PLANE AND SPHERICAL
TRIGONOMETRY
BY
GEORGE N. BAUER, PH.D.
N
AND
W. E. BROOKE, C.E.,M.A.
UNIVERSITY OF MINNESOTA
SECOND REVISED EDITION
D. C. HEATH & CO., PUBLISHERS
BOSTON NEW YORK CHICAGO
PHYSIOS
COPYRIGHT, 1907 AND 1917,
BY D. C. HEATH & Co.
H7
PREFACE
THE plan and scope of this work may be indicated briefly
by the following characteristic features :
Directed lines and Cartesian coordinates are introduced
as a working basis.
Each subject, when first introduced, is treated in a gen-
eral manner and is presented as fully as the character of
the work demands. This avoids the necessity of treating
the same subjects several times for the purpose of modify-
ing and extending certain conceptions.
Statements in the form of theorems and problems are
used freely to indicate the aim of various articles and to
define the data clearly.
Inverse functions are treated more fully than is customary.
The general principles governing the solution of triangles,
the solution of trigonometric equations, and the proof of
identities are carefully presented.
. Special attention is given to the arrangement of compu-
tations.
The sine and cosine series are obtained from De Moivre's
Theorem, thus completing the line of development which
leads to the calculation of the trigonometric functions.
The work on spherical trigonometry contains the devel-
opment of all the formulas that are generally used in
practical astronomy.
The right spherical triangle is treated from two points of
view : as a special case of the oblique triangle, and directly
iii
iv PREFACE
from geometric figures. The work is so arranged that
either view may be presented independently.
The solutions of the oblique spherical triangle by means
of auxiliary quantities, characteristic of astronomy, are
included as interesting mathematical problems and as
preparation for astronomical work.
PREFACE TO THE SECOND REVISED EDITION
The second edition embodies such modifications, rear-
rangements, and additions as have been suggested by experi-
ence in the classroom.
Due to numerous requests tables have been added.
Simple three-place tables have been included. By their use
the numerical work is reduced to a minimum, thus leaving
the student free to give more attention to the principles in-
volved in the solution of a triangle. Accordingly a few tri-
angles, suitable for three-place table or slide rule computa-
tion, have been introduced at the beginning of the list of
problems under the right triangle and also under the oblique
triangle.
In order to give a wider range to computational work four-
place tables as well as five-place tables have been included.
G. N. B.
W. E. B.
MINNEAPOLIS, MINNESOTA,
June, 1917.
CONTENTS
PLANE TRIGONOMETRY
CHAPTER I
RECTANGULAR COORDINATES AND ANGLES
ABT. PAOB
1. Introduction 1
RECTANGULAR COORDINATES
2. Directed lines .1
3. Lines of reference 2
4. Quadrants 2
5. Coordinates of a point 2
6. Signs of coordinates . . 3
7. Exercises 3
ANGLES
8. Magnitude of angles 4
9. Direction of rotation. Positive and negative angles . . 4
10. Initial and terminal lines 6
11. Sign of terminal line 5
12. Algebraic sum of two angles 6
13. Measurement of angles 6
14. Circular or radian measure 7
15. Value of radian 7
16. Relation between degree and radian 7
17. Relation between angle, radius, and arc . . . .9
18. Linear and angular velocity 9
19. Examples 10
v
CONTENTS
CHAPTER II
TRIGONOMETRIC FUNCTIONS
ART. PAGE
20. Trigonometric functions introduced 12
21. Definitions of the trigonometric functions .... 12
22. Values of the trigonometric functions of 30°, 45°, and 60° . 13
23. Values of the trigonometric functions of 120°, 135°, and 150° 15
24. Signs of the trigonometric ratios 17
25. Trigonometric functions are single valued .... 18
26. A given value of a trigonometric function determines an
infinite number of angles 19
27. Examples . 20
CHAPTER III
RIGHT TRIANGLES
28. Statement of problem 22
29. Application of the definitions of the trigonometric functions
to the right triangle 22
30. Trigonometric tables . . 23
