a,
x,
stands for tan A,
if /
2t
A
Use
cos
that
2
x and cos 3 # each in terms of functions of sin2 #.) Then cos x = \/i
sin 2
solve for sin x.
14.
18, then
A
_
^4
i
-
2 t
prove that
cos
A
A
,
2
sin
A
Fig. 145 to prove that
A = Vj
cos 4),
(i
(Suggestion, sin \
A =
,
RS*
=
M + OS 2
2
2
OS cos ^
O
,
2 r
17.
18.
If A + B + C = 180, show that tan A + tan B + tan C = tan A
tan4 =
2tan
i-
^
tan 2 !
.
tan 5 tan C.
Solve this equation for tan
4, and
.4
identify your result with (4), Art. 112.
that the equation a tan x c sin duced to the form (a b) cos 2 x 19.
Show
+
+ 2
x
b cot
=
a
x
+
= 6.
c,
may be
re-
FUNCTIONS OF TWO OR MORE ANGLES
113]
20.
A flagpole
50
ft.
high stands on a tower 40
ft.
distance from the foot of the tower will the flagpole
At what high. and tower subAns. 120
tend equal angles? 21.
A
river b
tower
an angle
is
wide.
ft.
of
situated at a distance of a
\/ 362
distance of 1000
contains a town
The
of 30'.
ground.
will the river
subtend
?
30
At a
dock the
-4a
2
from the foot
ft.
of
dial
center of the dial
-4a6.
is
Find the diameter of the
the clock
193.5
&
(Two of
solutions?)
a tower which
subtends an angle level of the
above the
dial.
Ans. 9 23.
The
ft.
from the banks of a
ft.
At what height on the tower
Ans. i b \/3 22.
219
height h of an object
AB
ft.
was computed from the
dis-
from the foot of the object a point 9 AB subtended at this point. which and the angle had been made in It was found that a.n error tance d
of
measuring
0.
Show
that h must be corrected
by o
an amount
Hg,
H<5.
dsine
""* .
ft
cos(0+e)cos0 113.
we
Sums and Differences Transformed into Products. From
obtain
by
sin
04
sin
+ B) =
(A
B)
cos (A
+ B)
cos (A
B)
addition
A cos B + cos A sin B = sin A cos B cos A sin B = cos A cos B sin A sin B =* cos A cos B + sin A sin B sin
and subtraction
+ B) + sin (A B} = 2 sin cos B = 2 cos 4 sin B sin (A sin (4 + 5) JB) - J?) - 2 cos 4 cosB cos 04 4- 5) + cos (A - 2 sin 4 sin 5, cos (A + B) - cos (A 5) = sin
(A
^4
[CHAP, xi
PLANE TRIGONOMETRY
220 Let us now put
+ B = x and4 - B = y, A = J (* + y)> -B = 4 (* ~ y)i A
from which
so that the preceding formulas
sin
a>
+ sin y =
become
2 sin
(a?
+ y) cos | (o& -
sinx- sin# cos
x+
cos
-
cos y
=
2 cos
+
| (a?
I/)
cos
| (as
W
-
cos2/=-2sin|(o?+$/)sini(a?-i
These formulas are frequently used in the further study of matheOne of their uses is that they enable us to replace sums or matics. formulas to comdifferences by products and thus help us to adapt putation by logarithms.
EXAMPLE
i
.
Transform
By
Solution.
sin 2
and by
(i), Art.
the <
+
first
+ sin 4 + sin 6
sin 2
<
formula
sin 4
(/>
= =
into a product.
(i)
2 sin
|
(2
+ 4 <) cos | (2
cos
2 sin 3
40)
^
in,
Adding sin 2
+ sin 40 + sin 6 = <
2 sin 3
^+
cos
^=
2
cos 2
(j>
+ cos<).
(cos 3
(i) to
next apply the third of formula theses on the right
We
cos 3
$
the expression in paren-
cos 0,
so that finally
0+ sin 6 = 4shi3 0cos 2
^+
sin4
<
2.
2
an
sin 2
=
2 sin
C=
2 sin
=
2
C cos (4C cos C
J5),
sinC cos 04
+ 5);
FUNCTIONS OF TWO OR MORE ANGLES
ii3l
221
hence sin 2
A
+ sin 2 B + sin 2 C=
sinC
2
but by the fourth of the formulas cos
(
A-
B)
-
cos (A
=
+ JB)
[cos
(A
-
B)
-
cos (A
+ JJ)];
(i)
2 sin
A sin B,
therefore finally sin 2
A
+ sin 2 J5 + sin 2 C = 4 sin 4 sin J5 sin C. EXERCISE
1.
51
State in words the theorems embodied in the formulas (i),
Article 113. 2.
Show
that
sin (30
cos (30
cos (30 3.
= cos y, = \/ 3 sin y, = V 3 cos y, = - sin y.
Show that
s+ sin 1 5
sin 7
sin 15
sin 75
cos 7
5+ cos 5= i 1
15=
cos
cos 75 4.
- y) - y) - y) - y)
+ y) + sin (30 + y) - sin (30 + y) + cos (30 + y) - cos (30
sin (30
or differences of two Express the following products as sums
functions, sin
io"
cos
5, cos 20
cos (i7r-.0) cos(iir
Ans. \ 5.
(sin
Show
15+ sin 5),
.
.
sin
10,
sin \
-0
sin 1 0,
+ 0).
.
.
} (cosi7r+ COS20)
=
that sin
16+ sin 14= 2 sin 15
sj n
iE
sin
22
(
si n
+
i)
*
-
=
2 cos
cos i,
x sin ->
+ sin (n -
2 i)
x
=
2 sin
nx cos *.
icos20.
PLANE TRIGONOMETRY
222 Prove the following
sin4
+ sinff
A- sin cosA + cosB
identities:
=
sin
cos
A
cos
cos
.4
+ cos
cos
/I
cos
tan
(A
tan
(A
+ B) -
B
.B
Show that
10.
sin (a
may
[CHAP, xi
+ x) sin x = w cos (a + #) cos x
be transformed into
+
cos (a 11.
12.
Show
that
I (cos
2
x
+ cos 2
;y)
J (cos
2
jc
cos 2
;y)
Show sin
Show
= =
+ y) cos (x sin (x + y) sin (#
cos (x
y), y).
show that
(x-y)+cos (y+z)
sin
(y-z)+cos (z+x)
sin
(z-^)=o.
that
ioo+
sin 2 6 14.
n
Using the results of
cos (x+y) sin 13.
+ cos a = m [cos (a + 2 x) + cos a],
2 x)
sin
40+ sin 60 =
+ sin 6 B + sin 8 6
sin 50 cos
4 cos 30
=
20,
4 cos B cos 3 B sin 4 6.
that sin
whence Similarly
x
+ sin (#
= sin x = cos # = = cos sin
x
2 sin (x 2
cos (x
2 cos (x
ff
=
2)
i) cos i,
2 sin (# i
sin (x
2).
i) sin i
+ sin (x
2),
cos (x
2),
i) cos
i)
cos
i
i) sin
2 sin (#
i
+
cos (x
2).
These formulas enable us to compute the sine or cosine of x the sines and cosines of (x i), (x 2), and i are known. 15.
sines
Assuming the functions of and cosines of 3, 4 and 5.
(Suggestion.
Apply the
i
results of
and
2
as
Problem
if
known, compute the
14.)
TWO OR MORE ANGLES
FUNCTIONS OF
ii3l
1 6.
By
223
the law of sines
a
__ sin
A '
b
Taking
this proportion
Apply formulas
by composition and
a+ b _
sin
a
sin
b
(i), Art. 113,
A A
division,
we
obtain
+ sin B sin
B
to this result and deduce the law
(This constitutes an independent proof of the law of This proof is the one generally given in elementary text-
of tangents. tangents.
books on trigonometry.) 17.
From
the law of sines
we obtain
readily
sinC
c
a
sin
b
A
sin
sinC
c
B
+ sin B
a
+b
sin
<*
+
cos%(A-B)
A
Hence deduce the double formulas c
s
a-b
sin(4-)
*>
In Fig. 83, let A' = angle PAC, B' = angle PBC, then by applying the law of sines to each of the triangles PAC and PBC, 18.
we
-
find ,
03
a sin 5' _ ---__ ;
:
sin p
,\
^
:
sin
(i)
>
a
from which sin
A'
a sin a b sin/?
By
composition and division and the identity in Problem 6 sin .4'
Now A' +
B'
+ sin # = asmct + b sing = ta&KA' + B')
''
is
known from
-
B'. Hence A' therefore (2) enables us to find A' of sines gives law the these with and be found known, may .
sin
>
a
,*
the relation
and
1*1
m
"*
a
sin p
This constitutes another solution of the three-point problem.
and B'
PLANE TRIGONOMETRY
224 19.
Prove the formulas
(i), Article
[CHAP.XI
113, geometrically
by means
of Fig. 147.
Suggestion.
OE = OD
i,
sin
x
Take the
+ sin y =
sin x
20.
sin
y
cos x
+ cos y
cos x
cos ;y
Show
radius of the arc equal to unity, that
then
= = =
2
GF,
2
LF,
2
OG, 2
Z,/).
that
tan (a
tan (a
+ x) + tan # __ + #) tan x
sin (a
+ 2 #)
sin
a
is,
CHAPTER XII TRIGONOMETRIC EQUATIONS Angles Corresponding to a Given Function. Every given In Chapter X it was sine, cosine, tangent, etc. shown how these functions may be found from the tables by first expressing them in terms of an angle in the first quadrant. Suppose now that one of the functions of an angle is known and it is required 114.
angle has a single
to find the angle.
If the
known
is
angle
once found from the tables;
to be less than
/?, it is
at
R, there also is no uncerin the case of the sine its (and tainty except reciprocal the cosecant), which may be either the angle given in the table or its supplement.
But
in case
no
restriction is
if
than
less
2
imposed on the magnitude of the angle,
the given function may belong to any one of an unlimited number of angles. We shall show how all the angles corresponding to a given function can in each case be expressed by a single fortheir separate values may be written down when
mula from which required.
115. Principal Value. Of all the angles which correspond to a given function, the one which has the least numerical value is called
the principal value (see Article 81). If there are two least values with opposite signs, the positive angle is taken as the principal value.
Thus,
if
sin 6
J, 6
creased or diminished angles, 30
=
30
or
1
50
by any number
or either of these values inof times
being the least numerically,
value of the angles whose sine
is
360.
Of
all
these
considered the principal
is J.
120 or either of these values \ \/3, 6 = 60, increased or diminished by any number of times 360. Here 60, having the least numerical value, is the principal value of the angles If sin
whose
sine
=
is
i_>/3.
= 45 or diminished by multiples of 360. principal value of 0. If cos 6
=
J \/2, 6
45 or either of these increased or In this case 45, not 45, is the 225
PLANE TRIGONOMETRY
226
Formula
Let
be required to determine
it
from the equation
6
=
sin
Let a
=
the principal value of TT
2
7T
3
TT
-
k.
Then the
6.
TT
a,
+
xn
Having a Given Sine.
116.
for Angles
[CHAP,
other values of 6 are
a,
~2 + - 3 - a, 7T
,
,
TT
a,
-67r
+a
Fig. 148.
etc.
etc.
It will be observed that the coefficients of i, 2, 3, 4, 5, 6, etc.,
-
i,
-
2,
-
3,
TT
are the integers
- 4, -
-
5,
6, etc.,
while the sign of a. is positive or negative according as the coefficient Now ( i) w is always positive when n is an of TT is even or odd.
even integer, negative when n is an odd integer. Making use of this property, we can express the whole set of angles by the single formula
where a
any
is the principal value of the angles positive or negative integer.
When n =
o,
we
get the principal value of
OTT+ (- i)a = 117.
Formula
If cos 6
=
k,
for Angles
whose
6,
sine is k,
and n
for
a.
Having a Given Cosine.
and a the principal value
of
0,
the other values of 6 are
x
a,
2T
+ a,
a,
6 ?r
a,
6r+a,
-6?r
+ a,
27T
a,
27T
+ a,
6 TT
etc.
2
etc.
TRIGONOMETRIC EQUATIONS
120]
Ail these values
and no others are expressed by the
= where
2
118. If
n
is
necessarily
Formula
tan 6
=
2
7T
3
7T
+ + +
Having a Given Tangent.
~ + 7T
,
27T+OJ,
Of,
3
7T
of
6,
the other values of 8
Of,
Oi y
+
Of,
etc.
All these values
119.
nw
Summary
formula
+ a,
of Results.
Summing up
the results of the last
we have sin (nir
+
cos (2nir
i)"a)
(
a)
tan (rnr -fa) is
single
positive or negative integer.
three articles
where a
etc.
and no others are expressed by the 6
n being any
single formula
an even number.
for Angles
are 7T
2 nir
and a the principal value
k,
227
=
sin a,
=
cos a,
=
tan a,
the principal value of the angle, and
d)
n any
positive or
negative integer. These three formulas should be thoroughly understood and orized by the student.
mem-
120. Trigonometric Equations Involving a Single Angle.
Equations involving one or more trigonometric functions of one or
more unknown angles are or
may not To solve
involve other
called trigonometric equations.
unknowns which
They may
are not angles.
trigonometric equations involving a single unknown express each of the functions which occur in the equation in terms of some one of them, and solve the resulting equation algeangle,
we
PLANE TRIGONOMETRY
228 braically,
function
this
considering
responding angles
The
unknown.
the
as
[CHAP,
xn
cor-
then be easily found from the formulas of
may
Article 119.
EXAMPLE Solution.
i.
3.
Expressing the cosecant in terms of the sine,
2sinx+
=
-7^-
consider sin x the unknown, call
we have
3.
x
sin
Now
+ esc x =
Solve the equation 2 sin x
it 5
we then have
say,
Clearing of fractions and transposing, 2 s
5
Solving
=
2
3 ^
i
3^3
2
=
o.
-4-2 = 3_J =I
that
sin
is,
from which
and by
The
=
x
or
i
J,
the principal values of x are
*
x
or
2
^> 6
Article 119 the general values of x are
general values of x just written are equivalent to the two sets
of values
222
2222
_*, "
or
4
4
= 7T
ftZ,
SJT,
liZ.etc.,
I37T
t^TT
I77T
7
.
>etC "
1 1 7T
7T
IQ7T
,
>etC" -6-' 6 esc x = i, and 2 sin # + esc x 2.1 + =3. Check. If sin If sin x esc # = 2, and 2 sin # + esc #=2.^+2=3. EXAMPLE 2. To find 6 in the equation tan 6 + 3 cot 8 = 4. 6'
T'
oc
Solution.
6' = i, = J,
Expressing
6
'
6
i
the
cotangent in
gives
=
terms of
the
tangent
1
TRIGONOMETRIC EQUATIONS
20]
where
t
=
tan
Solving for
0.
/,
2
2
that
=
tan0
is,
or
3
229
i,
from which the principal values of 6 are 8
=
'
1 33' 54" or tan'
71
1
=
45-
Article 119 the general values of 6 are
By
= n Check.
If
tan0
If
tan 6
EXAMPLE c
=
tan- 3
3.
=
Find
3,
cot0
i,
cot 6
, .
,
+
180
= i = i,
in tan
9= ,
,
tan
Solution.
33' 54" or
71
tan0 tan
w
+
180
+ 3 cot0 + 3 cot
= =
45-
+ 3 \ = 4. + 3.1 = 4.
3
.
1
-
+ 2 \/3 cos = o. cos
i
-
COSf/)
Vi
hence
cos2
0+2 N/3 cos
2
^=
o.
Transposing one term and squaring, I
COS2
12 cos 4
0=12 COS
+ cos
2
4
0,
0i =
o,
24 cos
The a
0=|,
-i
are not admissible, since the cosine of two values of cos angle cannot be imaginary, and the first two values each give
last
real
= Check, tan
cos (2
mr
-J=
+ 2 \/3 cos =
but not for the upper. tan
2 HIT
-
tan (2
|,
3
nir
+ 2 V$
Hence the general
0+2 N/3 cos = o
is
-J= ()
=
\/3
o for the lower sign
solution of the equation
= 2 HIT
"
o
PLANE TRIGONOMETRY
230
[CHAP,
xn
This example clearly shows the necessity of verifying the results them as the solutions of a given equation.
before accepting
EXAMPLE
+
Solve the equation sin x
4.
+ Vi
x
sin
First solution,
sin2
i
Squaring 2 sin
Solving
2
sin x
= sin2 x =
Vi #
2
=
2
x
N/2 2
cos x
=
\/2.
\/2,
sin x. 2
\/ 2
V 2 sin x + = i
sin
+ sin
#
2
x.
o,
4 Check. sm(n7r+(-i) V
according as
w
is
n
-W
4/
even or odd.
sin #
mr+ (- i) n - =
cos
\/2,
4/
Therefore
+ cos x = % V
\ \/ 2
'2
=
\/2
o
or
according as n is even or odd. We see that the lower sign does not satisfy the original equation, that is n cannot be odd; hence the general solution of the equation
+ cos# = \^2 x = 2 rnr + |
sin* is
TT.
We
have given the solution which would most Second solution. naturally suggest itself to the beginner. A more elegant solution of the foregoing equation is the following: Since cos x
=
sin
TT
(
the equation to be solved
x),
may
be
written sin x
+ cos x =
The middle member
may
x
sum
the
+ sin (^ of
two
TT
sines,
x)
=
\/2.
which by Article 113
be transformed into a product, thus sin
that
is
sin
is
or
from which
x 2
+ sin (J |
VJ
TT
=
x)
cos (x
-
cos (*
x
2 sin
J TT cos (x
JTT)
=
i*-)
J
TT
=*
=
\
IT)
=
N/2,
V^, i,
2
TT
o, or
#
=
2 rnr
+?
TT.
TRIGONOMETRIC EQUATIONS
120]
EXAMPLE a, by
a,y fty
being
known
o,
constants.
+ a) = = cos (* +
Solution,
+ a) + b cos (x + 0) =
Solve the equation a sin (x
5.
231
sin (x
ft)
sin
+ cos x sin a,
x cos a
cos # cos
sin
x sin 0.
Substituting these values in the original equation,
a
x cos a
(sin
+ cos x sin a) + b (cos x cos
Collecting the coefficients of sin x (a cos
and cos x
Dividing by cos
and solving
x,
tan x
separately,
we have
=
of
o.
o.
for tan x,
a sin a
=
=
# sin ft)
+ Dcos/3) cos
6sin0) sin;c+ (a sin a
a
sin
ft
+ b cos !
.
f
a cos a
sin
p
from which \
To check
Check. it is
+ 6cos0 \
asma tan- 1 /
=
a cos a
6 sin
#
= =
,
2,
b
=
i,
a
=
20
o ft
,
tan" 1 1.0181
WTT
=
=
a
o
15
^TT
,
sin
(
0.4308)
+
Thus
+ & cos - =
a
ft
'
.
a cos a
b
sm
if
Q
1.0181,
453i',
Substituting these values in the original equation, 2
/
complicated results like the one just obtained
best to assume arbitrary values for the constants.
a
ft
I
(=f 0.8615)
=
we have
O.
EXERCISE 52 In the following find both the principal and the general value the angle: 1.
tan0
=
2 sin0.
Ans. 6
=
mr or
2 nir
--; principal values o and
^
3 2.
2 3 sin
x
=
cos2 #.
Ans. x 2
3.
3 tan x
4
sin2
#
=
=
mr
6
J
principal value
6
i. 5.
a;
= mr
'-;
4
principal value 4
of
PLANE TRIGONOMETRY
232 4.
=
2 sin2
3 cos
+ cot y =
tan y
6.
tan2
*+ csc
7.
sin0
+
8.
csc*cot*
xn
.
Ans.
5.
[CHAP,
$=
2 WTT
=
irn
^4$. y
2.
principal value
~; 3
+\
o
TT;
-principal value
4 In the following problems find those values of the variable angle which are less than 360:
9.
10.
sin
2
2
1
4 3 sec
=
*
Ans.
3.
*=-,
44
cos0=i.
=
2
N3.
.
+ J = o. 0io sec + 8 = o. 2 cos
o
6
/
.
o
o
2
6446644 "
11.
tan*+ 4*w. *
12.
A
a
sec 2 *
=
=
63
+
2
6 cos *
3 71
IL
io826' 06", 243
*
=
13. 14.
sin* sin*
+ csc* cos*
=
=
19 28' 16",
26' 04'', 288
Ans. x
.
^4w5.
5.
If
a sin *
+ b cos * =
c,
show that
16.
If
a tan *
+ 6 cot * =
c,
show that tan *
17.
If sin (a
sin
+ x) = m sin *, show that sin
*
=
=
If
more simply cot * tan (a
If
tan (a
cos
=
sin
30, 150.
No
solution.
= = sin
Vm
2
a
m cos a + 1
a
a
+ x) = m tan *, show that tan*
19.
26' 06".
ac
2
1 8.
6
30, 150, 1 60 31' 44".
15.
or
LLZ.
Z-
7.
Ans. x
=
_E
TJL
7.
26' 04",
5 sin
SJE
- i)V(fit- i) -4wtan a 2 w tan a
+ *) tan * = w, show that ^ + w tan tan* = (
I
2
2
=
)
d + w) 2 tan
2
a
+4w
TRIGONOMETRIC EQUATIONS
i2i]
sum
expressing each product as a
By
233
or difference of functions
show: 20.
sin (a
If
21.
If
cos (a
22.
If
sin (a
23.
If
cos (a
= = x) cos x = #) cos x = x) sin # x) sin
x
m, then
cos (a
2 x)
m, then
cos (a
2 x)
w, then
sin (a
2 x)
w, then
sin (a
2 x)
= cos a T 2 m. = 2 w cos a. = 2 w sin a. = sin a 2 m.
121. Trigonometric Equations Involving Multiple Angles. equation involves multiple angles, as the equation
When an
cos 3 6
+ sin 2 6 =
a,
or tan 2 x
=
cot 5 x,
can frequently be solved by two or more different methods, and the answer will appear in various forms according as one or the other of it
these methods has been employed in the solution. Generally the different forms of the answers can easily be identified, but in some cases
considerable ingenuity is required to show that the different forms are Thus it is easy to see that 2 rnr really the same. \ TT and nir \w (n being any integer, but not the same integer in both forms) ex-
+
same general value, but
press the
7r
0=
and
it is
not so easy to see that
or
or
2W7r-|7r
10
5
are equivalent results.
EXAMPLE
Solve the equation sin 2 6
i.
First solution.
cos0.
Substituting for sin 2 6 its value in
(Article
single angle
=
in,
(i)),
2 sin 6
terms of the
we have cos 6
=
cos 6.
i)
=
o,
sin0
=
^.
Transposing and factoring, cos 6(2 sin 6
from which cos0
=
o,
or
Therefore
0=2W7Ti7T= *0=
27r-hj7r and 2mr is
}7r=(2w
OmTT+Ci)
every odd number, hence any integer, even or odd.
every even and
where n
WTT+^TT*,
2
n
i
W
00
l)
6
ir-flm Now we may write
2
n represents
=
rnr
-f-
i
T
PLANE TRIGONOMETRY
234
[CHAP,
xn
Check. sin 2 6
=
sin (2 rnr
w
according as
cos0
=
is
is
=
IT)
w
+(
o, or sin (2 nir
i)
-J=
even or odd.
cos (nir
according as n
+
+
|TT)=
or cosfrar
o,
+
=
w i)
(
|
-]
even or odd.
we have
Substituting these values in the original equation,
=
^l=
0, or
4^3-
Second solution. Since
cos 8 sin
from which
ffrf
20 2
solution.
= = =
sin (^
TT
0),
sin (%TT
0),
nir
M
+
i)
(
we have
(4 TT
0),
Transposing the second member of the equa-
we have
tion,
into a product of a sine sin
sin 2
sin (|TT
113, the difference of
Article
By
=
cos 6
sin 2 6
20
sin (!TT
and
two
=
0)
sines
o.
may be
tranformed
cosine, thus
0)= 2cos(|0+ J7r)sin(|0
ITT)
=
o,
from which
+ | = o, or sin (f J = o, = WTT+ ^TT, or ^TT |0 + i?r= 27r = 2 WTT + | or f WTT + ^
cos (i
and that
TT)
TT)
TT.
TT,
is,
|0
Identification of results. If in (2)
w
is
odd, say 2
0= and
if
w
is
(2
even, say 2 w,
m+ m+
i,
we have
i)ir
-|TT
=
2
mir
+ %ir,
we have
3
This shows that the results
(2)
and
(3) are
the same.
|TT
=
WTT,
(3)
TRIGONOMETRIC EQUATIONS
i2i]
235
Next consider the second value in (3). Every integer n is either of 3, say 3 w, or some multiple of 3 increased by unity. 2. say 3 m + i, or some multiple of 3 increased by 2, say 3 m + = In the first case, n 3 w, we have from the second value of (3)
ome multiple
--
3636
2W7T
-
7T
f-
=
=
In the second case, w
7T 2.3JW7T _^Z _-|- ,
3
w+
_
7T 2fW7T-t- ,
6
i,
363 In the third case, n
=
3
6
3
w+
2,
6
3
This last value, combined with the
0=2fwr+iir, or (2W+i)H-j7r, that
2
first
value of 6 in
(3),
gives
8=mr+%ir (n even or odd),
is,
and
=
2
WTT
+ |TT,
or
(2
w+
-
i) *
may be written
JT,
0=tt7T+(-l)'^. o This shows that the results
(3)
and
(i) are
the same.
EXERCISE 53 Find the general value of the angle in each of the following equa tions: 1.
cos2#+cos#+
=
sin 4 x.
=
cos 4 x.
2.
cos 5 x
3.
cos 5
4.
sn4* =
sns:r.
5.
tan
cot 26.
6.
sin 3 x
JC
i
=o.
Ans. x
Ans. x
=
=
mr
-
JTT, or ! 2 ^j
2 n?r
.
^^5.
.
.
oc
=, .
5^=
=
2
or
|TT. ,
TT *
2 n?r,
2 tnr
or
-
or
/
==
+ sin 2 x + sin # = o. =
mr
g
Ans. ^
Ans. x
2mr
or
WTT
i
^r
"T+~'
PLANE TRIGONOMETRY
236 7.
cos x
cos 3 x
=
cos 5 y
cos 3 y
xn
sin 2 x.
Ans. x 8.
[CHAP,
=
or HIT
nir,
+ sin y
=
or
|TT,
WTT+(
i)
JWTT+(
i)
n ~-
6
o.
^4w^.
y
=
WTT,
or
n
~24
9.
-
cos (60
+ cos (60 + x) = i
x)
4iw. x
mx =
10.
cos
ii.
cot$
12.
tan
2
=
OP
sin kx,
m
=
=
cos" 1 3
2 WTT
2
w i8o
70 32'.
and & being known.
tan^0, * being known.
tan x
=
4f.
i.
*
=
rnr
6
Solve each of the following problems by each of three methods, identify the results.
and
13.
cos
26
=
sin
Ans. d
0.
= 2mr
7r,orf 6
14.
cos 3 s
=
J
Suggestion.
15.
sin3^
=
(-
i
=
Ans. x
sin 2X.
+ \/5) =
sin
Ans. x
cos2o;.
i
,
10
J TT, or | WTT
2 HIT
(-
i
-
\/ 5~)
=
+ 10
sinf-
3Z\
=
122. Trigonometric Equations Involving Two or More VariWhen there are given two or more trigonometric equations, involving two or more variables, the solution, if it is possible
ables.
at
upon more or
less ingenious
combinations can be given. We shall illustrate the various methods and devices commonly employed by a few examples chosen from among those which most frequently occur all,
generally depends
of the equations, for
which no
in applied trigonometry.
definite rules
TRIGONOMETRIC EQUATIONS
122]
EXAMPLE
To
i.
find r
from the equations
and
= =
r sin r cos
a and
being
known
may
(i)
0,
(2)
be found,
equation by the i-econd,
first
tan0 from which
a,
constants.
Dividing the
First solution.
237
=
7
then found from either
r is
we have
a
(i) or (2),
b
cos
sin 6
Since the angles 6 as determined from the tangent differ by multiples will have two values each which are numerically sin 6 and cos TT, Therefore r will have two values which in but sign. opposite equal are equal and opposite in sign. If from the outset it is known that r can be positive only, which is the case frequently, then sin 6 must have of
the same sign as a and cos 6 the same sign as b. limited to the quadrants determined by these signs.
Second solution.
must then be
results Squaring each equation and adding the
gives r2
from which
6
=
a2
+
Then
r is obtained.
2 ,
is
found from either of the equa-
tions
sin0=-
cos0=-
r
r
Of the two methods the
first is
better adapted to the use of loga-
rithms than the second.
EXAMPLE
2.
Find
r,
and
from the equations
r cos
cos
r cos
6 sin r sin
First
obtain
solution.
(j>
= = =
a
(0
b
(2)
c.
(3)
Dividing the second equation by the
first
we
PLANE TRIGONOMETRY
238
from which
is
[CHAP,
being known, r cos 6
obtained.
xn
obtained from
is
either equation (i) or (2)
cos
From
and
(3)
and adding the
and
(4) r
Second solution.
9
r
The
is
first
be found as in Example
i.
and
(i), (2)
2
=
+p+C
a2
(3)
2
found.
found from the equation (3), and, found from either equation (i) or (2).
being known, 6
known,
9
results gives
may be
r
sin
Squaring each of the equations
r
from which
may
9
is
solution
is
preferable
r
and
8 being
logarithms are to be used through-
if
out.
EXAMPLE
Solve the equations
3.
r sin (a r sin 08
for r
and
B.
Applying the addition theorem for the
Solution.
Put
+ 0) = a + 0) = b
=
r cos
re,
= =
sine,
we have
r sin
a.
cos B
r sin
|3
cos 6
+ r cos a sin B + r cos sin
=
the equations then become
r sin
;y,
/3
a Z>.
+ y cos a = a #sin|8 + ycosj3 =
x sin
ot
b.
Solving these equations for x and y
=
o
= # and in
y,
that
is,
cos g
a cos |8
cos
a sin |3
_
a
Example
fl
sin
--
and
cos
sin
a sin g
r cos B
fr
a
y
we
a
_
sin ft
a a cos
6 sin sin
obtain
a cos ft
b cos
sin (a __ 6 sin
a
a
sin (a
r sin 6 being
known,
r
a
$) sin
ft
0)
and
6 are
found as
i.
EXERCISE 54 In the following equations, consider r positive. 1.
Given
r sin 8
=
8.219, r cos 6
=
12.88, find r
Ans. r 2.
Given
r sin
=
3, r
cos
=
4, find r
and
= 0.
and
6.
15.28,0
=
32
33'.
TRIGONOMETRIC EQUATIONS
122]
Given
3.
=
r sin 6
=
27.138, r cos 6
Ans.
Given
4.
r cos
cos
r cos
sin
rsin0 Ans.
Given
5.
r
= = =
cos
r sin
r cos B sin r sin
Find
r,
X and
r cos
7.
r sin
f
92.692, find r
-
=
r
=
96.583,
Find
and
r,
0.
39.062.]
=
208.16,
10 49' oo",
=
<
5,!
=
12,
h
84.
j
Find
r cos
4,
+ # J= ^3,
and 0.
r,
=
85, 6
81
<
12',
=
86
36'.
X
-
r sin (
sin
/*
=
=
+*
5,
i.
rsin\
= v/ 59.
Find
and
r
jt\
]
=
r
2,
to solve the equations
r cos
g.
74 42' oo".
from the equations
/i
X cos /i
Show how
0.
| ?
v4ws. 8.
and
163 40' 52".
59.953, 197. 207 V
By natural functions, r =
Ans. 6.
= = = =
239
+
a, r sin (#
a)
(jc
/3)
=
b y for r
and
x.
Solve the equations cos (*
y)
=
J
V2,
sin (*
+ y) =
giving both the principal and general values.
An
Principal values, x
General values, x
=
2-^or >y 24 24
=(w+
=
rc)-+(-i) 2
2
10.
w and w are
any two
integers.
Solve the equations cos 2 x
Ans.
x=2n*
cos 2 y
=
a,
cos-'
cos #
y
=
cos y 2 WTT
=
Show how
to solve the equations
tan (x
+ y)=
a,
tan x
tan y
b.
car'
V
ii.
24
m -->
68 T| y=(- )I + (-,)T68 2
2
where
or^ 24
=
6.
>
PLANE TRIGONOMETRY
240 123.
[CHAP,
xn
The
Solutions Adapted to Logarithmic Computation.
solution of a problem in trigonometry is not considered completed until it can be effected by the use of logarithms, in fact the adapta-
an important part in trigonometric equations whose
tion of formulas to the use of logarithms forms
Many
trigonometric investigations.
algebraic solution is exceedingly simple require further treatment from the trigonometric point of view. Equations 15 to 19, Exerwill show now how cise 52, are typical equations of this kind.
We
each of these equations
EXAMPLE
To
i.
may be
solved
by
logarithms.
solve the equation
sin x
a
+ b cos a? =
c.
Divide each term of the equation by
Solution.
-^ and
the coefficients -
vV +
squares equals
i,
-
2
we may -
--
Va +
r
Va + b
b*
=
Va + 2
are fractions the
sum
b
z
of
,
then
whose
2
therefore put
cos
=
(/>,
Va + 2
2 b'
2
(i)
.
sin
,
(2)
b*
and the given equation becomes sin
oc
cos
+ cos x sin
Va + b 2
2
or
sn
+b and from
2
(3)
a
(2) .
Having found
An
from
(4),
x
angle like the angle 0,
a problem,
is
called
is
found from
(4)
(3).
introduced to facilitate the solution of
an auxiliary
angle.
NUMERICAL ILLUSTRATION Suppose the given equation
then
a
is
- 0.9328 cos x = 1.3794, 3.4537 sin x = 34537, b = - 0.9328, c = - 1.3794-
TRIGONOMETRIC EQUATIONS
123]
241
Solution.
By
By
(4)
(3)
= 9.96979 n * = 0.13969 n log c = = 9.46172 a a log 0.53828 colog = = tan n cos log log 9.43151 9.98471
<
<
1
* Check.
By
=217
26' 04".
47' 40", 352
logc
(3)
colog b log sin log sin (x
EXAMPLE
To
2.
+
)
= 0.13969 n = 0.03021 n = 9.41621 # = 9.58611 ft
solve the equation
6 cot a?
=
c.
Expressing the tangent and cotangent each in terms of the sine and cosine, the equation reduces to (Problem 19, ExerSolution.
cise 50)
c sin 2
This equation
is
+
x
of the
b) cos 2
(a
By
+ b.
i.
Solve the equation sin (a
Solution.
a
+ b cos # = c,
which has been solved in Example 3.
=
form a sin x
EXAMPLE
oo
+
a?)
==
m sin
a?.
(i)
composition and division the proportion sin (a
+ x) = m
gives rise to sin (a
sin (a
+ x) + #)
-f-
sin
x
__ 2 sin
sin
x
2
(a
cos (a
+ \ x) cos ^ a + %x) sin | a
= tan (a+$ i *
This n indicates that the number to which the logarithm belongs is negative. is an auxiliary angle which is not retained in the end, any one of its values may be used. t Since
PLANE TRIGONOMETRY
242 or
tan(a
[CHAP, xil
+ Js) = m--tanJa-
(2)
i
From If
tan
=
i
m
a+$x and hence x can be found.
(2)
m=
--i:+ tan
i
i
tan %ir
-*, =
/
,
+f
\
i
jr
tan (0
,
tan J TT tan
TT)
=
,/j-
cot
i
%
f TT)
(
hence
+ 2 #) =
tan (a
Many
cot
J
(
tan | a, where tan
TT)
<
=
m.
(3)
computers prefer (3) to (2) in solving equation (i).
EXAMPLE
Solve the equation
4.
=
tan (a,+ &)
m tan &.
Taking the proportion
Solution.
tan (a
+ x)
__
m
tan x
by composition and
division,
tan (a tan (a
But by Problem
i
we obtain
+ x) + tana _ m+ i + x) tan x m i
20, Exercise 51,
tan (a tan (a
+ #) + tan = sin + x) tan # a;
(a:
+ 2 #)
sin
a
hence sin (a
we may
or
+ 2 *)
+ 2 x) =
Either (i) or (2)
EXAMPLE
5.
cot ($
(i)
i
may
| TT) sin a, where tan
be used to find a
+
$ = w.
# and hence
2
(2)
#.
Solve the equation
tan (a Solution.
5+J Sin a,
w
write the result in the form
sin (a
we
=
+
a?)
tan
a?
=
in.
Expressing the tangents in terms of sines and cosines,
find sin (a
which by Problem
cosa
+ x) sin x = m cos (a + x) cos #,
10, Exercise 51,
cos (a
+
2 #)
may be
written
= m [cos (a+
2
jc)
+ cosa],
TRIGONOMETRIC EQUATIONS
123]
243
from which
+ 2 x) = -
cos (a
m '^ T = m i
cot (0
hence
= ^-^W -p I
tan
and
i-^=^
cos a.
+m
i
(i)
=
I
By Example 3,
i
so that equation (i)
cos (a
From
+
(i) or
2 #)
from
+m
= -
may be
= (2)
tan
a
| TT), where tan
-
(<
tan
m,
J T),
(r/>
J *)
=
tan (J TT
-
0),
written
(J
+
x
2
=
0) cos a, where tan
TT
may
m.
be found and hence
(2)
x.
EXERCISE 55
= m cos x, = tan TT
+ a) + \ x) If sin (# + a) = m cos x, tan (| a + i + x) = cot (J 1.
If cos (x
tan (a
(|-
2.
?r
3.
tan 4.
TT
show that x <) cot
is
a,
show that # <) tan
is
TT
(J
by the formula
given
=
where tan
by the formula
given -|
a),
m.
where tan $=* m.
If cos (x -\- a) = m sin #, show that x is given by the formula = tan (J TT <) cot (^ TT + J a), where tan = w. (J a \ TT + #) If tan + a) = w cot x, show that = m. cos (a + 2 ^) = tan (| TT ^>) cos a, where tan If cot (# + ) = w cot (x a), then = m. sin 2 # = cot (0 i TT) sin 2 a, where tan If tan (a + #) cot x = w, show that = w. sin (a + 2 #) = cot (0 1 TT) sin a, where tan = w, show that If tan (a + x) tan (a x) = COt (| IT COS 2 # 0) COS 2 <
(ac
<
5.
6.
7.
Of.
8.
Find the angles between o and 360 which
4sin# + 3 cosff=
satisfy the equation
5.
Ans. x 9.
-
53
of 45".
Find 8 from the equation 2.76 cos 6
2.32 sin
=
1.91.
j.0=
17
S9 .6', 261
55'.
PLANE TRIGONOMETRY
244
Find the general solution x/3 sin x
10.
[CHAP,
of the equation
v
cos x
2.
TT,
or (2
Find the general solution of the equation ~ v's) cot * (i V^3) tans + (i
11.
+
x Find
12.
=
nir
xn
+ lir
y
nTr
n
=
+ I)T-
12
2-
--12
and # from the equations
r
r sin (a r sin (P
+ #) = w, + x) = w.
and difference of these equations, and Suggestion. Form the sum Divida into product by the formulas in Article 113. change each ing the
first result
by the second
gives
m or
tan
/J + \
a
cot f \ 14.
+ 0) tan^, where
2
If r cos (a
13.
x\= tanQTr
n
^~
-
+ x) =
w,
+x =
tan (i TT
)
2
r cos (P
+ x) =
n, then
+ <) tan -
->
tan x
=
by
=
m sin + m cos a of
i
to computation
where tan $
2
/
Adapt the formula
x is given by
logarithms.
the sine and cosine Express the tangent in terms of written be result the and clear fractions, may Suggestion.
sin*
which by Example 3 may tan
-
(*-!)
15.
Show
= wsin(a
#),
be transformed into
tan }
a= tan (0- JT) tan | a, where tan
that
a fl
-j.
&
5
>
cos 2
=
where
tan 2
=
->
^
2 a . cos2 <, where sin
^
==
&
=
m.
TRIGONOMETRIC EQUATIONS
i24l
These formulas enable us to find the logarithm Thus,
245
any sum or
of
difference.
log (a
+ 6)
log (a
6)
= =
+ 2 colog cos <, log a + 2 log cos <,
log a
log tan log sin
<
= =
? (log b
log a),
\ (log 6
log a).
Establish the following transformations:
16.
COS0
where tan
>>
=
a
cos0
124.
The two
Inverse Functions.
expressions (i)
a?
second that x
sin-
1
y
(2)
represent different views of the same relation. The first one states that y is the sine of x, the
Fig- 151-
is y.
=
is
the angle or arc (measuring the angle) whose sine y in terms of x, (2) expresses x in terms of y.
(i) expresses
Taking the
sine of each sin
x
and taking the inverse
=
member sin (sin
of (2) gives us
~l y)
sine of each
sin" 1 y
In precisely the same
=
it
^)=
!
tan (tan~ x)
=
may
=
x,
by
(3)
of (i) gives us
by
(cosjc)
1
x,
(i),
(2).
(4)
be shown that
cos" 1
x,
y,
member
sin- 1 (sin x)
way
cos (cos- 1
=
tan- (tan x)
= =
x, x,
*
etc.,
From
etc.
it appears that of each pair of operations, of that taking the sine and that of taking the arcexample either undoes the other, that is, if the two operations are persine, formed in succession, the result is that the quantity operated on is ~ left unchanged. This explains why sin l x is called the inverse sine 1 of x, tan" ^ the inverse tangent of x, etc., in fact in any pair of
these relations
say for
Viewed as operais the inverse of the other. the each the of members of relation tions, pair is like that of addition to subtraction, of multiplication to division, of involution to such functions each
evolution.
PLANE TRIGONOMETRY
246 In general,
[CHAP,
xn
= / (x)
represents any function of x, the inverse 1 represented by/" (x), and the relation between the two is always such that, considered as operations on #, each undoes the other, that is,
function
if
y
is
1 /[/- ()] =
.
/- 1 [/()] =
and
y = f(%) i s known, f~ (x) is found by solve) y = / (a) for 3, and by substituting in l
If
*.
(5)
solving (provided the result x for y.
we can Thus,
if
(6)
we
*
find
and
=
+ 4 = f~ V 3 *+4 = /\/3
l
(y),
3;
1
(7)
(*).
To (7),
verify the relations (5) we substitute for x in (6) the expression and for x in (7) the expression (6); thus,
3
There is this important difference between the trigonometric funcwhile each of the former has a single tions and their inverses, of the latter has an indefinite number of values. Thus, value, each if
x
=
~, sin x has a single value,
namely
6
may have any of
666
the values -> sin" 1
where a
Any
is
cos"
1
tan"
1
A"
JC
jc
= = =
SJL,
HTT
+
EXAMPLE
i.
A =
terms of inverse functions.
sin" 1
x
In general,
+ a,
We
i
^,
a,
may
be expressed
shall illustrate the
Express the relation cos 2
.
=
n
the principal value of the angle. between trigonometric functions functions.
x
if
i) a,
(
2 HIT WTT
but
I12E, e t c o
13JL,
relation
by means of inverse some examples.
in
\>
2
2 sin
A
method by
TRIGONOMETRIC EQUATIONS
124]
sin
A =
cos (2
sin""
Let
Solution.
m, then
A =
sin" 1 w,
247
and the given equa-
tion becomes
or
2
EXAMPLE
1
= i m = cos"
sin" 1
w?
2
m)
y
1
w
2
2
(1
).
Express the formula
2.
sin
+ B) =
(A
A
sin
B + cos ^4 sin B
cos
in terms of inverse functions.
Solution.
Let
then
sin
cos
A =
= Vi
yl
B = w, cos-B = Vi
sin
m,
m
2 ,
w2
,
and the given formula becomes sin (sin" 1
sin" 1
or
EXAMPLE
m+
m + sin"
3.
sin" 1
=
n
1
ri)
= m Vi
sin" 1
+ n Vi m w + w \/i w
n2
2
,
2
(m \/i
2
).
Express the formula tan
A
tan
B=
i
tan
/J
in terms of inverse functions.
Solution.
Let
tan
then
A = ^4
m,
=
1
tan" w,
Substituting
tan (tan" 1
w+
= m
tan" 1 n)
tan-
or
1
w+
tan" 1 n
=
tan" 1 w.
n
mn
i
w,
=
J5
,
W
tan' 1 -
Formulas involving inverse functions may be verified the process illustrated in the above examples.
EXAMPLE
4.
Show tan"
Solution.
Put
1
that
m + tan' tan"
then Substituting,
we
1
m= w=
1
n
=
tan" 1
A, tan" n 1
tan A,
n =
m+n mn
=
B,
tan B.
find
i
tan
^4
tan
B
by reversing
PLANE TRIGONOMETRY
248
[CHAP,
xn
or
tanU This latter expression is
we know
hence the original expression
is true,
also true.
EXAMPLE
Find the value of
5.
+
tan -1 Solution.
\ = A tan A = %,
Put
tan
then
+B=
A
EXAMPLE
B, J,
+ B.
^4
+ tan"
tan" 1
1
%
=
WTT
+-
(
f ).
Solve the equation
6.
sin" 1 2 x Solution.