31. Formulas used in the solution of right triangles ... 24
32. Selection of formulas 25
33. Check formulas 25
34. Suggestions on solving a triangle 25
35. Illustrative examples 26
36. Examples .......... 29
37 . Oblique triangles 30
38. Applications 31
CHAPTER, IV
VARIATIONS OF THE TRIGONOMETRIC FUNCTIONS
REDUCTION OF FUNCTIONS OF n90°±«
39. Values of functions at 0°, 90°, 180°, 270°, and 360° . . 34
40. Variation of the functions 35
41. Graphical representation 39
42. Periodicity of the trigonometric functions .... 40
43. Examples 41
CONTENTS vii
ABT PAGE
44. Use of formulas 42
45. Functions of — a in terms of functions of a . . . .42
46. Functions of 90° -f a in terms of functions of a . .43
47. Functions of 90° — a in terms of functions of a . .44
48. Functions of 180° — a in terms of functions of a . . .45
49. Laws of reduction . . .46
50. Examples 46
CHAPTER V
FUNDAMENTAL KELATIONS. LINE VALUES
51. General statement 47
FUNDAMENTAL RELATIONS
52. Development of formulas 47
53. The use of exponents 48
54. Trigonometric identities 49
55. Trigonometric equations 49
56. Examples .51
LINE VALUES
57. Representation of the trigonometric functions by lines . 53
58. Variations of the trigonometric functions as shown by line
values 55
59. Fundamental relations by line values 66
60. Examples 57
CHAPTER VI
FUNCTIONS OF THE SUM OF TWO ANGLES
DOUBLE ANGLES. HALF ANGLES
61. Statement of problem 61
62. The sine of the sum of two acute angles expressed in terms
of the sines and cosines of the angles .... 61
63. The cosine of the sum of two acute angles .... 62
64. Importance of formulas ........ 62
65. Generalization of formulas 63
66. Tangent of the sum of two angles 65
67. Cotangent of the sum of two angles 65
68. Addition formulas . 65
viii CONTENTS
ART. PAGB
69. Sine, cosine, tangent, and cotangent of the difference of two
angles 66
70. Exercises .66
71. Double angles 67
72. Half angles 68
73. Sum and difference of two sines and of two cosines . . 69
74. Equations and identities 70
75. Examples . .71
CHAPTER VII
INVERSE FUNCTIONS
76. Statement of problem 74
77. Fundamental idea of an inverse function .... 74
78. Multiple values of an inverse function . . . 75
79. Principal values .77
80. Interpretation of sin sin-1 a and sin-1 sin a . . . .77
81. Application of the fundamental relations to angles expressed
as inverse functions 78
82. The value of any function of an inverse function ... 79
83. Some inverse functions expressed in terms of other inverse
functions .81
84. Relations between inverse functions derived from the formulas
for double angles, half angles, and the addition formulas 82
86. Examples 82
CHAPTER VIII
OBLIQUE TRIANGLE
86. General statement 84
87. Law of sines 84
88. Law of tangents 86
89. Cyclic interchange of letters 85
90. Law of cosines 85
91. Sine of half angle in terms of sides of triangle ... 86
92. Cosine of half angle in terms of sides of triangle ... 87
93. Tangent of half angle in terms of sides 88
94. Area of plane triangle in terms of two sides and included
angle 89
95. Area of triangle in terms of a side and two adjacent angles . 89
CONTENTS
96. Area in terms of sides 90
97. Formulas for solving an oblique triangle .... 90
98. Check formulas 91
99. Illustrative problems 92
100. The ambiguous case 95
101. Examples 98
102. MISCELLANEOUS EXERCISES 102
CHAPTER IX
DE MOIVRE'S THEOREM WITH APPLICATIONS
103. Introduction to chapter 113
104. Geometric representation of a complex number . . . 113
105. De Moivre's theorem 114
106. Geometric interpretation 115
107. Applications of De Moivre's theorem 116
108. Cube roots of unity 116
109. Fifth roots of unity . . . ... . .117
110. Square root of a complex number 118
111. Any root of a complex number ...... 119
112. Sin n a and cos n a. expressed in terms of sin a and cos a . 120
113. Comparison of the values of sin a, a, and tan a, a being
any acute angle 121
114. Value of sm a for small values of a 121
a
115. Sine and cosine series 122
116. Examples 124
SPHERICAL TRIGONOMETRY
CHAPTER X
FUNDAMENTAL FORMULAS
117. The spherical triangle 127
118. Law of sines 128