= = tan B
+ 10-
tanfri
therefore
^.
tan" 1 1
,
and the given expression becomes
Now
tan- 1
-1
+ sin"
1
3 #
=
cos" 1
sin" 2 # = A, sin" 3 # = sin A = 2 x, sin B == 3 #, 2 cos A = Vi 4 x cos 5 = V i
Put
1
1
,
then
,
and the given expression becomes
A Take
+B=
cos" 1
(-
f).
the cosine of both sides, cos (A
and expressing
+ B) =
this in
Vi
cos
A
terms of
2 4#
Vi
cos
B
sin
A
sin J5
=
f,
x,
9#
2
2#3#=
f
.
Solving for x,
EXERCISE 56 i.
Find the general value of each of the following angles:
sin" 1 J, cos" 1
-, tan""
Ans.
1
1 1 \/3, cos"" o, sec""
tan" 1
oo
643 1,
mr+(-i) n -, 27T-,
.
nT
+ -,
2 9x ,
TRIGONOMETRIC EQUATIONS
124]
249
Considering principal values only, verify the following: 2.
tan"H- tan-H=
V
tan~"
4.
cos-Yr
5.
tan-4 + tan-4
1
w + tan-
+ cos-
1
w
4
=
2
H=
1
-
'
+ tan-
1
J
1
7 =-. 4
Express the following formulas in terms of inverse functions: 2
tan
6.
2
=
tan0
-
tan2
Ans.
m=
tan- 1
2
tan" 1 ( i
\
cos 04
7.
J5)
Ans. sin-
where
w=
sin 2
8.
1
w-
sin
x
=
Ans.
=
cos
sin" 1
A w= ic
w
tan
0.
/
A cos 5 + sin A sin #, n = cos" (tfw + V(i - m ) 2
1
(i
-
w2 ),
sin B.
,
2 sin
= ^-r-Y - 2 where m
cos x.
2 sin"
m=
1
sin- 1 (2
m Vi - w
2
m=
where
),
sin *.
=
3 sin 3 x 4 sin x. 3 sin x 1 = sin- 1 w sinAns. 3
9.
10.
Show
sin' 1
(3^-4
that
m=
V -w = ,
cos" 1
2
i
),
where m
m
.
tan"
3
i
=
sin
a;.
i
1
= csc n. Show 12.
Show
that sin" 1
+ cos-
tan" 1 i
+ 3 tan-
1
i*
=
tan- 1
f.
that 2
1
\
=
tan- 1 (-
3).
Find x in the following equations: 13.
14.
15.
sbr l 2x +
an-i
wr
l
x
=
-'
^an-i-o;=an-
-
Ans. .
-
5.
PLANE TRIGONOMETRY
250
*LJ
16.
Show that if / (x) =
17.
Prove that the inverse of
18.
What
is
then /- 1
,
y?
+4
the inverse of logio#
?
is
Of
(x)
[CHAP,
=
\/x i
x 4.
x
?
Ans. 10*,
Prove that
10.
x
~*~ I
*
i
is its
Find the inverse functions
own
of
xn
x.
i
inverse.
each of the following, and verify the
results: 20.
/(*)
=
21.
/()0
=
22.
/W =
I
i. (a) Define the sine, cosine and tangent of any the Give (b) signs of the principal functions in each of the angle, four quadrants, (c) Give the formula which expresses the periodicity = tan 6. (e) Which other of the sine, rnr) (d) Prove that tan (6
126. Review,
+
function has the same period as the tangent?
Follow the changes in the sine of an angle as the angle increases from o to 2 TT. (c) Do the (b) Do the same for the cosine, the in Follow the the same for reciprocal of changes (d) tangent, 2.
(a)
the tangent. 3.
(a)
Draw
the lines which represent the various functions of an its angle at the center of the circle) when the
arc of a circle (or of radius of the circle
is
taken as the unit of measure.
an angle in the third quadrant, (c) words secant, tangent, sine and cosine.
lines for
of the
=
+
(a)
t
sin
(
-
B)
=
sin
B,
cos (2
R+
B)
= -
Draw
these
Explain the derivation
= cos Prove geometrically that sin (R 0) cot 6 B being an angle in the first quadrant,
4.
(6)
6,
tan
(b)
(3
R-
6)
Prove that
cos B for every value of
B.
the principal functions of 30, 150, 210, (c) Give from memory the principal functions of 45, 135, from Give memory 330. (d)
225, 315. 5.
(a)
Show that
(- A) + tan (- A) -
sin
cos cot
(- A) _ (- A)
sin (QO
cot (180
+ A) + cos (270 - A) + A) + tan (360 - A)
TRIGONOMETRIC EQUATIONS
125]
+ *) cos R "" cos(2# + *)
^
Simnlifv P y
(c)
Find from the tables
(ft
tan (HIT
+
sin
(*
(
sin
*)
sin
~
(^
-I"*"
251
*) cos
+ *)
(*
+ *)
sin( 2
234, cos 342, tan 134
54', sin
967
45',
Y io/
\
Prove the addition theorem for the sine, for the cosine and (b) Give the formulas for sin 2 0, cos 2 6 and tan 2 0. ~~ Show that sin (x sin y) (sin # sin ;y). (sin x y) sin (# >0
6.
(a)
for the tangent,
+
+
(c)
(d)
Given cos #
=
22
j, find sin ~, cos
and tan - without the use
of
2
tables.
7.
Given the law of
8.
Express each of the following as a product: sin 7
By
9.
+ sin 15,
sines,
prove the law of tangents.
cos -
cos
2
^, 2
sin
A
+ sin 3 A + sin 4 A.
multiplying each side of the expression
5=i + cos x + cos 2 x + cos 3 x + and expressing each product on the show that
.
.
+ cos nx
.
y
by
right as a difference of
2 sin
2
two
sines
sin10. (b) (c)
(a)
What
meant by the
is
principal
Give the principal values of sin"" 1 1, sin" 1 Give the general values of the angles under
11.
(a)
What
is
(6).
meant by a trigonometric equation ?
the following equations, a sin x = b tan #. 12.
value of an angle? 1 1 i. 4, cos" J, tan"
sin
x
=
2 sin2
cos x,
x
Solve the equations: (a) sin (x (b) sin (a
(c)
sin (*
+ c)
cos x sin c
=
=
cos (a
+ x).
+ y) =
cos (*
-
x)
y)
=
cos
c.
Solve
(b)
+ 3 cos
2
x
=
5,
PLANE TRIGONOMETRY
2S2
[CHAP.XH
Solve the equations:
13.
+
sin 3
* cos x
=i
(a) sin 2
(c) sin
y
y
=
(b) i
3 sin y.
+ cos x =
(<0 sin
cos -
+ cos * = N/2. a sin x + b cos x =
x
c Show how to adapt the equation simultaneous the Solve of means (6) logarithms. to solution by = a, a;sina + jcosa = b. equations: #cosa + ysina
14.
(a)
15.
(a)
Define the inverse trigonometric functions.
(b)
Complete the following
=
16.
17.
tan"
1
(
)
(c)
1 Find tan (sin"
(d)
Show
(a)
Prove that
(b)
sin" 1 1
(a)
If/W
=
sec"
+
(
1
)
= 2X\/i -
x2
8
TT
+ sin"
1
tf
x).
= 45.
= I-
Solve the equation, cos" 1 *
+ cos"
1
(i
- x) =
cos" 1
(
=
= cor
1
tan" 1 % 1
csc- 1
(
)
sin (2 sin" x)
+ sin"
=
1
i), sin (tan"
1 that tan" 1
sin" 1 x
equalities,
x).
cos" 1 1
(
).
(
)
CHAPTER
XIII
TRIGONOMETRIC CURVES 126. Functions Represented by Curves. The student is probably already familiar with the fact that for every function of x,f(x), a
curve or graph may be constructed which is said to represent that function. This curve is merely the totality of all the points whose coordinates
x, y, satisfy
the relation y
f (x).
The
actual construc-
the curve representing a given function y = f (x) consists in plotting a limited number of points, using for abscissas properly chosen values of .r, and for ordinates the corresponding values tion of
of
= / (x). The
y determined by the relation y
smooth curve con-
necting the points thus plotted is said to be the curve or graph rep= /(#). resenting the function y
127.
The
y = mac
Straight Line,
Suppose the given function y
We of y.
+
c.
is
=
2
x
+
i.
give x certain values and compute the corresponding values Thus, if
-
-2, -3, -
i, i,
We now is 5
3, 5,
~
-
4, 7,
locate the point
and whose ordinate
5,
etc.,
9,
etc.
whose abscissa is
n, another
point whose abscissa is 4 and whose ordinate is 9, and similarly each of the points
(-
i,
-
(3, 7),
(2,
5),
(i, 3), (o, i),
i),etc.
We then connect the points thus located by a smooth curve, in this case This line is a straight line, Fig. 152. Fig. 152.
y
=
2
x
+
said to be the curve or graph repre2 x i, or, in senting the function y
=
i.
short, the straight line y
2
x
+
i.
253
+
PLANE TRIGONOMETRY
254
In the example just given, x
=
2
x+
i
may have one value as well
xin
as another,
consequently the line representing the will be indefinite in length and we must con-
say 1,000,000 as well as equation y
[CHAP,
2 or 3,
tent ourselves with drawing only a portion of it. What portion this is to be depends on the purpose in view, but unless there is a special
reason to the contrary, nearest the origin.
customary to construct the portion
is
it
In a similar manner every equation
form
= mx + c,
y
m
where line,
of the
and c are known numbers, is represented by some straight and c determining the direction and position of the line with
m
respect to the coordinate axes.
The
128.
v?
Circle,
9 y =
+
Suppose the given equation
a*.
is
then
y= = =
x
y ff=
5,
4,
3>
2,
o,
3,
4,
4v
-
=
3;
we
(4, 3), (4,
)
4-58,
4.89,
arate points (5>
-
-
-2,
i,
as before,
If,
V25-
for corresponding values of x
and we have
-
3,
-4,
4,
3,
and
y,
construct the sep-
3)> (3) 4), (3>
-
4),
and connect the points thus obtained by a smooth (2, 4.58), (2,
curve,
we
which
is
4.58), etc.,
obtain the circle, Fig. 153, said to be the curve or graph
+/=
representing the equation y? 2 y* or, in short, the circle x
+
=
25.
25,
In
x cannot be numerically greater for then y would be imaginary.
this case
than In
5,
like
where a center
is
Fig. 153.
a2
-f- 31*
=
25.
manner every equation of the form some known number, is represented by a at the origin and whose radius is a. is
circle
whose
TRIGONOMETRIC CURVES
130]
The Hyperbola,
129.
As another example,
as*
- y2 = a3
2SS
.
us construct the curve whose equation
let
#2
_
=
db
Z
y
_
is
25
Solving for y,
y
Corresponding values of x and
*
y
= =
6
5>
o,
y
=
10
8,
6.25,
-8,
-7, 4.90,
3.32,
25.
are
4.90,
-6,
+ o,
3;
2
7>
>
3.32,
*=-5,
V*
6.25,
In this case x cannot be numerically
less
~
748,
8.66,
9,
etc.
748,
etc.,
than
5,
for otherwise
y
is
imaginary. If we construct the separate points (5, o), (6, 3.32), (6, 3.32), etc., and draw a smooth curve connecting them, we obtain the two curves PQ, P'Q', Fig. 154. These curves constitute the two branches
known
of a single curve,
as the hyperbola,
more
specifically as the
equilateral hyperbola.
It
is
easy to see from the equation that the larger x is, the more x and y be equal, that is, the branches of the equilateral
nearly will
hyperbola approach the straight lines
PP' and QQ' drawn through the origin and making angles of 45 with the two directions of the #-axis respectively.
In
like
manner
be found that form
it will
every equation of the
some known number, is represented by an equilateral hypera determines the distance from bola. where a
the
is
origin
crosses
at
which
the #-axis.
the
This
is
Fig. 154.
& - y2 - 25.
hyperbola
known
as the semimajor axis of the
hyperbola.
The Sine Curve, y = sin a?. Let us now construct the curve representing y = sin x. Since x may have any value either positive or negative, 130.
the curve
representing the sine function will extend indefinitely in both direc-
PLANE TRIGONOMETRY
256
[CHAP,
xni
and left). Let us first construct that portion of the curve which corresponds to values of x between o and ^ TT, that is, to angles in the first quadrant. Referring to the table of natural sines,
tions (right
we
find the following corresponding values of x
In degrees, o, T
In radians, y
7T
=
-i
o,
18 o,
7T
7T
->
->
9
6
18
9
^
->
--i
>
7
7T
18
3
80, 90. 4 7T
7T
>
7T >
2
9
fractions,
we
will use
|
TT
i.
as the unit
jc-axis.*
- for a unit, 18 3 locate the points
With
O
7T
7T
0.17, 0.34, 0.5, 0.64, 0.77, 0.87, 0.94, 0.98,
In order to avoid awkward along the
2
y:
50, 60, 70,
10, 20, 30, 40,
j-
and
(o, o), />!
=
=
-J
unit,
-
=
J unit, etc., and
we now readily
9
0.17,
P2 =
.
.
(*, 0.34),
0.17),
Connecting these points by a smooth curve we obtain the curve Ri (Fig. 155), which is the sine curve for the first
OPiP^Pz
.
.
.
quadrant.
Fig. 155.
We may now
y
=
sin x.
easily continue the sine curve through as
additional quadrants as
we
many
choose.
Second quadrant. While x varies from \ TT to TT, sin x varies from to o. Moreover, since sines of supplementary angles are equal, ordinates equally distant from RI will be equal, that is, the curve will i
be symmetrical with respect to the ordinate at RI. This
will distort the
curve slightly, since 3
=
Hence, con1,0471+.
TRIGONOMETRIC CURVES
257
tinuing from RI, the curve will approach the #-axis, meeting (Fig. 156), whose distance from the origin is T.
it
at
#2
y
Fig. 156.
=
sin x.
Third and fourth quadrants. While x varies from TT to 2 TT, sin # negative, the numerical value being the same as when x varies from o to TT. Hence, continuing from R% the curve will descend below is
the tf-axis, reaching the lowest point at Ra, where x = J TT, and meeting the #-axis again at R*, where x = 2 w. The form of the portion of the curve below the #-axis will be like that above the #-axis when
revolved about this axis through an angle of 180. When x is increased or diminished by 2 TT, sin x has the same value as before, hence extending from RI to the right or from O to the left the curve repeats consists of
ORiRzRzRt
an is
itself indefinitely,
infinite
number
of
that
is,
the complete sine curve
waves or undulations
of
which
one.
OR*, the distance between two consecutive points at which the curve crosses the #-axis in the same direction, is called the wave
The greatest height of the curve, represented by the ordinate at RI, is called the amplitude of the curve. A curve like the sine curve, which repeats itself at definite inter-
length of the curve.
vals, is called
takes place
is
a periodic curve; the interval at which the repetition called the period. Likewise the function which such
a curve represents
The sine function 131.
is
called a periodic function.
is
a periodic function whose period
The Tangent Curve, y =
tan
is 2
TT.
a?.
To
construct the tangent curve for the first quadrant, we compute by means of a table of natural functions the corresponding values of x and v, as follows: f
In degrees, o, 10,
20,
30, 40, 7T
Iln radians,
y
=
o,
2
7T '
18
6'
9
5> 5?T *
18
70, 80, 90. 7T
3'
Z, *. 18
9
2
o, 0.18, 0.36, 0.58, 0.84, 1.19, 1.73, 2.75, 5.67, oo.
PLANE TRIGONOMETRY
258
[CHAP,
xin
Plotting the points obtained by using the #'s for abscissas and the corresponding values of y for ordinates, and connecting these points by a smooth curve, we obtain the curve OR, Fig. 157, which rep-
=
resents the equation y tan x for values of x in the first quadrant. Since tan | TT = oo the curve will not intersect the perpendicular at ,
\
TT,
but
will
it
approach
indefinitely.
y
Fig- iS7-
=
tan
DC.
Second quadrant. While x varies from \ TT to TT, the tangent varies oo to o, hence between 7^i and #2 the tangent curve will be from below the to those
Tan itself.
numerical values of the ordinates being equal
#-axis, the
and RI taken
between
+ *) =
in the reverse order.
x, hence beginning with R% the curve repeats (TT The tangent curve, therefore, consists of an infinite number
tan
of disconnected branches.
The tangent curve
is a periodic curve, the tangent function is a the function, period in each case is TT. periodic line like R\R\ or R-iR~\, Fig. 157, to which the curve approaches
A
indefinitely near without ever reaching
The
to the curve.
to the hyperbola.
lines
An
PP' and QQ',
asymptote
is
it,
is
Fig.
called
an asymptote asymptotes
154, are
a tangent to the curve at
infinity.
When the student has become familiar with the forms of the sine and tangent curves, he can readily sketch them from a very few points whose coordinates are known from memory. Thus, for the values 7T
7T
TT
" >
>
4
2
3
the corresponding values of the functions are sulting a table, namely sin
x
=
\/2 o,
0.5, 2
==
0.71,
\A* = 2
known without
0.87,
i,
con-
TRIGONOMETRIC CURVES
i3i]
=
tan *
^=
o,
0.58,
i,
A/3
o
The same remark
=
259
*-73>
to the sketching of each of
applies
the re-
maining trigonometric functions.
EXERCISE 57 1.
Construct the cosine curve.
2.
Construct the cosecant curve.
This curve
(Suggestion.
most readily sketched from the
is
=
curve by remembering that y
esc x
)
sin
3.
Construct the secant curve.
(
y
=
x I
sec x
5.
y
=
Construct the cotangent curve. ft Construct tan""
1
y
=
)
=
cot x
=
V
)
tan*/
equations
are
y
sin"" 1 x,
x.
If
(Suggestion. 6.
y
(
whose
curves
the
= cos x/
\
4.
sine
=
cos x
=
y
sin""
-
sin (
for every ordinate
1
x then x
+ x\
on the
=
y
From
sin y, etc.)
this relation it follows that
is an equal ordinate on the abscissa of the former dimin-
sine curve there
the cosine curve whose abscissa
is
ished by \ TT, that is, for every point on the sine curve there is a point on the cosine curve, the latter being ^ TT to the left of the former. Thus
we
see that the cosine curve
is
merely the sine curve shifted a
dis-
tance \ TT to the left. By a similar reasoning show that the secant curve is the cosecant curve shifted a distance \ TT to the left. 7.
y
=
cot x
=
cotangent curve distance of %
an angle 8.
y
of
=
TT
tanp + x\
may
to the
From
this relation
show that the
be obtained by shifting the tangent curve a and revolving it about the #-axis through
left,
180.
cos x
=
sin (-
=
x )
sin (x
^
From
this
relation
J.
show that the cosine curve may be obtained by
shifting the sine
PLANE TRIGONOMETRY curve a distance of
|T
through an angle
180.
of
Fig, 158.
132.
The
and revolving
to the right,
The
[CHAP,
it
xni
about the #-axis
Six Trigonometric Curves.
Sinusoidal or Simple Harmonic Curves,
y= a
sin
(k&
+
)
=
a sin #, a being constant. Each ordinate of the curve (a) y representing this equation is a times the corresponding ordinate of the curve y = sin x. The required curve is the curve obtained by lengthening or shortening (according as a is greater or less than
y
Fig. 159.
=
a sin
*.
unity) the sine curve in the direction of the
length unchanged.
amplitude
is
a.
The curve Fig.
is
a
;y-axis,
159 shows three sinusoids of
lengths and amplitudes J,
i,
2
wave whose equal wave
leaving the
sinelike curve (sinusoid)
respectively.
TRIGONOMETRIC CURVES
132]
y
(b)
=
a sin kx, a and
The curve
representing this equation has the same amplitude as a sin x, but the wave length differs, for it crosses
when kx
that
is,
The
being constants.
ft
=
the curve y the #-axis
261
=
when
o, O,
TT,
2
7T
27T
TT,
etc., .
o,
two consecutive crossings
distance between
the same direction
is 2 ir/k,
whose wave length have
is 2
and the equation
that
w/k.
If
X
=
f
y
=
2JC, *
is,
we denote
may
ft#
lengths
y
(c)
TT,
=
y
~
shows three sinusoids TT,
3
TT
be written
X
+
e), a,
ft
and
+ =
O,
The
(2 irx/\).
of equal amplitudes
and wave
being constants.
where
2
7T, 7T
is,
a sin
representing this equation crosses the oc-axis where ft#
that
we
respectively.
a sin (&#
The curve
X,
A
a sin
Fig. 160.
2
a sinusoid
rom
lA
1 60
is
wave length by
this
a sin^?, X being the wave length. X
y
Fig.
of the #-axis in
the required curve
~
6
2
*"-*'
ordinates of the highest
etC.,
7T,
7T ,
etc.
and lowest points are a and
a
respectively, and the distance between two points where the curve When x o, crosses the #-axis in the same direction is 2 ir/k.
=
y
=
a
sin
e.
The
curve
is
a sinusoid, amplitude
a,
wave length
2 7r/ft,
PLANE TRIGONOMETRY
262
[CHAP, xin
but instead of passing through the origin, it crosses the ;y-axis at a distance a sin above the origin. Fig. 161 shows three sinusoids of
Fig. 161.
equal amplitudes and wave lengths and e = o, \ The third of these curves has for its equation
y
=
a sin kx {
+
=
)
IT
and \ w
respectively.
a cos kx,
which shows that every cosine curve is a sinusoid. Sinusoids are extensively used in physics to represent the motion of vibrating strings, tuning forks, and other vibrating bodies emiting musical sounds. For this reason they are often referred to as harmonic, or, more strictly, as simple harmonic curves. 133* Angles as Functions of Time. In many physical problems which give rise to equations of the form y = a sin (kx c), the independent variable represents not an angle, but the time of an
+
action or motion.
Fig. 162.
Let /, Fig. 162, be the initial position of a point moving in the circumference of a circle (radius OP = a) with a constant velocity. Let P be the position occupied by the moving point, t seconds after leaving /, and let w represent the angular velocity of P, that is, the = otf, angle described by OP in one second of time. Then angle IOP
TRIGONOMETRIC CURVES
i33l
263
and angle AOP = coJ + e, where e is the angle which 01 makes with some fixed diameter as AOA' Furthermore, let represent the then projection of P on BB', the diameter perpendicular to A A
M
.
1
,
OM
=
a sin
(co/
+
The curve
c).
=
y
a
sin
(coJ
+
(i)
c)
obtained by using equal distances on the axis of abscissas to represent equal intervals of time, and using for ordinates the distances
OM, which
correspond to various values of
/,
enables us to see at a
M
at any given time t. glance the position of If T represents the time required by P to complete one revolution
we have o?r
so that (i)
may
also
=
2
TT,
from which
o>
=
>
be written (a)
the periodic time, or period of oscillation of P and M. represents the time required to complete one revolution, reciprocal i/T will represent the number of revolutions completed
where
T
its
is
T
Since
in a unit of time. If the
oscillation.
This value i/T is known as the frequency of the frequency is denoted by v, equation (2) assumes
the form
y = asin(2Virt+).