119. Law of cosines 129
120. Law of cosines extended 130
121. Relation between one side and three angles . . . 131
122. The sine-cosine law 131
CONTENTS
123. Relation between two sides and three angles . . . 132
124. Relation between two sides and two angles . . . .132
126. Formulas independent of the radius of the sphere . . 133
CHAPTER XI
SPHERICAL RIGHT TRIANGLE
126. Definition of spherical right triangle 134
127. Formulas for the solution of right triangles .... 134
128. Direct geometric derivation of formulas .... 134
129. Sufficiency of formulas 137
130. Comparison of formulas of plane and spherical right
triangles 137
131. Napier's rules 138
132. Side and angle opposite terminate in same quadrant . . 139
133. Two sides determine quadrant of a third .... 139
134. Two parts determine a triangle 139
135. The quadrant of any required part 140
136. Check formula 140
137. Solution of a right triangle 141
138. Two solutions 141
139. Examples 143
140. Quadrantal triangles 143
141. Isosceles triangles 143
142. Examples 144
CHAPTER XII
OBLIQUE SPHERICAL TRIANGLE
143. Statement of aim of chapter 145
GENERAL SOLUTION
144. Angles determined from the three sides .... 145
145. Sides found from the three angles 146
146. Delambre's or Gauss's formulas . . . . . .147
147. Napier's analogies 147
148. Formulas collected . . 148
149. All formulas excepting law of sines determine quadrant . 148
CONTENTS xi
ART. *AGE
150. Theorem to determine quadrant ...... 149
151. Second theorem to determine quadrant .... 149
152. Third theorem to determine quadrant ..... 150
153. Illustrative examples ........ 150
154. Two solutions ......... 153
155. Area of spherical triangle ....... 154
156. Examples .......... 154
SOLUTION WHEN ONLY ONE PART is REQUIRED
157. Statement of problem ....... .156
158. From two sides and the included angle, to find any one of
the remaining parts . . . . . ' . . 156
169. Two parts required ........ 158
160. Problems .......... 159
161. Each unknown part found from two angles and the included
side .......... 160
162. Each unknown part found from two sides and an angle
opposite one of them ....... 161
163. Each unknown part found from two angles and a side
opposite one of them ....... 163
164. The general triangle ........ 164
ANSWERS ........... 165
PLANE TRIGONOMETRY
CHAPTER I
RECTANGULAR COORDINATES AND ANGLES
1. Introduction. The word trigonometry is derived from
the two Greek words for triangle (r/otytuvov) and measure-
ment (/Aerpta). Originally trigonometry was concerned
chiefly with the solution of triangles. At present this is
but one part of the subject.
Certain preliminary considerations, concerning directed
lines and angles, are neqessary before introducing the fun-
damental definitions of trigonometry.
RECTANGULAR COORDINATES
2. Directed lines. A positive and a negative direction
may be assigned arbitrarily to every line.
If the direction from A to C is positive, the opposite
direction from C to A is negative.
If we let the order of the let- ^
ters indicate the direction in A c
which a segment is measured, it is evident that AC and CA
represent the same segment measured in opposite direc-
tions; hence
AC = - CA or -AC = CA.
Also, if B be a third point on the line, the segments AB
and BA are opposite in sign; likewise the segments BO
and CB. Hence
AB=-BA or -AB = BA,
and BC= - CB or - BC= CB.
1
2 TRIGONOMETRY
Then for all positions of A, B, and C on a line it follows
that
Thus for the following figures :
% K_i_f Ti P» Q
5
8
12 8 + f 4^ 12
-12
A -8
C
B
0
4-_6 + (4-l(r)-4
C-€
B
0
4
\ A
X'
3. Lines of reference. Two directed lines, perpendicular
to each other, may be taken as lines of reference or axes.
They are usually designated by X'X and F'F and are
called the X-axis and the F-axis respectively.
The X-axis is positive from left
to right and negative from right
to left.
The F-axis is positive upward
_j_ and negative downward.
X The point of intersection of the
axes is the origin. The origin
serves as a convenient starting
point from which to measure dis-
tances.
4. Quadrants. The axes produced divide the plane into
four parts called quadrants. The quadrants are designated
by number. The first quadrant is indicated by XOF, the
second by FOX', the third by X'OF', and the fourth by
F'OX.
5. Coordinates of a point. From any given point Px in
the plane, draw a line parallel to the F-axis intersecting
the X-axis in some point A.
Then the distance from the origin to the point of inter-
section, or OA, is the abscissa of the given point P±.
RECTANGULAR COORDINATES AND ANGLES 3
X'C
The distance from the
point of intersection to
the given point, or APlf
is the ordinate of the
given point Px.
The abscissa and the
ordinate of the point P2
are OC and CP2 respec-
tively.
The abscissa and the
ordinate of the point P3
are OC and CP3 respectively.
The abscissa and the ordinate of the point P4 are OA and
AP4 respectively.
6. Signs of coordinates. Abscissas are positive when
measured from the origin to the right, and negative when
measured from the origin to the left.
Ordinates are positive when measured upward from the
.X-axis and negative when measured downward.
Thus OA, APv and CP2 are positive; 0(7, CP3, and AP4
are negative.
7. EXERCISES
1. Locate the point whose abscissa is 4 and whose ordi-
nate is 7. This point is designated (4, 7).
2. Locate the point whose abscissa is 2 and whose ordi-
nate is 5, i.e. the point (2, 5).
3. Locate the points (- 3, 4), (- 6, - 3), (5, - 1), (0, 4),
(- 7, 0), and (- 8, 10).
4. Locate the points (1J, 3), (- |, 0), (m, n), (a?, y), (a, 0).
5. What is the locus of the points whose abscissas are 6 ?
6. What is the locus of the points whose ordinates are — 3 ?
7. What is the locus of the points whose abscissas are
twice their ordinates ?
TRIGONOMETRY
ANGLES
8. Magnitude of angles. In elementary geometry the
angles considered are usually less than two right angles ;
but in trigonometry it is necessary
to introduce angles of any magni-
tude, positive or negative.
To extend the conception of an
angle, suppose a line to revolve in
a fixed plane about a fixed point 0,
from the initial position OX to the
successive positions OY} OX', and
OP, generating a*, an angle greater
than two right angles.
• If the line continues to revolve,
making more than one complete rev-
olution, it generates an angle a which
is greater than four right angles.
Evidently by continuing the rotation
an angle of any magnitude may be
generated. Thus the size of the angle a depends upon the
amount of rotation of OP and is designated by an arc.
9. Direction of rotation. Positive and negative angles. As
a positive and negative direction may be assigned arbitra-
rily to a line, so a positive and nega-
tive sense of generation may be assigned
arbitrarily to an angle.
The direction of rotation indicated
— by the arrows in the figures of Art. 8
is the positive direction ; the opposite
direction, indicated by the arrow in the adjoining figure, is
the negative direction.
* Angles will usually be designated by Greek letters :
a ... Alpha 5 ... Delta
/3 . . . Beta 6 . . . Theta
7 ... Gamma 0 ... Phi
RECTANGULAR COORDINATES AND ANGLES 5
A positive angle is an angle generated by a line rotating
in the positive direction.
A negative angle is an angle generated by a line rotating
in the negative direction.
Negative angles, like positive angles, may have unlimited
magnitude.
10. Initial and terminal lines. The fixed line from which
both positive and negative angles are measured is the initial
line. It usually coincides with that part of the X-axis
lying to the right of the origin. Thus, in the last three
figures, OX is the initial line.
The final position of the revolving line, marking the ter-
mination of the angle, is the terminal line. Thus, in the
last three figures, OP is the terminal line.
Two or more unequal angles may PN
have the same terminal line. Thus
the positive angle a and the nega-
tive angles ft and y have the same
terminal line.
Angles having the same terminal
line, and the same initial line, are called coterminal angles.
11. Sign of terminal line. The terminal line, drawn from
Q f the origin in the direction of the ex-
/ tremity of the measuring arc, is posi-
/ tive ; the terminal line produced,
drawn from the origin in the oppo-
X site direction, is negative.