(3)
M
in Fig. 162, that is, any motion Motion like that of the point which can be expressed by an equation of the form y = a sin (bt c) The motion of vibrating tuning forks, of is called simple harmonic. water waves, of an oscillating pendulum, of a galvanometer needle, of alternating currents, of sound and light and magnetic electric
+
waves, are familiar examples of simple harmonic motions.
EXERCISE 58 Plot the following curves: i.
y
4.
y
sins*.
=
-I
sin 3 *.
2.
5.
;y
y
=
sin--
=
sin*
3.
+
-
6.
;y
y
= =
3sin-.
2
PLANE TRIGONOMETRY
264 Construct
7.
wave
sinusoids
the
having
lengths: (a) (b)
xm
[CHAP,
and
amplitudes
following
= 3.5. wave length = -
Amplitude =1.5, wave length Amplitude
=
Amplitude
=
0.25,
2 (c)
=
wave length
i,
STT.
Write the equation for each of these curves. Ans.
(a)
y
=
1.5
sin^^,
(b)
4? =
sin
4*,
y
(c)
=
sin--
4
7
Show
8.
that the equation
y
may
= /=
Plot the curves (a) y
9.
is
=
a cos
(bt
+ c)
be used to represent harmonic motion as well as the equation y = a sin (bt+ c).
10.
Plot the curves (a)
11.
A
sin2 x,
y
(b)
=
tan 2
=
tan
2
sin x,
(b)
y
x. x.
point moves in the circumference of a circle whose radius initial position whose angular distance from the
from an
8.5,
right-hand extremity of a horizontal diameter is 15, with a uniform velocity such as to complete a single revolution in 54 seconds. Write the equation between y and t, where y represents the vertical distance of the point from the horizontal diameter at
time
any given
t.
12. A piece of paper is wrapped around a wooden cylinder and then an oblique section is made by sawing the cylinder in two. If the paper is now unrolled and laid flat, its edge will
Prove
form a sinusoidal curve. Ans.
where
The equation
PR =
y,
OR =
it.
of the curve x,
AT =
a,
is"
y
and
=
a sin 2
OC =
r
>
=
the radius of the cylinder.
Fig. 163.
134. Composition of Sinusoidal Curves.
Curves representing an
equation of the form
y = a t sin (6 t a? + may
c t)
+ a2 sin (6 2 + c 2 + a?
)
etc.,
be readily constructed from the component curves yi
=
ai sin (bi
x
+
ci),
y*
=
a* sin (b*x
+c
2),
etc.,
TRIGONOMETRIC CURVES
265
for since
=
y
yi
+ y*
the ordinate of any point on the required curve is found the corresponding ordinates of the component curves.
EXAMPLE
Plot the curve y
i.
Plot separately the
Solution.
=: sin x, yi
y
2
=
=
+
sin x
by adding
cos x.
two curves
[curve (i), Fig. 164],
cos x, [curve (2), Fig. 164].
Fig. 164.
=
+
The required curve y [curve (3), Fig. 164] is then y
+
.
portant to remember that for points below the z-axis the ordinates are negative; thus, if x OG, y GQ GQi GQ*. GQi is
=
=
=
+
Now
GQ2 being the longer of the two, their positive but GQ% is negative. will be negative. that is, sum, GQ, algebraic It should be noticed that certain points, as, for instance, those for After the shape which y = 2, or o, may be located at sight. i, known, these points
of the curve is
suffice to
sketch the required
curve.
Each of the sine curves y\ = sin x and yz = cos x has the period 2 w and the resultant curve y = sin x + cos x is another sine curve with the period 2 TT. This may be shown analytically as follows: y
=
sin
2
x
+ cos x = sin x + sin
sinsn
4 2
cos \4
cos (-
x\ by
[
-
(-
x J,
formulas in Article 113,
/
x\
=
x/2
sin(#
+'Y
PLANE TRIGONOMETRY
266
[CHAP,
xm
=
sin x + cos x is a sinusoid having an ampliand crossing the #-axis at the point x = ITT. The method employed in constructing the curve in the preceding example applies equally well to the compounding of any number of
that
is,
the curve y
tude N/2, period
curves.
braic
The
2
TT,
ordinates of the resultant curve are always the algecomponent curves.
of the ordinates of the
sums
EXAMPLE
=
Plot the curve y
2.
sin
x
+ sin 2 x.
Construct separately the two curves
Solution.
y2
= =
y
=
3^1
then
[curve
(i)],
sin 2
x [curve
(2)],
yl
y%
sin#,
-f-
y
=
sin X
yields the curve (3), Fig. 165.
+ SHI 2 #
This curve has the period
2
TT.
Fig. 165.
EXAMPLE
=
Plot the curve y
3.
sin
x
Solution.
yi
=
SB ;y2 SB 3>3
y
=
sin a
gives curve (i).
sin 3 x gives curve (2),
i sin 5 x gives curve yi
+y +y 2
z
(3),
gives curve
This curve also has the period
V
2
(4).
TT.
,C
Fig. 166.
+ % sin 3 x +
sin 5
:
TRIGONOMETRIC CURVES
135]
267
136. Theorem. The resultant of two simple harmonic curves having equal wave lengths is another simple harmonic curve having the
same wave
length.
Let the given curves be
Proof.
X being the common wave length.
(Art. 132.)
Now y\
*
y%=
0i sin ( ~
02 sin
+ a\ cos ( ^~ x
x\ cos c\
(~ x
+ 02 cos
cos c2 J
j
#
sin c\ 9
sin J
[
<^,
therefore
y
=
=
yi
+
3^2
(0i cos
ci+
02 cos c2 ) sin
a cose sn sin
(^ ^ + J
^ J
[
+ a sin c cos
f
^
(ai sin c\
+ 02 sin c
2)
*
cos
J
[
jc
j
(i)
where a cos c a sin
c
= =
a\ cos ci
ai sin c\
+ 02 cos + 02 sin
Dividing the second of the equations
01 COS
and taking the sum 2
a (cos
2
c
2.
by the
first,
gives
(3)
Ci+
of their squares
+ sin c) = a? (cos ci + sin Ci) + 2 a^ (cos c\ cos c + sin c\ sin ^ + 02 (cos c + sin c -c a = 0i + 02 + 2 aiOz cos 2
2
2
2
2
2
2
2
2
2
2)
or
(2)
^2,
2 ),
2
(ci
2 ).
(4)
Equation (i) shows that the resultant curve is a simple harmonic curve whose wave length is X. The amplitude a of the resultant curve
is
given
by
(4),
and the constant
c
by
(3).
PLANE TRIGONOMETRY
268 136.
Fourier's
Theorem.
[CHAP,
Before leaving the subject of
xm
har-
monic curves, we will state in simple language and without proof a most famous theorem, which in the language of Thomson and Tait * " is not only one of the most beautiful results of modern analysis, but may be said to furnish an indispensable instrument in the treat-
ment
of nearly every recondite question in
modern
mention only sonorous vibrations, the propagation of
To
physics.
electric signals
along telegraph wires, and the conduction of heat by the earth's crust, as subjects in their generality intractable without it, is to give but a feeble idea of
Any
its
importance."
arbitrary periodic curve can be considered the resultant of a
sum
of simple harmonic curves, and can therefore be expressed by an equation of the form
y
=
#1 sin (bix
+
Ci)
+ 0% sin (b& + c + etc. 2)
The importance of the theorem lies in this, that every periodic phenomenon whose changes can be measured can be represented by a periodic curve.
nomenon can be
Fourier's theorem shows
how every such
phe-
resolved into a series of simple harmonic motions.
EXERCISE 59
=
sin x cos x. Plot the curve y analytically that this curve is a sine curve, whose amplitude \/2 and which crosses the #-axis at the distance \ TT to the right of 1.
Show is
9
the origin. 2.
Construct the resultant curve of which cosfjc
+ -J>
and
yz
=
cosf x
)>
are the components, and show that the equation of the resultant cos x. curve may be written y
=
Construct the following curves: 3. 5.
7.
= sinac + $sin3#. y = sin# + Jsin4#. = cos2# + 4cosff. y x = tan~ (i + Vy) + y
l
9.
*
4.
y
6.
y
8.
y
tan"
1
(i
= sin2#+ = 2 sin* = sin*
-
sin
33.
sin 2x.
sin3&
+ isin5#.
Vy).
Elements of Natural Philosophy, Second Edition, Chapter
I.
TRIGONOMETRIC CURVES
i37l
269
two curves
Plot the curve resulting from compounding the
10.
yi
=
2
+ gj,
sinfs
and show that the equation
=
y
where
a
= V'I,
Show
ii.
=
3 sinf*
|Y
of the resultant curve is
asin(#+c),
=
tan" 1
f
--
3
.
that
+ & cos = Va +
a sn in
2
2
12. The force by the equation
where
c
yz
(inertia force)
sin
+ tan"
f
-
1
]
on the piston of an engine
which the crank arm makes with a
6 is the angle
is
given
fixed direc-
R
the length of the crank arm, L the length of the connecting tion, rod, and FQ a constant (the ideal centrifugal force at the crank pin center).
13.
5,
F
Plot the curve, showing the value of
tion of the crank,
when F =
600,
R=
L=
15,
for
any given
the distance of the piston of a steam engine from
=
treme position corresponding to 6
o,
is
posi-
45. its ex-
given approximately by
the equation s
where R,
= R (i -
cos0
+
sin
2
A
L and 6 have the same meaning as
curve showing the relation between
s
in Example 12. Plot the and 0, when R = 10 and L = 20.
The Logarithmic Curve, y = Iog 10a?. With the aid of a table of common logarithms and 137.
of the fundamental properties of logarithms, corresponding values of x and >:
*= y
=
i,
o, 0.30,
*= y
= -
i,
0.30,
In any case,
4,
3,
2,
if
0.48,
0.60,
i
i
- 0.48, x
=
~ y
n
,
0.60,
=
io
5>
0.70,
0.70,
log n.
2,
o.i,
-
i,
-
a knowledge
find the following
loo,
i,
i,
-
we
.,
.,
oo.
.,
oo
.,
o.oi,
.,
.,
2,
.,
.,
.
o.
-
oo
.
PLANE TRIGONOMETRY
270
ICHAP.
xin
These values enable us to plot the portion of the curve shown in They also give us a definite idea of the shape of the curve
Fig. 167.
beyond the
To
limits of the figure.
the
right the curve diverges more and more from the #-axis as x increases, below
the
the
#-axis
curve
approaches the as x ap-
more and more nearly
y-axis
proaches o; that is, the logarithmic curve y = Iogi # has the j-axis for an
asymptote. Fig. 167.
The Exponential Curve, y =10*. Taking the common
138.
side of the equation y
logio?
y
= =
y
(a)
A
by interchanging the #'s and by using the ordinates of
is,
in Fig. 168.
e^", where k is any constant, the natural of base the system of logarithms. 2.718+,
=
e
= log?.
The Exponential Curves, y =
139.
and
orx
x,
the logarithmic curve for abscissas and the abscissas for ordinates. The result-
10*.
shown
is
logio#
that
y's,
ing curve
logarithm of each 10*, we obtain
=
This shows that the exponential curve iox may be obtained from the curve y
(0,1)
Fig. 168.
=
y
=
e*.
Let us
first
consider the special case f or k
=
i.
y=
e? may be gathered general idea of the shape of the curve sets of corresponding values of x and y:
from the following x
=
y=i, =
x y
From
=
J,
o,
(2.7)*
=1.6,
|,
-~ = 1.6
2,
i,
2.7,
(2,7)2=7.4, 2,
i,
=
0.61,
oo
.,
=
0.38,
0.14,
,
.,
oo,
.,
oo.
.,
o
.
7.39
2.7
be sketched as in Fig. 168. If the curve is to be plotted with greater accuracy than the above figures will permit, it is best to employ logarithms. For, on these values the curve
may
TRIGONOMETRIC CURVES
139]
member
taking the logarithm of each
of the equation
have logio
y
=
oc
e
log
=
x
logio
271
2.718+
=
log
y y
The
= = =
.,
.,
o,
o.i,
0.2,
.,
.,
i,
2,
o,
0.04343,
0.08686,
.,
.,
0.4343,
0.8686,
i,
1.1052,
1.2214,
.,
.,
2.718,
7.390,
intervals for x
=
e*,
we
0.4343 x.
Assigning to x in succession the values o, o.i, 0.2, we obtain by means of a table of logarithms,
x
y
i, 2, 3, etc.
3, etc.,
1.3029, etc., 20.085,
etc.
be chosen as small as we please, and the
may
on the accuracy desired and the
table extended at will, depending
extent of the region for which the curve is to be plotted. The values of y corresponding to negative values of x are the reciprocals of the values of y when x is positive. (6)
=
y
When
x
e
kx
k positive.
,
=
o,
=
y
i,
no matter what value k has, the
therefore,
curve passes through the point (o, i) Fig. 169. x Let k > i. So long as x is positive, e > i, and therefore d* x x k > e that is, to the right of the y-axis the curve y = e?x , k > (e*) ,
lies
above the curve y
=
e*,
and
it
will diverge
from
it
= i,
the more, the
= (ex) k e*, For negative values of x, k i, greater the value of k. kx that is, to the left of the y-axis the curve y = e , k i, will lie between the curve y = e? and the #-axis and will converge the more
>
rapidly to this axis the greater the value of of k the curve
may
chosen points.
If
k.
#x >
<
For any given value
be roughly sketched by means of a few properly greater accuracy is required, we first compute a
values from the relation logio y = 0.4343 kx. Considering in like manner the cases when k i, we find that the
frable of
<
=
between the curve y = ex and the straight line drawn parallel to the #-axis through the point (o, i) and that the curve will approach this straight line more nearly the smaller the value of k. Fig. 169 shows the curves for the values k = k = J, k = i, k = 2 and k = 3. curves y
e
kx
,
k
<
i, lie
,
(c)
y
= #x
,
k negative, or y
The curve y -
e~ kx (i) For y = e' kx
y = eP (2). a positive value of x
x,
=
e~ kx k positive. ,
most readily obtained from the curve = (0 )~ x from which it is plain that for the ordinate y of (i) will be the same as the is
fc
,
PLANE TRIGONOMETRY
272
[CHAP,
xni
value of y in (2) for the corresponding negative value of x, and vice versa, the y in (i) when x is negative will be equal to the y in (2) when x is positive. This means that the two curves y = e~ k * and y = **, k being the same in both cases, are symmetrical with respect to the y-axis, so that either one being given, the other may be traced from it without computing anew the coordinates of its points.
Two and
such curves have the same relation to each other as an object a mirror. For this reason either of the two is
its reflection in
said to be the reflection
the curves y
=
*J
on the y-axis
for the values *
Fig. 169.
of the other.
=
y
3,
=
2,
i,
Fig. 169 J,
shows .
eky .
140. The Compound Interest Law. In physics, chemistry and various branches of engineering, related quantities occur which are subject to laws which may be expressed by the formula
y or
by the
=
= k
fact, it
(i)
log*. a
(2)
equivalent formula
* In
aek *,
can be shown that whenever two quantities are so related,
that the ratio between their changes is always proportional to one of the quantities, the relation between them may be expressed by
where x and y are the quantities in question, and k and a constants which depend on the rate of change and the initial values of x and y. The amount of money due at any time on a sum of money put out at compound interest, the interest being added to either (i) or (2),
the principal not at stated intervals but as fast as it accrues, varies with the time according to this law. For this reason the general law
TRIGONOMETRIC CURVES
HO]
273
by (i) or (2) is commonly known as the compound The student of science will meet numerous examples
expressed law.
compound of
A
interest law.
interest
of the
few simple examples are given in the set
problems which follows.
EXERCISE 60 Plot the following curves: X i.
y
=
log e x.
2.
y
-4.
y=io
7-
y
=
io.
y
=
13.
Given the
'.
3*2~~
y
5.
.
io
=
x Jloge--
2 .
= 3-*. x ii. y = 2 e curve y = ekx trace ,
computing the coordinates of
14.
=
io~*. -
6.
9-
12.
the curve
= 3.io 2 y = 2 *. x 2 e y = &x without y = j
.
;y
.
its points.
The required curve (Suggestion. the given curve.)
is
the reflection on the #-axis of
a principal of P dollars, put out at / years, the interest being added
The amount A due on
compound
y
3.
y
-
2>e
=
interest at r per cent for
to the principal as fast as
it
accrues,
is
given
by
the formula
rt
A = Pe. Plot the curve showing the P = loo and r = 5. (Suggestion.
Use
responding values of 15.
The work
amount due
different values of
A
/
at
any given time, when
for abscissas
and the
cor-
for ordinates.)
W due to the expansion of steam in the cylinder of
a steam engine, while expanding from a given volume to a is given by the formula
W = a log
e
F-
volume 7,
6,
where a and b are two constants depending upon the initial volume and pressure. Plot the curve showing the relation between V and
W
6
for
=
any volume from
V=
i
to
V=
4,
when a =
15,000,
and
0.
Newton's law of cooling. The difference 6 -between the temperature of a body and the temperature of the medium surrounding it is given by the formula 16.
-
be-**,
PLANE TRIGONOMETRY
274 where a and
and the
=
any time 141.
t
up
=
when
If
temperatures.
60 seconds, 6
Put
(
6
=
50, and when
+ e ~~) in which e
consider the special case c = = ~x e*
3/1
o,
to 60 seconds.
The Catenary, y =
first
=
t
25, plot a curve showing the temperature 6 at
the natural system of logarithms and
We
and
^2
,
the base of
any positive constant.
c
=
then
is
3;
i,
y
=
=
^
\ (y\
(e*
+
+ e~*).
3/2),
which shows
that any ordinate of the required curve is equal to half the x the corresponding ordinates of two curves y\ = e and y% If,
therefore, these
two curves are drawn
Fig. 170.
first,
is,
= "
each point
P2
of
e~
x .
222
AP, equal
.
be observed that
y
Pi and
=
The Catenary.
AP
that
sum
as in Fig. 170, the
required curve is easily drawn by taking each ordinate, as to half the sum of the ordinates A PI and Z It should
xni
depending upon the nature of the body
b are constants
initial
[CHAP,
'"=y* +
^~
P bisects the line joining
two corresponding points
.
We now pass to the
general case,
y
=-c
\e
c
-{"
e
C
J
may be written
X ~
y
which becomes where
e
c
(2)
TRIGONOMETRIC CURVES
142]
From
this
we
see that
if
the coordinates
27S
#', y'
of
any point on the
(2) be multiplied by c, the resulting numbers are the coordinates x, y of some point on the curve (i). The curve (i) is therefore
curve
merely some magnification of the curve (2). The curve in Fig. 170 may therefore be taken to represent any equation of the form (i), provided the proper scale is employed in the construction. It is only necessary to let each unit of length along the coordinate axis represent c units. I *
c The curve y = -{e c 2 \
formed by a rope or
+e
_ *\ c
the catenary. )is called
It is the curve
/
flexible cable
suspended between two points. weight of a unit length
c is the ratio of the horizontal tension to the
of the rope or cable.
142.
The Curve of Damped Vibrations, y = ae"** sin
Put
=
ae,-kx
then
y
= :
y\ 3>2 3/13/2
=
sin (ex
(ex
+ a).
+ d),
+
kx ae~** ae~ sm (ex + d), sin
(2)
(3)
of is, any ordinate of the required curve is equal to the product the corresponding ordinates of the two curves (i) and (2).
that
Fig. 171.
Let the curves Fig. 171.
Now
(i)
The Curve
and
of
(2) first
Damped
Vibrations.
be constructed separately, as in
observe:
Since y2 is less or at most equal to i, y of (3) is less or at most equal to yi, that is, the required curve lies below the curve (i). zero also, hence (b) Whenever yi or y^ equals zero, y of (3) equals (a)
the required curve crosses the #-axis at the points NI, N*, Nt 9 etc., at which the curve (2) crosses this axis.
N*
9
PLANE TRIGONOMETRY
276 (c)
Whenever
(d)
Whenever y2
H
=
[CHAP,
xra
s i, as at H\, etc., y of (3) equals yi, hence the required curve touches the curve (i) (since it cannot cross it) at the points K\, K*, etc.
y*
=
i,"as at
K
us to locate the points z K, The shape of the curve (3) is ,
,
=
Hz, #4, etc., y y\. of the required curve.
This enables
now apparent. It is a wave curve of constant wave length, but the amplitude of the successive waves The rapidity with which the amplitude decreases diminishes. depends on the value of the constant k in
(i).
This constant k
is
known
as the logarithmic decrement of the curve. Like the exponential curve and the sine curve, this curve finds
frequent applications in science. While the sine curve represents free vibrations, that is, vibrations not retarded by friction or otherwise, the curve
=
ae~ k * sin (ex
+
d) represents damped vibrations, vibrations suffering resistance of some kind. A pendulum vibrating in air or water, waves propagated in a viscous fluid and oscil-
that
y
is,
from an
latory discharges of
electric condenser, are familiar
examples
vibratory motion.
damped
EXERCISE 61 Plot the curves: j.
y
=
4.
y
=
written y
~*cos _e *~~ e
=~
x.
_
2.
(
.
y'
=
y
?__
s-
e
\~
y
written y
e
.
ex sin
_
-*
x.
The given equation
may be
\ )
/
See suggestion under Problem
4.
e
""
e 6-
*
2
y
^
=
Suggestion.
\
where
,
y
x
=
C
e
j
l,
*""*
C
-x
-
(Suggestion.
The given equation may be
\
where yi
_
=
(Suggestion.
This equation
is
the reciprocal of
that in Problem 6.) 8.
The
six functions, x
e*
-
e~ x
TRIGONOMETRIC CURVES
are
known as the Show that
277
hyperbolic func-
tions.
yf
-
2 3>6
Compare tions,
Fig.
-
2
3>i
yJ
=
i,
=
i.
the graphs of these funcThe graphs are 172.
F
in correspondence with 72. The tethe suffixes of the /s in their equations above.
numbered
9.
two
A
J
Six Hyperbolic Curves,
one pound to the foot, is suspended between under a tension of 100 pounds. Plot the curve which it piers cable, weighing
forms. 10.
The displacement
equilibrium, at the time
where a
of the t
end
seconds
is
of a spring,
given by
from
its
position of
the equation
were no
the amplitude of the vibration of the string if there friction, k represents the effect of friction retarding the vi-
bration,
and
is
T
is
the time of a single vibration. Plot a curve showmoment during the first ten seconds, if
ing the displacement at any
CHAPTER XIV TRIGONOMETRIC REPRESENTATION OF COMPLEX QUANTITIES Numbers.
143. Imaginary
we obtain x
V
=
_
#+
is
any
real
i
+
a2
V
i as
V
iz
=
$= that
i, it i,
i
solve the equation
o,
form
O,
its
=
solution
a
,
a factor.
This new number denoted by the letter * and Since
=
number, has for a
which contains
we
=
Similarly, every equation of the
i.
#2
where a
If
i,
is
whose square
is
is
i,
commonly
called the imaginary unit.
follows that
4
i,
$ =
^,
i
6
i,
j
7
=
/,
**
=
i,
etc.,
is,
Zfoery integral
i, according as i, or power of i is equal to i, i, power when divided by 4 leaves the remainder o, /,
the exponent of the 2, or j.
In symbols
^n^ Numbers
4n+l x>
like
i,
22',
^^ 5^',
4n+2
= -
r,
f4n
- ^.
-> ^/a-i, etc., are called imaginary 2
numbers or
Every imaginary number multiplied by some real coefficient.
quantities.
consists of the
From the rule imaginary unit i for the powers of i it follows that every even power of an imaginary number is a real number, every odd power of an imaginary number is again an imaginary number. .44. Geometrical Representation of Imaginary Numbers. Every positive or negative real number x may be represented geometrically by a distance on the x axis measured to the right or left according as x is positive or negative. Equal distances measured
278
COMPLEX QUANTITIES
J 44j
279
same direction represent equal numbers, but equal distances measured in opposite directions represent numbers which are equal in magnitude but opposite in Thus, in Fig. 173, if the points sign. in the
X-2, -X-i, O, Xi, X% are taken at equal intervals, and OXi represents unity, then each of the segments
OX and each
i,
XiXz, X-zX-i,
X-iO
represents
+i,
X-iX-z, X^Xi, XiO represents
i.
of the segments
OX -i,
In general, if the segment represents the number a, QP repre= a ( a) = o. It sents the number a, and therefore PQ
PQ
+ QP
follows that a line segment sent the same real number
PQ
on the
+
continue to repreif it is moved along the #-axis from one position to another, so long as its direction remains unchanged. If all the line segments have the same initial point O, as OXi, Z ff-axis will
OX
,
O-Y-i, OJf-2, Fig. 173, the terminal points Xi, X* X-i, X- z will represent the numbers quite as well as the segments themselves. 2 are represented Thus, if OXi= i, the numbers i, 2, i, equally well
Xi,
by the segments OXi,
X
2,
OX
2,
OX-i,
OX- Z
,
and by the points
X-i, X-2, respectively.
Likewise every positive or negative imaginary number iy may be by a distance on the y-axis measured
represented geometrically 1
upwards or downwards according as y is positive Equal distances on the y-axis meassame direction represent equal imaginary numbers, equal distances on the y-axis measured in opposite directions represent imaginary numbers which are equal but opposite in sign.
or negative. ured in the
Thus, in Fig. 173, if the points F_ 2 F-i, 0, FI, are taken at equal intervals, and OY\ is taken ,
F2 in
length equal to OXi, then each of the segments
+
and i, OFi, FiF2 F- 2 F_t, F_iO represents each of the segments OF_ F-iF_ 2 F2 Fi, FiO represents - i. In general, if the segment RS represents the number ai, SR repre,
t,
,
= o. ai + ( and therefore RS + SR ai) to continue It follows that any line segment on the v-axis will represent the same imaginary number if it is moved along the y-axis from sents -the
number
ai,
one position to another so long as
its
direction remains unchanged.
PLANE TRIGONOMETRY
280 If all the line
segments have the same
[CHAP, xiv
initial
point 0, as OFi,
OF-i, OF_2, the terminal points FI, F2 F-i, F- 2 will represent the imaginary numbers quite as well as the segments themselves. 2 i are equally i, Thus, if OFi represents i, the numbers i 2 i, well represented by the segments OFi, OF2 OF-i, OF- 2 and by the
OY2
,
,
,
,
F2
points FI,
,
F_i,
F- 2
,
respectively.
,
the points on the jc-axis and ^-axis are used to represent and imaginary numbers respectively, these axes are referred to as the axis of reals and the axis of imaginaries respectively.
When
real
The
reason for choosing the axis of imaginaries at right angles to is found in the following considerations: 2 = i may be put in the following form, i equation
the axis of reals
The
i. Now the a mean proportional between +i and the perpengeometric construction for a mean proportional gives dicular OFi, erected at O, to the semicircle constructed on X-iXi
that
is, i is
as a diameter.
Hence,
OXi
if
V+
OFi will represent Or, we may reason as i,
successive multiplications
by
Now
i.
OXi = XiO
should have the is,
i
represent
145. If
we
X
= iX
i*
have the same
i
i
represents
=
that
i,
is,
two
effect as multiplication
multiplying any line segment = OX-i that is, multiplying by
OXi by
y
OX\ through an
of turning
that
follows:
by
+i, and OX-i
represents i or i.
X
i
i
i
gives
has the
effect
angle of 180, hence multiplying by i OXi through half this angle or 90,
effect of turning
OXi = OFi,
so that
if
OXi
represents
i,
OFi must
i.
Geometrical Representation of Complex Numbers. 2 solve the equation #
we obtain
x
that
x
is,
= =
4#
3 V
2 2
+
+
3
i,
13
=
o,
i,
or 2
-
3
i.
Each of these numbers consists
of
two parts, a real part
+
A
number
or
like 2
+
+
2,
or
2
and an
3 i, 3 i, 3 3i imaginary part which consists of two parts, one which is real and the other imaginary, The general form of a complex number is called a complex number. real numbers, a and b may be positive b are a and where is a bi, i.
+
or negative, integral or fractional, rational or irrational.
COMPLEX QUANTITIES
H6]
A complex number a segment joining the origin
+ bi
281
represented geometrically by a line whose coordinates are a
is
to the point
+
and
b. Thus, the complex number 2 3 i is represented by the line segment OP lf Fig. 173, obtained by joining the origin O to the point Pi, whose coordinates are 2, 3. Here also the direction of the line
segment as well as 3,
,
4
the same numbers. resents 2
-
3
2
+3
length is to be considered. The line-segments are equal in length, but they do not represent OPi represents the number 2 3 i, OP* rep-
its
OP2 OP OP
OPi,
+
OP
3
i,
2
represents
3
and
i,
OP 4
represents
*
Two directed line segments which have the same length and the same direction are considered equal, and may be taken to represent the same number. It follows that a line segment will continue to represent the same number if it is moved parallel to itself. Thus, OP i, X- 2 F3 P3O, -3X2 being parallel and equal in length, all represent the same complex number 2 + 3 i. "If the directed line segments all have the same initial point O, as ,
OP OP
OP2 3, 4 the terminal points PI, 2 3 P 4 may be taken to represent the complex numbers as well as the directed segments. In this sense we may speak of the points PI, 2 , 3 4 as representOPi,
,
P P
,
,
,
P P P ,
ing the numbers 2
+3
2
+
3
2
3
2
3 i, respectively. possible real numbers a ; if a o, we get all possible imaginary numbers bi, so that the complex numbers a bi include as special cases all real and imaginary numIf in
a
+ bi
=
we put
*,
b
=
o,
we
*',
get
i,
all
+
bers.
'146. Trigonometric Representation of Complex Numbers.
P (Fig. 174) represent the point x iy. denote the length OP and B the angle which makes with the #-axis. Then
+
Let
Let
OP
r
as
Fig. 174.
ar?
r (cos
+ i sin 6)
number x+iy,
= r cos 6,
y = r sin 8,
is
+ iy =
called the
which
r (cos 6
trigonometric
+ i sin 6). form
complex
is
P
and
called the argument or amplitude of the
is
of the
(2)
always positive, being the distance of from the origin 0, is called the modulus or absolute value, r,
the point 6
(i)
and
complex number
PLANE TRIGONOMETRY
282
From
(i)
[CHAP, xiv
we obtain as
The equations (3) enable us to express any complex number in = 3 4 i, we find the trigonometric form. Thus, if x iy
+
+
8'
+ *sin 53
so that
3
+ 4* =
5 (cos 53
+ iy 3 4 i, = 3, y = 4, 2 r = VV+C-4) = 5> cos0 = |, = 306 52' or from which If
=
#
#
8').
and we have sin0
= -
tan0
f,
= -f,
53^',
3-41=5 (cos 306 52' + i sin 306 52'), 3-4^=5 cos 53 8/ + i sin - 53 8'),
so that or
(
=
8'
5 (cos 53
-i sin 53
8
;
).
Conversely, if a complex number is given in the trigonometric iy. form, equations (i) enable us to express it in the form x i sin 30), r = 2, 6 = 30, Thus, if the given number is 2 (cos 30
+
+
and we have from
(i)
=
x
2 cos
=
30
x/3, y
=
2 sin
30
=
i,
so that 2 (cos
30
+
i
sin
30) =
V + '3
^.
"
Geometric Additionand Subtraction of Complex Numbers. Let OP and OP' (Fig. 175) represent any two complex numbers x + iy and 147.
x
r
+ iy'
respectively.
allelogram
two and
Complete the parOP and OP' for
OPQP', having
of its sides.
Draw QC
PD parallel to OX. angles PDQ and 0P' are
perpendicular
The
right tri-
equal (Why?),
A
B
Fig. 175-
hence
PD = OB =
DQ
x',
and therefore
OC = OA
+ PD = * +
so that the directed line (a 4- x')
OQ
*',
CQ = AP + DQ =
represents the complex
+ i(y + y'}
=
(*
y
+ y',
number
+ iy) + (*' + */)
C
COMPLEX QUANTITIES
148]
283
The sum of two complex numbers represented by drawn from 1 gram formed with OP and OP as sides; or, spectively is represented by the diagonal
OP
and OP'
re-
of the parallelo-
To find geometrically the sum of two complex numbers represented by and OP' respectively, move OP' parallel to itself so that its initial The directed line, drawn to point falls on the terminal point P of OP. connect the initial point of OP to the terminal point of OP in its new
OP
1
position, represents the required
In
its
second form this rule
sum.
may be
easily extended to the
sum
of
complex numbers. To add geometrically the numbers represented by OP, OPi, OP*, OP*, etc., we first add any two
any number
of
and OPi, to their sum we add any third, as OP2 to the sum of these three we add a fourth, as OP3 etc. Leaving out the lines which are not needed to obtain the final result, we obtain
of them, as
OP
;
,
the following rule: Construct a broken line equal and equal and
PQ OP RS is joining O to the with OP, to
is
2,
OPQRS
.
.
OP
terminal point of
this
such that
.
parallel to OP\, 3 parallel to
QR ,
is
broken
coincides
equal and
The
etc.
OP
parallel
directed line
line represents the
required sum.
length of the line representing the sum of two or more comwill be less, or at most equal, to the sum of the lengths of the several lines representing the numbers added, hence
The
plex numbers
The modulus of
the
than, or at most equal
sum to,
of
the
any number of complex numbers
sum
is less
of their moduli.
Geometric subtraction follows from geometric addition. Let OQ and OP (Fig. 175) represent any two complex numbers whose differ-
OQ - OP is to be found geometrically. Since OP + OP' = OQ, OQ-OP = OP' = PQ. PQ, or its equal OP', represents then the ence
required difference, in words, OP between two complex numbers, represented The difference OQ
by
OQ
and
triangle of
148.
OP
respectively, is represented
which the other two sides are
Application of
OP
by the third side
PQ
'of
the
It
is
and OQ.
Complex Quantities
to Physics.
shown in physics that if OP and OP' (Fig. 175) represent in length two forces acting in the directions indicated by the arrows, then OQ, the diagonal of the parallelogram of which OP and OP' form the sides, will represent their resultant, both as regards magnitude and
PLANE TRIGONOMETRY It follows that
direction.
OQ
if
OP
tude and direction, and
its
PQ =
ICHAP. XIV
represents a given force in magnicomponent in the direction OP,
OP' must represent the other component. The laws of composition and resolution of forces are then precisely those which govern the geometrical addition and subtraction of complex numbers; we
may
therefore
or resolve forces
compound
by adding
or subtracting
the complex numbers which represent them.
Suppose n forces
/2 /3
/i,
,
,
176) act in the same plane on the same point with the respective in ten.
.
.
fn (Fig.
P
...
sities r i, r2 , ^3,
and at angles with the are
0i,
rn
pounds, On
2 , 0s,
The
z-axis.
forces
then represented by the
Fig. 176.
complex numbers:
/3
= =
fn
=
/2
Adding
F=
these quantities
all
TI cos 0i
+ = R where
R
+r
2
(cos^ is
cos
^'( f i s i n 0i
(cos
2
rz (cos
3
*2
rn (cos
2
n
+ i sin
we obtain
+f
2
+r
+ i sin + i sin
3
cos
sin$2
+
3
+r
3
2), 3 ),
n ).
for the resultant: .
+ ?n cos
.
sin03
+
.
the modulus and
equal and parallel to n,
O
.
+rn sin0n
)
+ i sin), <
the argument of the complex
bers representing the resultant force F. The resultant may be found geometrically parallel to r3 ,
.
A B -equal and
and so on.
Then OD,
by
num-
constructing
OA
BC
equal and the directed line joining the parallel to r2 ,
extremity of the broken line thus conforce F both as regards magniresultant the structed, represents origin
to the terminal
tude and direction.
We will
apply the method just explained to the solution of Prob26. Taking AP for the direction of the #-axis we
lem 16, Exercise have
U9
COMPLEX QUANTITIES
1
ri=i5> ^2=6,
0i=o,
= /i
2
= 12
fsS^i = 31 30', 3
15 (cos o
+
*
sin
+ fz=* /*= 5.7 (cos3i2i + / = 7.9 (cos 47 46' + /5= 12.3 (coss8io' + = io (cos 72 i8 + /e 6 (cos 12 30'
f4
2i',0 4
= 7-9, f = = 47 46', 06 5
=
o) i
sin
1
2
= = = ~
/
f
sin3i2i
i
sin
^
sin58io
i
sin 72
)
47 46') 7
;
)
18')
= io',0 6 re
12.3,
58
io.
72
18'
+ i o.oooo 5.8578 + 1.2984 4.8678 + 2.9657 * 5-3*o4 + 5-8492 6.4882 + ^10,4501 3.0400 + 9.5270
15.0000
~=
30')
/
4
285
^*
^"
3*
Adding
F = R (cos
<
= 40.5642 + i 30.0904 2 = = R V^5642 + 30.0904 50.51, o ^ cos-'^ = sin -i3^29 = 3fi
+ i sin 0)
2
/
50.51
F=
whence that
is,
50.51 (cos 36 34'
the resultant force
is
50.51
50.51
+
i
sin
36
34'),
pounds and
acts at
an angle
34' with AP.
Historical Note.
149.
The method
of representing complex numbers by points in a plane is often referred to as Argand's representation, after J. R. Argand, a French mathematician, who gave a discussion of the method in 1806. This is another misnomer, for it is
now known
method
that Caspar Wessel, a German, published the same The method was, however, completely as early as 1799.
forgotten until in 1831 it was rediscovered and applied by the great Gauss, to whom is generally conceded the credit of having established the true theory of imaginary numbers. To Gauss we owe the term " complex numbers" and the symbol i to represent the imaginary unit. The term " imaginary " was first used by Descartes, The choice of the
term was unfortunate, for imaginary numbers as now understood " " are no more in the ordinary meaning of that term imaginary than are negative numbers. In view of their geometrical interpre" " has been suggested in the place of lateral numbers tation the name " numthe name imaginary." The trigonometric form of complex mathematician French the Cauchy. bers was first used by great complex numbers and their geometrical representation forms the basis of many of the branches o higher mathematics,
The concept and
is
of
indispensable as well to the study of theoretical physics.
[CHAP, xiv
PLANE TRIGONOMETRY
286
EXERCISE 62 of the numbers, Represent geometrically each ~% 3 *> *> 2 i, i i> i 2, -3, 4 i,
1.
~
+
+
*
each of the numbers, Represent geometrically ^ sin 30), 2 (cos 30
2.
+
cos
6o+*sin 60,
120+ i
3 (cos
sin
240+ *sin
120), % (cos
~
+ i sin o, cos 90 + i sin 90, N/2 (cos - + i sin
cos o
240),
j-
in thp trigonometricjorm: Express the following numbers
3.
ii,
-1,1.
*,
+ *smoo in Problem 2 in the form x + iy. Express each of the numbers
4ns. cos 45 4.
+ * sin 45, cos 60 + * sin 60,
.
.
.
coso
,
o
-
,
'
Ans. of the following
Compare the moduli
5.
numbers:
-I-
2f. 2, *V3, 2, 2*, can What conclusion you draw with
reference to the location of
the points representing these numbers ?
Add
6. i
+2
<,
i, f, 2,
7.
geometrically
* + f and i - f ,
and
Add
i
i
;
+ i and 2 + -2+3 i + 2 i i
i;
,
i, i
-
-3-
i,
*,
and 3
-
3
*;
f.
+
2 <, 2
+
i,
and
i
-
geometrically, in three different
<
ways: third, By adding the first and second and then then the second, and third and first the By adding the first. By adding the second and third and then
the
(a) (b) (c)
8.
Subtract geometrically
3 + 2 i from 9.
i
2
+3
i
from 3
+4
i;
i',
i.
Prove the rule
for geometric subtraction separately
of the relation
x
2-1 from i +
+i -
(*'
+
(*
- *0 +
'
(y
- /)
by means
COMPLEX QUANTITIES
ISO]
287
Prove that the sum
10.
of the complex numbers representing the polygon taken in order equals zero.
sides of a
Three forces of
11.
make
and 61 horsepower
106,
151,
-
respectively,
96 respectively 50 04' 30", 211 20' 305", and Show that the resultant is zero, that is, that the
of
angles with the #-axis
forces are in equilibrium.
Multiplication and Division of
160.
Zl
represent !
s
= = -
= n (cos di + i sin 0i),
rirz (cos 0i
#s
+
n?2 [(cos 0i cos
r^
[
z\
cos (61
by
z2
+ 0)
we
2)
2
2
~r
^
Q [cos (81 - 8a)
2)
2 )]
2
2)
cos 0i
cos02 2)
+
(cos 0- * + i sin (81 2
2
=
is
)].
__
i
sin
2)
+ i (sin 0i cos + cos 0i sin i sin + (81 + 6 2
+ 3sin0i) r\ r + sin sin cos + 0i sin __ n (cos 0! 2
i
Let
(i)
obtain
y i (cos 0i r2 (cos
+
2
sin 0i sin
2
+ i sin
2
Their product
any two complex numbers.
Dividing ?!
r2 (cos
z2
i sin 0i) (cos
Complex Numbers.
2
+ i sin + zsin0
0i
i (sin 0i
sin2 0)
cos
=
cos ^
2
cos02
2
2
i
sin
2
$sin02
cos 0i sin
2)
i
82)].
(2)
^a
From
equations (i) and (2) it appears that, The product of two complex numbers is another complex number whose modulus is the product of the moduli and whose argument is the
sum
of the arguments of the numbers.
The
numbers is another complex number modulus of the dividend divided by the modulus of and whose argument is the argument of the dividend dimin-
quotient of two complex
whose modulus the divisor,
is the
ished by the argument of the divisor.
Corollary
i.
Since
= = i i = i
cos o
cos 90 cos 180
the modulus in each case being unity,
+ i sin o, + * sin 90, + i sin 180, it
follows that:
number by i leaves it unchanged, multiplying any complex number by i increases its argument by 90, i increases its argument multiplying any complex number by multiplying any complex
by 180.
PLANE TRIGONOMETRY
288
[CHAP, xiv
case the line segment representing the complex number unchanged, in the second case the line segment is turned in
In the is left
first
the positive direction through an angle of 90, in the third case the line
segment
is
Corollary cos 6 i sin
2.
+
reversed. (cos 6
+
and cos 6
reciprocal of r (cos 6
Two numbers
+
i
such as
sin 0) (cos 6
i
i
sin 0)
-
is
+
r (cos
i
(cos sin 0)
i
Each
said to be conjugate to each other. the other.
ZiZa
=
=
r ifi [cos (0!
Let us multiply Zi32S3
+
r\ (cos 0i
this result
i sin 0i)
r (cos
By (i), Article
Xr
2
(cos
+ i sin
2
+
by a
third complex
i
sin (0i+
= f ir2 [cos (0i + 2 + i sin (0i + 2 )] X = iWs [cos (0i + 02 + ) + * sin (0i + )
we
3i
=
fi
obtain for the product of
(cos0i+
isin0i),
z<8
Zn
*n
*lfc
=
r l r2
-
n
number
r z (cos 3 2
+
(i)
we
we have
z3 , thus:
+
i
sin
3)
3 )L
factors .
.
.
+ On + + + ;sin(0i + + +001 2
.
.
.
)
2
From
150, 2)
= r2 (cos02 + *sin02), = ?n (cos n + i sin n )
r n [COS (0i
-
are
2 )].
3
Similarly
i sin 0)
called the conjugate of
+0 ) 2
therefore
sin 0).
and is
i;
in general the
/
Powers of Complex Numbers.
151.
and
are reciprocals,
sin
=
i sin 0)
.
.
.
(i)
see that,
of the product of any number of complex numbers is product of the moduli of the factors, and the argument of
The modulus equal
to the
the product is equal to the
Now
sum
of the arguments of the factors. n factors in (i) are all equal, each
let us suppose that the
factor being 2
we then have In particular,
2
if r
=
n
= =
r r
(cos0 n
+
(cos n
i
sin0),
+
i
sin
n 0)
(2)
.
i,
I
+ i sin 8) n = cos n 9 +
i
sinn
6.
j//
(3)
COMPLEX QUANTITIES
i 5 2]
Equation
embodies one of the most famous theorems of modern is known as DeMoivre's* theorem, after its discoverer,
(3)
It
analysis.
289
and may be stated thus: The argument of the nth power of any complex number is equal to n times the argument of the number. The theorem may be shown to irrational or even of-tf, negative, fractional, i sin n is integral cos n n unless but represents but imaginary, n one of the several values which (cos + i sin 0) may have. To illustrate the use of DeMoivre's theorem, we will employ it to
hold for any value
+
rai se 1 4. i \/3 to the Qth power.
22
trigonometric form
=
r
22 i
=
.
*
sm0 sin0
~, i,
=
V^,
=
hence
= + V*Y 6 2 )
the
(cos? \
+
3
i
sin^ 3/
cos
->
3
2
and (I \2
to
we have
cos0
i,
+ -\/3 22
-
Changing
$2 +
i sin
21 =
-
t.
3
3
Similarly
^ + * sin &\
s s
Complex Numbers. Suppose it is ^th root of the complex number z - r (cos
'152. 'Roots of find the
Denote the root
Then
n (z')
[/ (cos 0'
-
*>
Z/z
required to
+
i sin 0).
by
and we have by DeMoivre's theorem
+ i sin 6')]* = r
"*
(cos ntf
+
i
sin
n 0') =
+ i sin 0),
r (cos
from which
/
-y^,
0'
=
0,
+ 2T, + 4T, ^ +
*
e
when an angle is increased by any number of times and cosine remain unchanged. It follows that
since sine
67T,
DeMoivre (1667-1754)
+ 2
2 *1F
T both
the
created a large part of that portion of trigonometry His death has a curious psychological in-
which deals with complex numbers.
each day, until Shortly before his death he slept a little longer limit of twenty-four hours was reached, he died in his sleep.
terest.
when the
[CHAP xiv
PLANE TRIGONOMETRY
290 v
, '
n
n
n
n.
and
-
--
!/ 0+- 27T,h*sm M-2T\ = rfcos n I n \
x
2 :
where k
is
any
rn
)
'
+
(cos
^
sin
integer.
These values are not
when
all different, for
k
=
n we have
/cos0 r 1
n\
n
<
= rw [cos-
+ w
\
and
i
sin- )> w/
^^
similarly
n
n
\
I
n
/
and generally
n
.^ ^4-2 ^ MU7T,+ sm w (y
s
so that
z'
has only w distinct values corresponding to the values k
that
=
o,
2,
i,
.
.
n
i,
is,
+ i sin 6}
r (cos Every c0w/>fe# number the
.
formula
/KM
.
.
.
n
f.
has
n
nth roots given by
COMPLEX QUANTITIES
152]
EXAMPLE
To
i
find the fourth roots of
In the trigonometric form the formula (i) above
i
Solution.
by
i.
=
z
*i, to, to
>
=
cos^
2>
=
cos -*
+ i sin 2=i,
44 ~l~
EXAMPLE
zi
==:
to
i.
=
cos
=s
cos
find the fifth roots of
=
cos
2w +
*
.
o, i, 2, or 3.
we have
44 44 "
~
hence
+ i sin
=
sin
==
"r* ^
i,
i.
i.
In the trigonometric form
Solution.
[
To
2.
~
i sin
=
respectively,
44
z
+ i sin o,
cos o
where k
),
Denoting the four roots by
291
i
^Tj
=
cos
+ i sin
TT
where ^
=
TT,
hence
o, i, 2, 3, 4.
5
S
This gives the five values
55^ 55 55
55
cos-
3
The
last
sin-,
21
cos^ +
==
=
cos^"
i sin
2^,
+
s4
sin 3^, z2
i
=
z3
= cos^
i sin 2-^
-,
Z4
=
COSTT+
^sinir,
+ ^ sin 2^"-
cos
two roots may be written equally
=
well
cos-
^'
sin->
5
5
for
cos U=cosf27r-
A
1
S s
V
i5
cos3jr > sin
S /
5
21= sin (2 7TS
\
and
S
so that finally
555
V~= cos-
i sin -,
cos^
i sin 3JT, COSTT H- i sin 5
TT=
- 1.
PLANE TRIGONOMETRY
2 92 '
163.
zn -
To Solve the Equation z
and the n
=
n
=
i
=
i
+
cos o
[CHAP, xiv
If z
o.
n
-
i
=
o,
then
i sin o,
roots are
=
ZQ
cos -
w 2
+ i sin n2 = 7T
2
.
i,
T '
w Z2
=
zn _2
=
+
cos
w
>
w
2i2
cos
sin
i
?2J
^
+ ^ sin
:
Msm
:
Now ^-i)7r =
cos
w
-
I
sin
)^
=
/
27r ^.27r\
\
n)
S in (2
w
TT
-
\
2
5> n
COS
^= n
2
sin
I
-7r n
,
hence 2n _ i
=
i sin
and
.
n
n
similarly zn - 2
is, zi
and
zn - 2 , zs
cos
*~ n
i sin
^n
,
Let n
=
w + i, an odd number. m pairs of conjugate roots,
=
Then
2
S ^;sin^, i,cos^/sin^,co n n n n
Let n
2
w, an even number.
+
i
.
.
that
.
is,
besides the
first
the roots are
cos^ n
i
sin^n
Then
sin^ZE
and the roots are i,9
cos
i sin
n
%
,
root Zo there are
(b)
=
and zn -i are conjugate complex numbers, and likewise and zn - 3 etc., are each pairs of conjugate numbers.
that
(a)
-
cos
n
,
cos^
i sin
cos
4^, n
n ^
i sin
n
n
,
i.
COMPLEX QUANTITIES
i$4]
293
=
roots of the equation z n i o are represented geometrically by the n lines (or by their terminal points) drawn from the origin as center to the circumference of a circle, radius unity, so as
The
to divide the circumference into n equal parts, one of these lines coinciding with the positive direction of the
Z
n
=
I
zn
+ i = o. If zn + = = COS7T + i SUITT,
To Solve the Equation
164.
i
o,
then
hence the n roots are zo
=
cos 2"
n 2l
+ i sin-n
= Cos^ + i
sin
n
n
n
sn
n where (a)
zo
and
=
Let n
so that the
zn -i, z\
n
2
and
m+
z n - 2 , etc., are pairs of
an odd number.
i,
i
conjugate roots.
The middle
root
roots are
nun
cos-
n
sin^, cos
'
2-2!
t si n
d
n
cos
n
n (b)
Let n
=
cos-
w, an even number.
2
i sin ->
cos
^
Then
i sin
2^>
the roots are
.
.
.
,
is
PLANE TRIGONOMETRY
2 94
n roots of the equation z
The
[CHAP,
xw
+ =
i o are represented geometriterminal their n lines the points) drawn from the (or by cally by radius unity, so as of a to circumference as center the circle, origin to divide the circumference into n equal parts, the positive #-axis
being taken to bisect the angle between a pair of consecutive
lines.
EXERCISE 63
Compute the
following expressions by DeMoivre's theorem, then by expanding the binomials by the binomial
verify your results
theorem. 1.
2.
(i
+
2
Ans. 2i.
.
i) 4
(i
i)
Ans.
.
4.
3-17" Ans.
6.
Show
that each of the following expressions equals
cos \ 5
i sin
(
7.
>
cos-
V
+
i sin
^
i sin
> )
(COST
i shiTr)
5 .
5 /
5
\
i sin -]>
i,
cos^iisin
77
cos^
,
Problem
etc., in
6.
5/
5
Find the seven seventh roots of
7
9.
^
-V (cos5/
5
ing these roots.
Ans.
(cos
V
5/
i,
Construct the points representing the expressions (
8.
~\
i.
Find the seven seventh roots
of
and plot the points represent-
i
77
sin^,
i
i
and
cos^iisin^. 7
plot the points represent-
ing them. 10.
If z
,
3i, zz,
.
.
.
z n -i are
the roots of the equation z n
i
=,o,
show that
11. cally.
Find
all
Ans.
the values of
^2
(cos 9
+
Vi i
+1
sin
9),
and represent them geometril f/2 (cos 81
+ i sin 81), etc.
COMPLEX QUANTITIES
iS6| * }
295
The Cube Roots
156.
cube roots of
i,
=
uQ
of Unity. Let WQ, wi, u^ represent the three then by the preceding article
cos o
+ i sin o =
i,
Now / cos
Z+;
V
3
3 /
3
3/
also
that
sin
^1^2
?JEY = C o S
=
=
3 i
+ i sin 4Z = 3
3
3
3
=
3 2
i,
is,
The square of either of the imaginary cube other, and their product equals unity.
We may
then denote the cube roots of unity by I, CO, CO
where
f
to is
'166.
Let
roots of unity equals the
either
one of the roots
The Cube Roots
+
r (cos
i
2 ,
HI, Uz.
Any Real
of
or
Complex Number.
any number and
sin 0) represent
3
,
*i, 22 its
cube
roots.
We
have z
=
r*(cosV
+i
sin-\ 3/
3
3
3
ri/cosV
=
r* ( cos
V
where
to
+ * sin
+ i sin-Vcos
3
3/\
-
+ i sin -} (COB & +
3
3/ V
has the meaning given
it
co2o
3
3
i sin
3
in Article 155.
^}= 3 /
co%<>,
d)
PLANE TRIGONOMETRY
296
[CHAP, xiv
Furthermore, by applying the results just obtained
= =
2
22,
<
30,
C0
= W3 = = %=
2l
2
2fo
Z2
(2)
2fo,
=
O)Zo
(3)
l.
From (i), (2) and (3) it appears that any two cube roots * a 2 number may [be obtained by multiplying the third by w and
from
Thus: from
(i),
= <03o, = W2^,
21 22
=
where w
1
22
,
=
(2)
2
Every cubic equation can
3 are given numbers. can be replaced by another in which the second
ai, 02,
Equation (i) term is missing by putting
x
On making
Z2,
t
157. Solution of Cubic Equations. be expressed in the form
,
(3),
=
Zo
COZi,
'.
2
where a
from
=
-^>
z
(2)
this substitution, equation (i) reduces to
& + 3 Hz + G - o, where
H = 0o02
2 <*i
>
G=
Now put Z
then
(3)
2
a3
""
(3)
.
J ^o^i^a
4- 2
3 i
(4)
-
- U + V,
($)
becomes o.
If
furthermore
(6)
we put T
+
wi>
=
o,
that
is,
uv
-
ff ,
(7)
then (6) becomes
-G, and substituting s
tt
for v in (8) its value
_ ?! = - c,
or
w6
from
(8) (7)
+ Gw 8
fl 3
=
o.
(9)
COMPLEX QUANTITIES
i S 7l
(9) is
a quadratic equation in ,
=
so that for either sign
From
(7)
v
Solving
-G
we obtain
_ -rr
=
*.
297
so that (10)
,
may be
written z
if = u --
>
u=
u where
of every number has three values, which by Article be denoted by u, cow, co 2w respectively, where u is any one The three values of z which satisfy the equation (3) of these roots.
But the cube root 156
may
are therefore
=U
H -U
9
cm
= am -
(n)
,
u
u u being either one of the cube
-4
roots of
(j 2 )
In applying this method to the solution of any equation of the and G from (4), then u from (12), then the (i), we first find
H
form
(n) and
three values of z from
values of x from
EXAMPLE Solution.
From
i.
finally
the
three corresponding
(2).
Solve the equation 8 a?
Here a
=
8,
fli= 4,
+
=
12 x* 14,
42 x ^3 =
95
=
95-
(4),
= = G
ff
OQOz 2
ao a3
01*
=
3 a ai02
128
=
+ 2 d^
27 , 3
=
4608
=
2 93 2 .
o.
[CHAP, xiv
PLANE TRIGONOMETRY
29 g
From
(12),
+ 4# = vV 3 - 4-2 - 7-2 '/-G + N/g+g- //a a + 7.a 8
3
9
21
4
>
>
>
9
-
2 4.
From (n),
From
(2)
_
2
Check.
or
168.
The
Irreducible Case.
preceding example, root ^ f
its
When G"+4 H3 is positive, as in the
square root
-G + VG3 + 4g
is real,
and
,
which
is
the cube
3
c .)n be
found by the rules
of arithmetic.
2
be imaginary, and But if <7 + 4 H* is negative, its square root will is the so-called This u. we must employ DeMoivre's theorem to find "
irreducible case
Since
G2
"
+4H
we may put
of the cubic equation. is
negative, _
-
(G
2
Its solution is as follows:
+ 4 &)
will
be positive, and
COMPLEX QUANTITIES
158]
whence, by Article 146,
299
(3),
4 (2)
V=
r J /cos
f
\
3
+
sin
=
-"\
V I~ff
3/
( cos
-
\
3
l+u +
i sin
3
3
V^H
f cos \
3/
3/
3
=
+ i sin -Y
-
+
sin
3
3
2
M
Hence,
and from
Article 157, (2), (4)
EXAMPLE
i.
a
Solution.
G2
+4^
From
3 ==
Solve the equation 3
=
3,
<*i
=
i,
207, hence
#2
=
we
3
ac
i,
+3#
2
03
=
3 #
2
=
are dealing with an irreducible case.
(2),
cos
-
2V-
0.
2.
64
3' 20".
PLANE TRIGONOMETRY
300
From
[CHAP, xiv
(3),
20=
VffcOS
2
=
-
4 COS 21 2l' 7"
=
3-7256,
o 2
3
2=2 V-#COS From
7r
=
4 COS 26l 2l' 7" = -0.6015.
(4),
=
=~
0.9085,
=-
1.3747*
0.5338.
EXERCISE 64
+ * ^3
"" I
_
*
Show 2.
2
show that
actual multiplication,
By
+
also that i
Compute
^3
*
2
coi
+
o>2
=
o.
the cube roots of 27; of 27
all
Ans. 3
i.
,
-3^3 + 3^ _ 2 3.
Compute
all
the cube roots of
+
i
*sn--
2cos-
Ans.
2
12
i.
sn
cos
i2/
V
4
4
12
12
Solve the following equations: 4.
a3
-
6
x
-
9
=
^3.
4*w. 3,1:3
o.
2 5. 6.
7.
^ - 24 + 45 * s - 12 x - 10 = o. 4
a;
2
25
8
a?
=
Ans.
o.
4ns. 3-8232,
-
2.9304,
i, 2.5, 2.5.
-
0.8928.
+ 9 x + 24 # + 19 = o.
Ans.
z
By
use of 4-place tables,
1.4680,
4.8794,
2.6527.
COMPLEX QUANTITIES
i59l
8.
301
A man
invests $5000 and two years later $2000 more. The to the principal at the end of each year. At the of the third year the total increase was found to be $2058.16.
interest
was added
end Find the rate per cent of profit. = i (Suggestion. Let x rate.)
Ans.
n %.
+
9.
than
The inside it is
of a tank
wide.
is
3
volume
Its
ft.
wider than
is
100 cu.
deep and 3 ft. longer Find the dimensions of Ans. 2.284, 5-284, 8.284.
the tank.
A loan
of $500 is to be repaid in three equal annual payments each without further interest. Find the rate per cent. $190
10.
of
it is
ft.
11.
In determining the deflection of a beam, uniformly loaded its two ends and points of trisection, the follow-
and supported at
ing equation occurs: z
24 x
20 y?
Find
+ 3 = o. Ans. 0.4460, 1.0687,
its roots.
~~
-3 I 47-
159. To Express sin n8 and cos n6 in Terms of sin 8 and cos 6. DeMoivre's theorem enables us to express the sine and cosine of in terms of powers of the sine and cosine need only compare separately the real parts and the
any multiple angle nd of
0.
We
imaginary parts of cos nS
+ i sin nS =
+ i sin 0) n
(cos 6
(i)
expanding the right-hand member by the binomial theorem. For shortness sake let us put
after
cos 6
=
c,
sin 6
=
s,
the right-hand side of (i) then becomes
+ nicn ~
=
c-
l
s
+ nc- s -
-
i
since
P=
i,
9=
i,
V _ (-!)(.
2 i
4
3 -4
=
i, etc.
a)
PLANE TRIGONOMETRY
302
The
[CHAP, xiv
must equal cos nd,
real part of this expression
real part of the equivalent left-hand
member
=
and
for
a
like
we have
reason the imaginary part must equal sin nS, hence
cos n6
since this is the
in (i),
- *(*-') cw - V
c
1-2
-
+ n(n
(n
i)
l)
-
(n
a)fr
?)
*
^.4
,
etc<
(a)
_
etc
(.)
1-2-3-4
1-2-3
^ (n
i
-
2) (n
-
3) (n
4)
c
*t-5
^a
1-2-3. 4-5 Thus,
if
=
W
=
cos 2
2,
c
2
2
"" x ^
^2
=
^2
<:
c
2
-
s
=
2
sin2 0,
cos 2 ^
2
I
=
sin 2
2
cs
=
cos 6.
2 sin
H~3, =
cos s
""
3 ^3
3
c
*-/-
a2 =
c
3
3 cs
1-2
2
=
cos3 (?
=
=
sin 1
2
^ c s
-
iiSJI-lillZLi)
3 s
c
1-2-3 results
and
2 3 cos0 sin 6
3 4 cos ^
=
3 cos
=
3 sin 6
2
3cos^,
-
sin0
4 sin
in Exercise 50,
sin3
3
0,
Problems 10
11.
160* of
which agree with those obtained
-
To Express
cos 6
By
Multiple Angles. cos 6
Put
cos0
z
n
z
+ i sin 6
=
+
cosfl0
f
+
and
sin 8 in
Terms
of
Sines and Cosines
Article 150, cor. 2,
and cos 6
sin0,
i
sin
then
are reciprocals.
~ = cos0
i
sinn0,
~=
cos w0
21
hence
=
isin0,
z
2 cosn0,
n
z
-
^-
-
COMPLEX QUANTITIES
160]
Now
/
let
us expand ( z H \
i
\
n
)
z/
/
i\ w
\
z/
and [z
)
303
by the binomial theorem,
(1)
where in (2) the upper signs in the last terms are to be used when n is even and the lower signs when n is odd. Let us group together the first and last term in each of the expressions (i) and (2), also the second and second last, the third and third last, etc.;
we may then
write
(3)
-
,-" . f
where
in (4) the
t
*-'
upper signs are to be used when n
is
even and the
lower signs when n is odd. The total number of terms in each binomial expression is one more than the index n, and since we have
grouped the terms in pairs, there will be one term left over in case n even. This term will not contain z at all, for since the exponent of z diminishes by 2 for each successive term, it will be o, and z = i.
is
Let us now substitute for 2 cos nQ, 2 i sin nQ, etc.,
gives 1
2*
n
+ zn
,
z
n z
n
,
etc.,
and divide out the common
" 1 cos n 8 = cos nO '
z
+
n
(
+ wcos (n
?
"
I}
cos (n
their values
factor
2.
This
2) 6
- 4)6 +
etc.,
(5)
PLANE TRIGONOMETRY
304 2*- *
r sin" 9 = cos n9- n cos (n - 2)6 + 2^LZjO cos (n-4)e-etc., =
n n" (
1
)
2
I
factor
n sin (n
n0
i [sin
+ The
ICHAP.
(n odd)
etc.,]
disappears in either case, for
i
(6)
2)6
- 4)6 -
gin (n
(neven)
xiv
when n
(60
.
is
even, say
w which is +i or n 2m = i according as w, we have i = i i) ( w is even or odd, and when n is odd, say 2 m i, we can divide both sides of the equation (6') by i and have left on the left side 2
,
+
i
2
m
=+
i
or
EXAMPLES. 2 cos
as before.
i
n
If 2
2 ;2 sin2
2,
=
cos 2
+
=
cos 2
-
=
If
=
a? sin
=
i,
or
cos2
i,
or
sin2
=
I
+cos20
>
3,
* (sin
39
-
3 sin 0),
cos'tf
=
sin3
=
3 sinfl
-
sin 3 g .
4 If
n
=
4,
8
= COS40- 4COS20 +
6,
sin'fl
=
6
- 4 coS2 g + cos 4 g. O
EXERCISE 65 1.
By
powers of
= sin40 =
Ans. cos 4 2.
and
method of Article 159 express cos 4 and cos 0. sin
the
Show
cos4
6 cos 2 3
4cos 0sin0
that
4tan0(i-tan2 0)
i-6tan2
'
sin
sin2
each in
4
+ sin 8
4cos0sin
0.
4
0,
COMPLEX QUANTITIES
i6o]
3.
Show
that
cos50 sin 5
= = = =
+ scos0sin 20 cos cos0 (i6cos + 5). cos sin 10 sin + 5 sin cos 2osin + 5). sin0(i6sin e tan4 ~ I0 tan2 e + 5) tan = 5tan 0- iotan + iocos8 0sin2
cos5
4
5
4
2
3
4
2
4
By the method COS50
=
Sm60 =
of Article 160 express cos
and
2QCOS0
+ 5COS30 + COS50
20 sin
~
5 sin 3
+ sin 5
16 that 6
=
20
+ ig cos 2 + 6 cos 4 + cos 6
g __
20
~
32 .
6
0.
i
16
Show
0.
2
functions of multiple angles.
5.
4
2
(
tan 5
4.
305
15 cos 2
+ 6 cos 4 32
cos 6
sin
in terms of
CHAPTER XV TRIGONOMETRIC SERIES AND THE CONSTRUCTION OF TABLES '
;
161. Definition of Infinite Series.
An
.10
9
indi-
Infinite series are of
Thus
in
.
looo
100
an
infinite series is
sum of an endless number of terms. common occurrence in arithmetic and algebra. :ated
.
10000
,
(i)
the right-hand member is an infinite series. Similarly, every recurring decimal may be expressed as an infinite series. Again, if we divide i
by
i
x,
according to the rule for long division X I
we obtain
- =I+* + 3? + 3 + * +... X 4
(2)
member of this is an infinite series. The square root number which is not a perfect square and the cube root of every number which is not a perfect cube may be expressed in the
The
right-hand
of every
form of an series,
infinite series.
impossible to write down all the terms of an infinite the series is not completely determined until we know the law
Since
it is
or rule according to which the various terms are formed. When In this law is known we can write down any term that is needed.
the
first series
series is
above, the law for the general term
completely expressed thus
manner the
series (2) is
and the
+-T+ I0n
3
like
->
:
-+-,+-,+ 10* I0 10 In
is
completely expressed by
In each case n stands for the number of the term. When the law of formation is clearly apparent from the first few terms, as in the above examples, the general term is not always expressed. 306
TRIGONOMETRIC SERIES
162]
307
A
series is frequently expressed by writing the Greek letter 2 before the general term; thus the second series above may be written n~ l " S z*" 1 which is read "
summation
,
The
2 un = where
ui, u*, u$,
.
.
ui .
first
ww
n terms
,
+ un +
.
-
are the terms of the series.
Let
Sn
stand for
of the series, thus
Sn = As n
+ uz + us +
is
Convergent and Non-Convergent Series.
'162.
the
of x
general expression for an infinite series
Ui
+
1*2
+ Uz +
increases indefinitely (approaches
'
'
ex),
one of three things must
happen,
Sn may approach some finite quantity as its limit. Sn may become larger than every assignable finite quantity. Sn may neither approach a finite limit nor become infinite,
(a) (b) (c)
but fluctuate between two or more different values.
No
other alternative
is
conceivable.
In the
first
case the series is
said to be convergent, in the second divergent, in the third oscillating. Divergent and oscillating series are grouped together under the
term non-comergent If
a
series
series.
series contains
may
a variable, as the series
(2),
Article 161, the
be convergent for certain values of the variable and non-
convergent for other values.
EXAMPLE
i.
Consider the series
x (a) If
x
=
^,
+ x* + x*+
...
+xn +
...
the series becomes
24
4
i+i + |8 + 16 ^+ 2 4
248 +JL-Xn 2
PLANE TRIGONOMETRY
308 As n approaches
oo
comes under
series
Next
(b)
let
-^ approaches o and
,
(a)
Si
=
is
convergent.
=
2,
then the series becomes
+ 2
(b)
3
,
if
x
=
i,
5i
n
is
as
limit
=
=
oo
hence the
,
series conies
+ (S4 =
=
i)
+
-
-
-
Sn =
i or o according o, Ss i, Si i, o, In this case S n neither approaches a finite odd or even. nor becomes infinite, hence the series is oscillating.
Divergent be avoided.
x
14,
the series becomes
-i+i-i+ =
hence the
.
n approaches oo S n approaches and is divergent.
Finally,
(c)
xv
3
2
Plainly, as
under
i,
+ 2 + ... + 2" + ... + 4 = 6, S = 2 + 4 + 8 = n 2 + 4 + 8 + ... + 2
22
5 = Sn =
2,
approaches
and
us put x 2
Sn
[CHAP,
2,
the
series frequently lead to absurdities,
left
For instance,
if
and must therefore
we put i, while the right member
in the series (2), Article 161,
=
-
member becomes i
2
becomes 1
.
;
=
hence for the value x
two members
+ 2+4 + 8+
of (2) can
2, which makes the series divergent, the no longer be considered equal.
163. Absolutely Convergent Series. A series which remains convergent after all its terms are made positive is said to be absolutely convergent.
Convergent are
made
series
positive
which become divergent when all the terms be semi-convergent or conditionally
are said to
convergent.
For instance, becomes
if
in (2), Article 161,
i 2
+ 2 ~~~* 2
28
we put x =
+
|,
the series
i
which is convergent, for it remains convergent when all its terms are taken with the positive sign. The series is therefore absolutely convergent.
TRIGONOMETRIC SERIES
i6aJ
The
series
i
the terms are given series
?
+i
J+
is
309
convergent, but
when
all
made
positive the resulting series is divergent. The therefore a semi-convergent or conditionally con-
is
series.
vergent
Absolutely convergent series are subject to all the fundamental laws of algebra,* that is, they may be added, subtracted, multiplied and divided, like expressions consisting of a finite number of terms.
This
is
results
not true of semi-convergent and divergent series. Curious may be arrived at if this is not kept in mind. For example,
take the divergent series
S=i+3 + 5+7+9 +'" + 13 + i5 + 0=1
!+ 1+ + I+I 5=2 + 2 + 6+ 6+10 + 10 + 14 + 14 + = + 28 + +20 +12 4 -4 d +3 +5 + 7+
Adding
=
1
I
1
45, which of course
is
)
absurd.
Or take the semi-convergent series
5=I _i + I_I + I__E +
23456
which
is
known
to be equal to log 2
=
...
We may write
0.69315
66 2 4
2345 = is
absurd.
series like
o,
that
is,
a constant 0.69312
2
+ 26 + |8 3
4
/
equal to zero, which
These examples show that we cannot treat an we do other expressions until we know whether it
infinite is
abso-
lutely convergent or not. * For the proof of this statement we must refer the student to textbooks on XXVI. higher algebra, such as Chrystal's Algebra, Chapter
PLANE TRIGONOMETRY
310
A
which
series
[CHAP.XV
absolutely convergent will of course remain conall of the terms are made negative, but all its terms are positive it may become
is
vergent when some or even if a series is divergent when
convergent when a certain proportion of
terms are
its
made negative.
s 1
The Sum
164.
called the
sum
When a 5n approaches as n
an Infinite Series.
of
not otherwise, the limit which
series converges,
approaches
Thus when we say that the sum
of the series.
oo is
of the
series
!
!
+ + +. 248
we mean
is i,
is i
+
JL+... n
2
that the limit of
n approaches
as
..
oo
A divergent series has no sum in
.
the proper
sense of that word.
165. (a)
The Limit
Let
=
r
i
i.
the multiplication (b)
Let
>
r
is
n Approaches
multiplied repeated.
We may
i.
oo
.
by i equals i no matter how often n Hence the limit of r = i.
=
write r
i
+ d,
where d is some posiwe have
applying the binomial theorem
By
tive quantity.
rn as
of
n
r
Now no matter how
-
(i
+ dY = + fid + i
n
-
-
-,
taken sufficiently large nd may be, can be made larger than any assignable quantity, hence we see that as
n approaches (c)
Let
,
r
n
oo as
n approaches
oo
if
>
i,
n approaches
and oo
is
oo also.
approaches
r
approaches
4$ n
oo
small d
,
since
by
therefore r
and
(a)
n
(6)
0V
=
=o
approaches
approaches oo
The
,
r
n
approaches
Infinite
o,
i or
as
oo according as r is less
Geometrical Series.
+ ar + ar* + = ar + ar + rSn Sn
then
m
.
than, equal to or greater than f.
166.
I
a
2
Let .
.
.
+ arn ~ +
,
1
TRIGONOMETRIC SERIES
167]
Sn - rSn = Sn =
Subtracting
from which
(i
'* "" r
fl
Let
r
n
than
less
approaches o, hence
oo r
approaches
and the
be numerically
'
By
i.
Sn
Article 165 (c) as
+a=
+ + +
approaches
n approaches
divergent. (c)
or
Letr
a,
=
i.
according as
when
oo
a + a Then 5n = a n is even or odd. In
+(-
all its
na, which
hence in this case the
;
n
a
approaches the limit
Since the series converges
series is convergent.
as
- r),
(i
.
terms are positive, it is absolutely convergent, = i. Then Sn = a a a (b) Let r oo
a
r
i
(a)
311
- r) Sn = a - arn =
series is
n ~l i)
a
=o
this case the series
is
oscillating. (d)
Let
approaches
and the
The
An
be numerically greater than
r
n
oo r
approaches
oo
,
By Article
i.
hence S n =
a
(I
~~ r
)
165
(b)
as
approaches
n oo
series is divergent.
results
may
be
summed up
in the following theorem:
infinite geometrical series is absolutely convergent if its ratio r is
numerically to or greater
than
less
than
i,
i.
non-convergent if
When
convergent
r
i is the first
term and r the
167. Convergency Test. Wl
+ W2 +
3
ratio is numerically equal
sum
a
o
where a
its
its
ratio.
Let
+
'
-
+Un +
-
-
any infinite series of positive terms. Let rn represent the any term un to the preceding term un -i. Then un = un ^rn) U2 = Wir2 u* = ^3, W 4 = W4,
represent ratio of
-
,
and consequently
Wl= M2
Un
Ml,
-
PLANE TRIGONOMETRY
312 Adding Ui
+ u* +
+u
3
4 -\
=
----
Ui
(1+
r2
>u
(i
l
[CHAP,
xv
+ r^ + r^r^ H----)(i)
+ R* + R*+
.-.)
+ r + r* + + r*
(2)
(3)
)
R is greater than the greatest, and r less than the least, of the ratios r%, r3 r 4 rn
where all
,
Now is
is
(2)
provided
an
.
.
,
infinite
R is less than
.
.
divergent provided r
is
.
geometrical series
and
i,
.
,
(3) is
an
which
is
convergent
which
infinite geometrical series
equal to or greater than
hence the
i,
infinite series
+Un +
'
is
convergent provided
to or greater than
The ratio
S#n
the series
A
rn
.
=
R
is less
than
divergent provided r
i,
-^
is
w n -i
called the ratio of convergency or
We have
R which is
test ratio
itself less
than
always
A
is settled
this case other tests
of
in case the
must be
less
than
i.
series is divergent if the test ratio is always greater than ber r which is itself equal to or greater than i.
Nothing
equal
then the following theorems:
series is absolutely convergent if the test ratio is
some number
is
i.
test ratio
is
some num-
ultimately equal to
i.
In
applied.
These theorems remain true
if,
not from the
first,
but after some
particular term, say the &th, the test ratio has the values stated. For the sum of k terms is finite, hence the whole series will be con-
vergent or divergent, according as the series beginning with the &th term is convergent or divergent.
The convergency test-ratio
test.
test established in this article is
It is
sufficient for all
known
as the
the series treated in the re-
maining portion of this chapter and the chapter following. In fact, the test-ratio test will answer most purposes of elementary mathematics; the cases in which the test fails, that is when the test ratio approaches i in the limit, form exceptional cases which can usually be avoided.
There are a great many other convergency
tests
by
TRIGONOMETRIC SERIES
168]
313
which the convergency of a series can be settled in doubtful cases. of series forms a separate subject of study, to the development of which many famous mathematicians have devoted
The theory
their best efforts.
168.
Convergency of Special The exponential series*
(a)
Series.
i+* + 4+4+ 2!
'
+4 + n(
'
3!
where n\
(factorial n) stands
Herewn +i =
,
=
un
X2X
f or i
*" ,
n\
(n
3
X
hence the
'
-X
n.
test ratio
i)!
"
un
w
If
is
n\
(n
i)!
n
taken sufficiently large, -will become and remain
less
n
i
no matter how
the series
is
The
(6)
large x
may
be, provided only that it
absolutely convergent
is finite;
than
hence
for every finite value of x.
cosine series V4 2!
4
6!
1
Here the general term
(2*)!
x is
.
N
,
the preceding term
hence the
is
than
i,
values of
x.
less
(c)
*
The
The
shortly.
7^
-
>
2)!
test ratio is
(2 ^)
When n
=F
(2n
(2 w)!
!
(2^2)!
2n(2n
i)
taken sufficiently large, this ratio becomes and remains therefore the series
is
absolutely convergent for
all finite
sine series
reasons for the
names given
to the series in this article will appear
PLANE TRIGONOMETRY
314
In this case the test ratio
(2
When #
n
(2
!
The
(d)
wn -i
= T
xn ~ l
n
^
n
n
The
greater than
than is
i.
(e)
,
n
i,
n hence the test ratio
n
i
i
as
n approaches
.
i.
The
convergent so long as x is less any co elusion in the case
to lead to
test ratio fails i.
binomial series
n
(n
-
-
i)(n
2)
.
.
.
-r+
(n
i)
|
The
oo
approach a number less than i, equal is less than i, equal to i, or
series is therefore absolutely
The
is
according as x
equal to or greater than
"*
absolutely convergent for
a proper fraction which approaches
or greater than
The
series is
Ts*- 1 = ~(n-i)x
test ratio will therefore
i,
+ i)
i
'
- is
n
value of x.
xn
-
(2
234 ,
n
2
!
logarithmic series
xn
un =
x
i)
taken sufficiently large this ratio becomes and remains
is
finite
every
to
n
than unity, therefore the
less
xv
is
"*"
+ i)
[CHAP,
general term
is
n
(n
i)(n
-2)
.
.
.
(M
- r 4-
1)
f
[
r
rl
The
preceding term
is
n (n
i)(n
-
2)
.
.
.
(n
r +"2)
>,t '
fr-x)I Dividing the general term by the preceding term and canceling the common factors, the test ratio reduces to *
1
1
x.
TRIGONOMETRIC SERIES
i68J
As
x as
approaches is
H "-4~ r
r approaches oo,
315
i
approaches o; the test ratio therefore
its limit,
and we may conclude that the
absolutely convergent so long as x
is less
than
series
i.
EXERCISE 66 Write down the
first six
terms of each of the following
Write down the general term of each of the following
*-
e
5*
6
*
I
1 i
4-5
3-4
2-3
.
x 7
~r
I i
I
series:
*
*
"T
1-2
series:
i
^ 3579 + - +
.
2---+.
8.
Show
.
2Z
3*
3*
3*
+ ^-
n\
,6!
Examine the following '
14.
~l
I -
4!
i0
2n
23
22
3 II.
r
that the following series are absolutely convergent:
2 IO.
2
series as to
convergency:
'
'
2!
1+!*
3
22
!
2
3
2 .
15.
x
Conv.
for
ac
<
i.
+ a! 5.
Div.
PLANE TRIGONOMETRY
3l6 16.
i
+ -++-+ 123 2
+? +
n
<
Ans. Conv. for x
6
2
it is
20
12
(Suggestion.
Adding 5'
hence 28'
+ S"=o.
where
Show
.
.
,
.,
x
>
i.
S*
= (-
-
-
i)
-
#
From
this
the following reasoning:
7+ 9+ 3+ 5 + 10+ + 8+ 6+ 4 + + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + i+ 2
2
,
is
3
w3
-
>
'
'
'
l
4
'
> wm
>
U*
!!
and wn approaches o as n
certainly convergent.
The Number
Consider the infinite series
e.
l+iI + -i; + i+^+ 2! 3!
The
f or
that the series
Ui ~ U2 > u\ >
approaches oo 169.
div.
is oscillating.)
What is wrong with LetS'=i +
S"= + S"=i
19.
,
,
i,
42
30
Find S2 S3 5 4
seen that the series
1 8.
xv
[CHAP,
4- -
test ratio is
nl
4!
= i
-
(n
i)!
-
+-.+ n\
which
is
less
d) than
i
for all
n
n greater than i, hence the series is convergent. Let the sum of the series be denoted by e. The sum of the three terms alone is 2.5, hence e is certainly greater than 2.5.
values of
will
show that
e is less
than
2.75.
= 4!
3U 1
Si
3U-5
5F" jiU-e
S'3
<4-
first
We
TRIGONOMETRIC SERIES
169]
317
hence r H
3
f -
The
r
H
r
4l series
whose
T
^
"
'
6!
on the
ratio is
'
77
'
S'
;
r~~
right is a geometric series its
sum by
~>
H
r 2
"~;
313
3!
!
whose
r
!
3 3
3 3
3
+
!
first
term
is
and
Article 166 is equal to
Y^i-iirJ-;-"* A + i +i + i +
thereto,
and
since
i
+ - + 2!~ =
.
.
.
,., s ,
2.5,
i
therefore
e=i+~ + 2! +
+
i
that
is
To
terms of the
2.5
determine
e
more
series e
by
exactly, ui,
=
++
+
nl
~
as follows:
=
i.oooooooo
=
1 .000
000 00
i
W3
=
^= 0.500 ooo oo 2
W4
=
?*?
=
0.166 666 67
=
0.041 666 67
3
^5 =
-J
4
w6
= ~5 =0.00833333
w7
=r 6
^8
=
I
=
14*
W9
8 WlO
=
= ==
-
<2.7S,
2.75.
we proceed
#3, etc.
2,
u\ 2
-^-
4!
3!
001 3 88 8 9
o.ooo 198 41
0.000 024 80
O.OOO OO2 76
Denote the
PLANE TRIGONOMETRY
318
Ull
*
[CHAP,
_
xv
= O.OOO OOO 28
10 13
=
=
II
This
an
is
sum
the
__
=
S 12
Adding
O.OOO OOO O3 2.718 281 84.
of the first twelve
terms of the
There
series for e.
owing to the neglected part of each of the decimal fractions added, but this error cannot exceed 10 0.5 or 5 in the last decimal place. Besides this error there is the neglected is
error in the last figure,
X
portion of the series, which
Now
=
is
Adding
ia
+
tti4
+
i6
+ u\\ + u\i +
is u<&
-^-
+
<
-^rf 1
13
From
the computation above
portion of e
is less
12
e
170.
<
than
the ninth decimal place.
=
u&
<
1
+ -+
+
2
i3
\
)
/
!3
o.ooo ooo 03, hence the neglected
o.ooo ooo 003, that
is,
less
than 3 in
Therefore
2.718 281
The Exponential
...
correct to six places.
Series.
By
the binomial theorem
,
3! If
n
less
is
greater than
i,
each term of
(i) is
numerically equal to or
than the corresponding term of e
=i + i + i + 2!
i+ 3!
.
.
(2)
TRIGONOMETRIC SERIES
i;o]
319
series (i) is convergent for every value of n > approaches oo, each term after the second approaches as the corresponding term of (2), hence
hence the
the limit of
(
i
+-
J
as
,
i
n approaches
+
oo
,
equals e
=
2.718
i.
As n
its limit
.
(3)
by the binomial theorem.
-) (-r\nx
n
n2
21
nx (nx
t)(nx
2) i_
,
'
3!
.
If
ft
is
greater than
i,
each term of
(4) is
equal to or less than the
corresponding term of the series
2
3
which in Article 168 'was shown to be convergent for every value of hence (4) is convergent for every value of x. As n approaches oo,
x,
each term of
(4), beginning with the third, approaches as the corresponding term of (5). Hence, as n approaches oo,
n/ Finally,
by the law
2!
3!
its limit
nl
of exponents
no matter how large n and x may be. As n approaches oo, the expression on the right approaches the series (6) as its limit, while the limit of the expression within the brackets on the left equals e, hence in the limit
The
known
as the exponential series because equivalent of the exponential function e*. series (8) is
it is
the
PLANE TRIGONOMETRY
320
[CHAP,
xv
The exponential series may be used to compute the number corresponding to any given natural logarithm. Assign to x any given be the number which is obtained by substituting k value k y and let
N
for
x in the
Then
series (8).
e
=
k
N, and
natural system of logarithms, Article 36,
since e is the base of the
we have by
the definition of
a logarithm, that
is,
N is the number which has k for its natural logarithm. The Logarithmic
171.
and
By
Put
Series.
i
+y = e*, then k = log(i +y),
+ y) x =ekx = eK(H-v>.
(i (8), Article 170,
(i
+ yy -
* <>+>
=
i
+ x log. (i + y) + I*l2giL ..og.i
y
.
3!
By
the binomial theorem
By
the preceding article the first of these series is absolutely conand so is the second provided y < i, hence we may treat like algebraic expressions containing a finite number of terms
vergent,
them
(Article 163).
Equating the two
and dividing out
x,
we 2!
subtracting
series,
from each side
i
obtain
3!
y^ + ** ^
y*
+
(*
2\
-*)(*3
2
)
y+ ^^
.
.
.
!
This equality holds for every value of x. As x approaches o, the series within the brackets on the left approaches i, and the series on the right approaches y
2L 2
4- 2L 3
+ -+ 234 Similarly,
if
y
is
.
.
.
t
-
'
hence in the limit
+(-l)- 1 ^+n
negative, but numerically less than
i,
we
obtain
(I)
TRIGONOMETRIC SERIES
172}
321
The series (i) and (2), which differ only in the sign of y, are known as the logarithmic series because they are equivalent to the logarithms of
i
+ y and
i
y respectively.
the aid of (i) and (2) the natural logarithm of any given number between o and i may be computed, but the actual computation
By
of logarithms will
be very much shortened by means of the method
explained in the following article.
Calculation of Logarithms.
The
series (i)
and
(2) of
the
preceding article have been shown to be absolutely convergent (Article 168, (d) ); we may therefore subtract the second from the first
and obtain
If in this series
l
oge
5+J =
we put y = *
2/
-
2U+ 1
1
(i)
,
we
obtain
I
.+.
1
1
--)
.+
\ m
.
or loge (n
The
+ i) = log n + 2 ( e
-
convergent so long as y is less than greater than o. If n is greater than i the series converges rapidly, so that but a few terms need be taken to obtain the first five or six decimal places of the number to which the i,
that
series (2) is absolutely is,
so long as
n
is
The natural logarithm of any number may thus be computed provided we already know the logarithm of the next lower number. But the logarithm of i to any base is o (Article Knowing log c 2 26), hence the logarithm of 2 may be computed. we may compute loge 3, thence logc 4, etc. Of course, only the logarithms of prime numbers need be computed by means of the series, for the logarithm of any composite number is equal to the sum of the logarithms of the factors of the number. series converges.
readily
To compute
the natural logarithm of 2
we put
in (2)
n =
i,
thus (3)
7'3
PLANE TRIGONOMETRY
322
[CHAP,
xv
The actual calculation to five places of decimals may be conveniently arranged a follows. Denote the terms of the series in the parenthesis by ifi, W2, 1*3, etc., then
=
0.000 000 O
=
0.666 666 7
= =
o.ooi 646 i
=
o.ooo 130 6
2.0000000
=
u\
=
4
= o.ooo on 3 ii = o.ooo ooi o ~ 13 = o.ooo ooo i =
HI
0.666 666 7-7i
-f-
0.008 230 5
-f-
0.074 074
i
3 5
o.ooo 914 5-7-
7
o.ooo 101 5-72
< .ooo 3 o
on
3
log e
I
9
-4-
o.ooo ooi 3
Adding
0.024 691 4
+ Ui + W2 +
'
+ W = 0.693
J 47
7
2
There is an error in the last figure, due to the neglected parts of the fractions added, but this error cannot exceed 7 0.5 or 4 in the last decimal place. Besides, there is the neglected portion of the series, consisting of the terms
X
15
15
'
3
'
3
17
IS
' )
Adding w 8
+ #9
-,
-u,
3
19
2
15
3 ^
*5
-
,
3
2
^
_2^
'
15
'
3
.,,
-
whose
series
9 2
,
18
I5'3 V
The
1^ 4
3
within the brackets on the right
ratio is ^2 3
-
.
The sum
of this series is
9
neglected portion of the series 6
is
less
than
4 4
,
3
3
a geometric
is
=
-
series
2 hence the
i-48' ^-716
15 -3
2 8
From the computation above we already know that-^ = o
=
?
120 -3 18 o.ooo ooi
3,
TRIGONOMETRIC SERIES
172]
-
hence
=
.
The
o.oooooooi.
120- 3 13
323
which the above sum
error, to
0.693 147 2 is subject, may therefore cause a difference of at most It follows that Iog e 2 = 0.693 X 5 i in the sixth decimal place. > correct to 5 places of decimals. -
we
Similarly decimals:
Ioge3
log.2
obtain the following, each correct to 5 places of
+2
+
3'5
\S
Ioge 4
+
+ -- + -;
-
5-5
7
7'5
3-9
log, 6
=log e 2+log3
log e 7
= loge 6+2
(
3
5'9
3
'
3
\
H
i3
5
.
=
1.38629
.
.
.
=
I.6o 9 4 4
-
-
-
=
1.791 76
=
1.945 91
= = =
= 3 log* 2
=2loge 3 10= loge 2 +
.
... -
-
13
log e Q log
.
7
7'9
--
+ -1 i3
5
1-09861
9-SV
= 2 loge 2 \9
loge 8
3
loge 5
2.07944
.
.
.
2.197 22
...
2.302 $9
...
and so on. It will be observed that the number of terms of the series, which need be computed to obtain the logarithm correct to a given number When n is 43 of decimal places, grows smaller as n grows larger. or more, the
first
term
of the series suffices to give the first five
places of the logarithm. When the natural logarithm of a
logarithm
easily obtained
is
from
it,
number for
by
known, the common
is
Article 35
10
Now
was
just
log*
N=
loge 10
found to be 2.302 59
-
ogg
...
2.302 59
The number called the
0.434 29
.
modidus * of the
.
.
,
.
.
0.434 29
.
,
hence
...
loge
N
more accurately 0.434 294 48
common system
of logarithms.
(4)
.
.
We
.
,
is
have
then the following simple rule. *
of
The base
2,
3
and
5,
the modulus of the common system, and the natural logarithms have been calculated to more than 250 places of decimals.
e,
PLANE TRIGONOMETRY
324
[CHAP,
xv
RULE: To find the common logarithm of a number, multiply the corresponding natural logarithm by the modulus of the common system, Thus for the numbers from 2 to 10 inclusive we find the
Common
each correct to
From
(4)
Logarithms
five places of decimals.
we have
log c
W=
2.30259
.
.
N.
logio
(5)
By means of (5) we can readily find any required natural logarithm from a table of common logarithms. It is therefore not very important to have separate tables of natural logarithms. Thus, if at any time we needed to know the natural logarithm of 237.3 we would look up the common logarithm of 237.3 an d multiply the result by 2.30259 and obtain log* 237.3
=
2.30259 logio 237.3
=
2.30259
X
2.37530
=
546933
173. Errors Resulting from the Use of Logarithms. In Article 44 were given certain rules governing the accuracy to be expected in the results obtained by the use of logarithmic tables containing a
number
certain
of decimals.
Some
of these rules
we
are
now
able to
verify.
Let
N
be any number obtained by logarithmic computation, and the number due to the use of logarithms, so that the
d the error in
true result positive.
and the logio
is
N+
The result
(#
5,
where of course
difference
5 may be negative as well as between the logarithms of the true result
N is
+ )- logio N - logio
~
=
logio
i
+
TRIGONOMETRIC SERIES
i73l
+
Y
325
=
where M
043429
(By Article 172,
(4) )
- JL-j- J!__ (By Article 169,
=
>
/*
provided This error M
-
(i) )
approximately,
small as compared with N.
5 is
in the logarithm of
N
is
the result of the errors in the
logarithms as given in the tables.
In a four-place table the value of a unit in the -iio 4
in a five-place table it is o.ooooi
value of the unit in the last place
=
>
last place is o.oooi
in
io5
an w-place table the
The neglected part
-
is
=
of
any one logarithm does not exceed one-half of a unit in the last place; when several logarithms are added the positive and negative errors will tend to offset each other, but in special cases the errors
may be
We
accumulative.
do not exceed a unit
will
assume that the combined errors This gives for an w-place
in the last place.
table Iog 10
(N
i=
+ B)- loglo tf
JL,
from which IO n /Z
Putting n successively equal to
^ ^ 6 ^
I0n 3, 4, 5, 6, 7,
we
find:
or less than J of
%
In a
3-pl. table, 6
0.002 3
In a
4-pl. table, 5
o.ooo 23 N, or less than J of j^
In a
5-pl. table,
In a
6-pl. table, 8
In a
7-pl. table, 5
jV,
o.ooo 023 N, or less
of AT,
% of N, than J of T fa % of N, than } of J^TT % of N, than i of % of #
=
o.ooo 002 3 N, or less
^
o.ooo ooo 23 N, or less
It follows that the third figure of a
i
-rvinsv
number found from a
three-
place table, the fourth figure of a number found from a four-place table, the fifth figure of a number found from a five-place table, etc.,
cannot be relied upon with certainty.
PLANE TRIGONOMETRY
326
xv
[CHAP,
EXERCISE 67 1.
Compute the
natural logarithms of
2.
From
n,
Ans.
places of decimals.
101, 257, each to four
2.3979, 4-6151, 5-549 1
'
the results already obtained compute the natural logarJ, each to four places of decimals.
ithms of 12, 15, 0.05,
Ans. 2.4849, 2.7081, 3. From of n, 101,
the results of Problem
i
compute the
2.9957,
1.0986.
common
logarithms
257, each to four places of decimals, and compare results with those given in the table of common logarithms. 4.
By means
of a table of
common
logarithms, find the natural
logarithms of 7, 341, 0.0473.
Ans. 5.
Prove that
6.
Compute
lem 7.
r
%,
log, 341
=
= -
5.4848, Iog 6 0.0473
3-05I3-
- to five places of decimals, using the series in Prob-
6
Ans. 0.36788
5.
Show
your
.
.
.
compound amount A on P dollars for / years at be added to the principal as fast as it accrues, is
that the
interest to
A = Pf. (Suggestion.
when the period
Set up the expression for the compound amount is the nth part of a year and find the limit which this
expression approaches as 8.
Sum
the series
i
n approaches
oo).
+L n 4. e
+ ie + -+ e 2
.
.
.
Ans. e
174. Limiting Values of the Ratios S12L2,
**,
a? 7
x being the
a?
radian measure of the angle. Let # be the radian measure of any angle less than With the vertex O as a center and any radius 7T/2.
OA =
o
i
r
c A angle in
describe an arc cutting the
A and
B.
Join
A and
B.
sides of
the
From # draw a
Fig. 177perpendicular to AO cutting AO in C. At the perpendicular to AO cutting OB produced in T. Then
A
erect
TRIGONOMETRIC SERIES
*74J
327
AB = rx, AT - r tan x. ^ X C^ = r* sin *. The area of triangle OAB = CB =
r sin x,
arc
2
The
2
_*.
OAB -OAXticAB
area of sector
2
The area
of triangle
Also
triangle
tha tis
,
22
OAT = OAX AT =
^ 222 x
sin
x
<
x
< tan x
by the
(i)
,
hence
i
x approach
which always Similarly,
lies
if
we
(i)
positive quantity sin#,
sm x
let
)
than -
less
Dividing each term of
Now
*.
OAB < sector OAB < triangle 04 T, ^sm < < rHano;
from which for every value of
2
-
^ >
sin
x
x
cos x
--
^ >
cos x
/
then cos x approaches
o,
between cos x and
i,
i,
therefore
x
must approach
i also.
by tan x
divide (i)
cos x
<
-~ <
(3)
i,
tana;
from which
it is
seen that
\
(2)
i
^nx i
approaches
x
as x approaches o.
Summing up, roaches As x approaches
o,
the limit of 5JL5 equals
As x approaches
o,
the limit of
i.
x
(4)
oo
As n approaches
oo
,
the limit of
equals
x
Corollary.
As w approaches
-
sm
Wn
)
equals
i.
i.
X/H ,
the limit of
tan (*'*'
x/n
|
equals
i.
(5)
PLANE TRIGONOMETRY
328
176. Limiting Value of cosn -and
n
[CHAP,
as
x/n
\
xv
w approaches oo.
/
Put y then
=
logy
2
-log(i-sin -) 2 \ n/
= - -/sin2 - + I sin n
2\
The
4
2
~
n
3
series in the parenthesis is less
~ n which
is
a geometrical
+ sin
hence, log y
Now
let
is
- sin2 (x/n)
=
'
00
+ sin
6
2
n
-, whose
n
sin2 (x/n)
=
i
sum
is
(Art. 166)
tan2^,
n
cos 2 (x/n)
=
w o,
n
approaches
Art. 171, (2).
-
then - approaches
,
by
/
2
n \'-
n
-\
- tan2 numerically less than
n approach
n
-
4
+
than
series, ratio sin
sin 2 (x/n) i
and
i sin6 -
+ n^
(Article 174, (5)) (5)),
2
tan tan -
n
^n
)
#/w
tan - approaches
o.
hence in the limit
x/n log y approaches o,
and consequently y Also from Article 174,
=
n cos - approaches
(2),
1> ^W5) >C08 2
>
n
x/n
and
i.
n
therefore
provided n
is
taken so large that-
n
<
-, that 2
is, if
n
is
taken suf-
TRIGONOMETRIC SERIES
1 76]
329
sm ficiently large f
V latter
has just
/sin \x/n)\
n fo^V lies always ' between i and cos *, but the n' x/n I been shown to approach i as n approaches o hence ,
aj go
approves
Summing
l
As n approaches
oo
,
up:
j
As n approaches
oo
,
The
176.
and
the limit of (
cos
nB
and Tangent
Sine, Cosine
(3), Article 159,
=
n
cos 6
H--,
i.
(i)
sm
W nA
x/n
equals
i.
(2)
I
In the equations
Series.
namely,
- **fc~i) cosn-
-
2!
^
w(tt
equals^
.J
\
(2)
-
the limit of cos n
-^
2)(n "
1)(
2
*)
sin2
n_ 4
sm 44/l(?-.. .
,
cos
B
r
4-
=
n
n
putw0
cos
n- J
^ sin 6
- *("""
I )(
n
- -- - -
(n ^
=ic, so that
2)(w /v
i)(w iii
0=
^)(w sliv
,
cos
6
rt
^
sins
sm5K n .
n
n
4!
n
-
4)1
cos n-s
-, then
n
n
"" 2 )
3
2)(n
n
n
!
.Q(n -
n
4)
n
5!
n
An
examination of the test ratio shows that both series are absoAs n approaches <*> lutely convergent for every value of n. ,
cos
n
-
n n
1
cos'
"1
-
n
sin
-
n
x cos'1
approaches -1 -
n
sm
^
x/n
i,
n)
approaches
x,
PLANE TRIGONOMETRY
330 -
-
sin'
i
-
2i
and
similarly
(*-i)
n
ft
3!
(-)(-
3) cos
-4
2
3!
Sin
4
n
4!
Hence as n approaches
oo,
2 approaches *, approac n 4!
the above series become
These
may
series being absolutely convergent (Article 168, divide (2) by (i) and obtain
3
The law (2).
The
(b),
(c) ),
we
3iS
IS
of the exponents in each of the series (i), (2)
obvious, and so
etc.,
and
(3) is
the law of the coefficients in the series (i) and of the coefficients in the series (3) is too complicated law is
to be worked out by beginners.
177.
We know that
Computation of Natural Function Tables.
the functions of any angle whatever may be expressed in terms of the functions of an angle less than 45. Moreover, the sine and cosine of any angle between 30 and 45 may be expressed in terms of the sine
and cosine sin#
of angles less than
+ siny =
cos x
cos y
=
2 sin
for
30,
by
2LJ12 C os 2LZJ! 2 .
2 sin
x
Article 113, (i),
,
2
22-
(l)
+ y sin x-y *
.
*
+
we put for x, 30 6, and for y, 30 0, and transpose the second terms on the left to the right-hand side of the equation, we obtain If in (i)
sin
(30+0) =
2 sin
30 cos 6- sin(3o-0) =
cos(3o+ 0) = 2 sin 30
sin
0+cos(3o-0)
cos 6- sin
(30-0),
- sin 0+cos (so
-*),
^
TRIGONOMETRIC SERIES
i77l
331
from which it appears that the functions of angles between 30 and 45 may be obtained from the functions of angles less than 30 by giving to 6 successively the values from o to 15. Thus, if = 5, sin (30
cos (30
A
+ 6) = sin 35 = cos 5 - sin 25, + B) = cos 35 = - sin 5 + cos 25.
complete table of natural functions
we know
structed provided
may
therefore be readily con-
the functions of angles between o
and
30.
The sine and cosine of any angle less than 30 may be easily computed by means of the series in Article 176. Suppose we wish to compute the sine and cosine of 10 correct to four places of decimals.
The
radian measure of 10
of logarithms
we
je
4
=
-^ 18
=
0.174 53
.
.
.
,
and by means
find
= =
x2
x
is
0.030 46
o.ooo 93
.
.
.
= =
x9
,
1
.
.
.
x
,
0.005 32
.
.
.
o.ooo 16
.
.
.
Substituting these values in the series for the sine
,
and cosine we
have cos x
sin
i.ooo oo
~= 2
0.015 23
x
=
... 3
!
= + o.ooo 04 ...
+
=
x
0.174 53
...
o.ooo 89
...
!
= + o.ooo oo ...
H
4!
5!
cosio=
0.98481
.
.
.
sin 10
=
0.173 64
...
In either case the error due to the neglected parts of the decimals In either case the added, cannot exceed a unit in the fifth place. error
The Its
due to the neglected terms of the
series in parenthesis is
sum is
i
less
than
x
,
series is less
a geometric
series
than
whose
hence the neglected terms of either
ratio is
series
x
<
i.
add up to
6
-
PLANE TRIGONOMETRY
332
i
!
- 22--
I a- ^(p 74
-x
53 6
-) .
=
0.825 47 ...
!
xv
[CHAP,
...
0tOOQ 000 05 D
In neither case, therefore, can the combined errors affect the fourth decimal place, and we have
=
cos 10
0.9848
.
.
.
=
sin 10
,
0.1736
.
.
.
,
each correct to four places.
As a check we have
+ sin
cos 2 10
2
=
10
+
(0.9848)2
(o. 1 736)2
=
i.oooo.
from the above computation that if the angle is less than the sine and cosine are given correct to four places by the
It appears
10
formulas
cos
a?
=
i
These approximation formulas are remembered.
We
will
next show
how
cosines for intervals of
to
If
=
(i)
=
x
6
+
+ = cos (0 + i') = sin (0
Putting
now
6
=
i'
sin 2'
=
cos
=
=
Next put
2'
2',
sin 3' \ cos 3'
Similarly,
i')
if
$
=
sin 4'
cos 4'
we
.
sufficiently
compute a table
o.ooo 290 888 2
now we put in
a?
(3)
important to be
of natural sines
and
compute the
sine
i'.
By means of the sine and cosine and cosine of i'. We find sin i'
&=
sin
>
i',
.
.
.
cos
,
=
y
we
series
i',
2 sin
cos
2 sin
sin i'
;
i
first
=
0.999 999 957 7
we obtain
sin (6
i'),
+ cos (0 -
i').
i'
,
x
(4J
find
2 sin i' cos i'
sin o'
2 sin i' sin i' -f
cos o
;
= o.ooo 581 776 = 0.999 999 ^3 X
.
.
.
.
.
.
then
= =
2 sin 2'
cos
i'
2 sin 2' sin i'
sin i'
+ cos
i'
= o.ooo 872 665 = 0.999 999 619
.
.
.
3'
= =
2 sin 3' cos i'
sin 2'
2 sin 3' sin i'
+ cos 2'
= =
o.ooi 163 553
.
.
.
0.999 999 322
.
.
.
,
etc.
TRIGONOMETRIC SERIES
178]
To
333
construct a table of sines and cosines for intervals of 10", we first compute the sine and cosine of 10" and then make use
should
of the formulas sin (B -f 10")
cos (6
178.
+ 10")
= = -
2 sin
6 cos 10"
2 sin 6 sin
10"
sin (0
10"),
+ cos (6 -
10").
Approximate Equality of Sine, Tangent and Radian of very Small Angles. In Article 174 it was shown that
Measure
the ratio of the sine to
its
angle expressed in radians, as well as the
ratio of the tangent to its angle expressed in radians,
approaches
i
as
the angle approaches o. This means that for very small angles the sine, the tangent and the angle expressed in radians are approximately equal. Thus, if we actually compute the sine and the tan-
gent of i" by means of the series in Article 176, and compare the results with the radian measure of i", we shall find that the results
The sine, tangent and radian measure when the angle is as large as i the and even places, and measure radian are equal so far as the first five tangent
agree to 15 places of decimals. of i' agree to sine,
n
It follows that when the angle is places of decimals are concerned. or we i small, say less,* may replace either or both the sine and the tangent by the angle expressed in radians, without affecting the first five
places.
EXAMPLE
i.
Find the smallest value of x that
equation 2 sin
#
will satisfy the
+ 3 # = 0.0513.
Solution.
3#
hence
we may
= <
replace sin x
2# x
0.0513
2 sin
0.0513, that
by
x.
x
is,
<
x
0.0171 (less than i),
Then
+ 3#=5* = 0.0513, =
0.01026 radians
= o3s'
17".
NOTE. The methods described in this chapter are not the methods that were used in calculating the tables now in use. The methods actually used were clumsy and laborious as compared with those we have studied. If the tables had to be calculated anew *
In
fact, the first five places are the
same up
to
i
59',
that
is,
practically
2.
PLANE TRIGONOMETRY
334 still
more
[CHAP,
xv
methods would be used, methods based on the
refined
calculus of finite differences, a branch of higher mathematics which
cannot well be explained at this point.
and cosine
of the differential calculus the sine
By means
series
easily derived than has been done in this chapter. In the differential calculus all the series of Article 168 and many
can be
much more
by means
others are derived
theorem known as Taylor's
of a single
Theorem.
EXERCISE 68 1.
Calculate the sine and cosine of 5 correct to five places. Ans. sin 5 = 0.08716, cos 5 = 0.9962.
2.
Using the results of problem
gent, secant
and cosecant
of
5.
Ans. 3.
By means
cosine of 4.
By means x
and
sine
sin 10" =
of the sine
+ cos x = 2
5.
sin
7.
Find the
An
less
o.ooo 048 481
and
i
is
to
and
Find 8 expressed
A
sine
and
straight
A
and
ing a curve arc of some
4,
cos 10"
=
0.999 999 998
8.
cosine series verify the relations: 6.
i.
sin 2
x
=
2 sin
x cos x.
in
rail
B
ACB.
=
=
^ cos x
i
+2 +
be corrected by an amount is
known
26=
9.
compute the
1.4737
three terms of the series for sec x.
first
angle
than
tremities
1
cosine of 10" correct to 10 places.
Ans. sec x
8.
=
1.0038, esc 5
5.
Ans.
be
=
sec 5
of equations (4), Article 177,
Compute the
2
calculate the tangent, cotan-
i,
5
cos 8
+ sin
which
Ans.
mile long,
whose
the distance
which the middle point of the
known
to
rail
CD
6=5' 3O//
B
^
Ffe- 1 ? 8 -
through
moved during
-
ex-
are fixed, expands i inch, formAssuming the curve to be the
circle, find
is
8.
minutes and seconds. i
h
24
to satisfy the equation
1.001599
AB,
**
the expansion.
Ans.
12
ft.
10.14
in.
INDEX (References refer to the pages.)
Common logarithms,
defined, 57
Abscissa, 117 Absolute value, 281 Absolutely convergent series, 308 Accuracy of results, 43
Complement, arithmetic, 56 of an angle, 178
Addition theorems,
Complex numbers,
sine, 210,
211
cosine, 210, 211, 215
tangent, 215
Amplitude, 257, 281 Angle, definition of, 177
measure
of, 179,
181
of depression, 5 of elevation, 5
Antilogarithms, 75 Arc sine, 18 Arc tangent, 18
tables of, 61
Conditionally convergent series, 308 Conjugate complex numbers, 288
Convergency test, 311 Convergent series, 307 Coordinates, 117
Cosecant, 10, 118, 191, 338 hyperbolic, 342 Cosine, 10, 118, 192, 338 curve, 260 of sum, 210, 2ii
Archimedes, 182
Argand,
J. R.,
defined, 280
trigonometric form, 281 Compound interest law, 272
of difference, 210
285
of double angle, 216
Argument, 281
of half angle, 216
Arithmetic complement, 56
of
Asymptote, 258 Auxiliary angles, 240
i8\ 24, 218
hyperbolic, 342
law
of,
128
series, 313, 3 2 9
Binomial
series,
314
Briggsian logarithms, 57 Buergi, Jost, 87
Cotangent, 10, 118, 192, 338 hyperbolic, 342 Coterminal angles, 178 Coversed sine, 10
Catenary, 274 Cauchy, 285 Centesimal measure, 180
Cyclic substitution, 127
Characteristic of logarithms, 58
Decimal measure
rules for, 59
Circular functions, 10 measure, 181
Cofunctions, defined, 19
Cologarithm, 56
Damped
vibrations, curve of, 275 of angles, 180
De Moivre, 289 De Moivre's theorem,
289 Departure, defined, 121, 170 Depression, angle of, 5 Descartes, 285
359
INDEX Divergent
307
Logarithm, defined, 53
131, 223
Logarithms, applications
series,
Double formula,
computation Elevation, angle
mantissa
of,
series, 313* 3 l8
decrement, 276 series, 314,
Fourier's theorem, 268 oscillation,
Function, definition of, of an acute angle, 9 of
any
58
Logarithmic curves, 269
equations, 71
an obtuse
72
characteristic of, 58
Exponential curves, 270
of
of,
321
directions for use, 67
of, 5
Euler's theorem, 338
Frequency of
of, 73,
263
320
trigonometric functions, 78 Ludolph van Ceulen, 182
o,
Mantissa
of a logarithm, 58 Mariner's compass, 101
angle, 118
angle, 191
Modo-cyclic functions, 343
of imaginary angles, 337 Fundamental laws of logarithms, 54 relations, 24, 119
Modulus of
of
common
logarithms, 76, 323
complex numbers, 281
Mollweide, 131
Gauss, 285 series, 310 Goniometric functions, 10
Natural functions, computation
Graphic solutions, 3
Natural logarithms, 76 Natural system of angular measure, 181
Hansen's problem, 150 Harmonic curves, 260
Napier, John, 54, 76, 87
Geometric
Hero
7,
330
Napierian logarithms, 76
of Alexandria, 133
Hipparchus,
of,
tables of, 14, 35
Newton, 131
So
Non-convergent
series,
307
Hyperbola, 255
Hyperbolic functions, defined, 342 formulas, 34S, functions, curves of, 277
Ordinate, 117 Origin, 117 Oscillating series, 307
logarithms, 76 sector, area of,
350
Period, 193, 257 of oscillation, 263
Identities, 32
Periodicity of trigonometric functions,
Imaginary numbers, 278
192
unit, 278
Infinite series, definition, Infinite,
symbol
for,
306
22
Interpolation, 37
Inverse functions, 245 sine,
17
tangent, 17
hyperbolic functions, 348
Periodic curves, 257
fun
t'ons, 193,
257
time, 263 Pitiscus, 51
Pothenot's problem, 153 Principal value of angle, 178, 225 Projection defined, 127 theorem, 126
Latitude, defined, 101, 170
Proportional parts, principle of, 38
Laplace, 53
Ptolemaus, 50
INDEX kadiah, 181 measure, 181 Ratio of convergence, 312
Rectangular coordinates, 117 Reciprocal, denned, 9 relations, 24
Reflection of curve, 272
Rheticus, 50 Schuitze, J. H., 180
Secant, defined, 10, 118, 192, 338 hyperbolic, 342 Semi-convergent series, 308 Series, denned, 306 absolutely convergent, 308 binomial, 314 convergent, 307 conditionally convergent, 308 cosine, 313, 3 29
divergent, 307
exponential, 313, 318
geometric, 310
361 law
Sine,
of, 1
26
329
series, 313,
Sinusoidal curves, 260 Snellius, 153
Small angles, 44, 49, 83 Square relations, 24 Subtraction theorems, sine, 210 cosine, 210 tangent, 216
Supplement of an angle, 178 S and T tables, 84 Tables, of
common
S and T, 84 Tabular logarithmic Tangent, defined,
of half angle, 217
with complex terms, 336 Significant figures, 93
Simple harmonic curve, 260 motion, 263 Sine, defined, 10, 118, 191, 338 curve, 255 Of SUm, 210, 211 of difference, 210 of double angle, 216
18, 24
hyperbolic, 342
law of tangents, 131, 223 series, 329 Test ratio, 312 Three-point problem, 150, 223 Trigonometric functions, 10, 118, 337 formulas, 345 Trigonometry, defined, 8
Versed
sine,
10
Vlacq, Adrian, 87
of half angle, 216
of
18,
24,
218
hyperbolic, 342
338
of difference, 216 of double angle, 216 of
329
79
curve, 257 of sum, 215
non-convergent, 307 oscillating, 307 tangent, 329
sine,
10, 118, 192,
logarithmic, 314, 320
sine, 313,
logarithms, 61
of log. trig, functions, 78 of natural functions, 35
Wave
length, 257
Wessel, Caspar, 285