Thus the terminal line OP, of the
, angle a, is positive, and the terminal
line produced, OQ, is negative. As
the terminal line OP revolves it retains its positive sign.
12. The algebraic sum of two angles. To construct the
algebraic sum of two angles, conceive OP to rotate from the
6
TRIGONOMETRY
position OX} through the angle a, to OP'; then from this
Pf
0 X
a positive
R positive
O X
a positive
R negative.
a negative
positive
position let it rotate through the angle fi, whether positive
or negative, to OP. Then XOP is the desired angle a + ft.
13. Measurement of angles. Various units may be em-
ployed in the measurement of angles. In elementary geom-
etry the right angle is frequently used. Two other units in
common use are the degree and the radian. The degree
is generally used in practical problems involving numerical
computations, while the radian is essential in many theoret-
ical considerations.
The degree is defined as one ninetieth of a right angle.
The degree is divided into sixty equal parts called min-
utes. The minute is divided into sixty equal parts called
seconds.
Then 60 seconds (60") = 1 minute.
60 minutes (60') = 1 degree.
90 degrees (90°) = 1 right angle.
The angle 26 degrees, 39 minutes, and 57 seconds is
written 26° 39' 57".
RECTANGULAR COORDINATES AND ANGLES 7
14. Circular or radian measure. A radian is an angle of
such magnitude that, if placed with its vertex at the center
of any circle, it will intercept an arc equal
in length to the radius of the circle.
Thus, if the arc XP is equal to the
radius OX, the angle XOP is a radian,
or Z. XOP = lr, r being used to designate
radian.
15. Value of the radian. Since, in the same circle or in
equal circles, angles at the center are proportional to their
intercepted arcs, it follows that
/.XOP
ZXOY
one radian
or
one right angle i (2 irr)
2
Therefore, one radian = — • one right angle.
7T
or TT radians = 180°.
It is clear that the value of the radian is independent of
the radius, depending only upon the constant TT and the
right angle, and hence is an invariable unit.
16. Relation between degree and radian. From the pre-
ceding article it follows that
i-=^, a)
'
Also, from equation (1),
r±jf_ (3)
180'
or 1° = 0.01745''. (4)
Equations (1) and (2) are used to convert radians into
8 TRIGONOMETRY
degrees, and equations (3) and (4) are used to convert
degrees into radians. Thus,
fromd),
o Y TT
from (2), 4r = 4 (57° 17' 44") = 229° 10' 56",
from(3), 20°=
from (4), 3° = 3 (0.01745') = 0.05235r.
Using equation (3) to convert degrees into radians intro-
duces TT into the numerical value of the angle. Thus TT
becomes associated with radian measure. Since the radian
is the angular unit with which TT is commonly associated,
no ambiguity arises by omitting to state with each angle,
expressed in terms of TT, that the radian is the unit.
Thus, 90° = - radians, 180° = TT radians, 29° = radians
are usually written
90° = -, 180° = TT, 29° =
2 loO
It must be especially noted that when no unit is specified
the radian is always understood. The constant TT is always
equal to 3.1416, and can never equal 180°, but TT radians
are equal to 180°.
17. Relation between angle, radius, and arc. It is evident
from the definition of a radian that if
any arc of a circle AB be divided by
^^ the radius, the quotient indicates the
number of radians contained in the
central angle subtended by the given
arc, hence
radius
where angle AOB is expressed in radians. Representing
RECTANGULAR COORDINATES AND ANGLES 9
the length of the arc AB by a, the radius by r, and the
angle AOB by 0, the relation above may be written
or
a=r9 (1)
It should be especially noted that the angle 0 is expressed
in terms of radians.
In the above figure 9 is approximately equal to 21 radians.
Equation (1) expresses the relation between a, r, and 0,
and determines any one of these elements when the other
two are known.
PROBLEM. What is the length of the arc subtended by a
central angle of 112° in a circle whose radius is 15 feet ?
Solution. Reducing the angle to radians it is seen that
112° =?^-r = 1.95'".
45
Hence 0 = 1.95.
Substituting in Equation (1) we have
a = 15 x 1.95 = 29.3
Hence the length of the arc is 29.3 ft.
18. Linear and angular velocity. Equation (1) of Art. 17
leads directly to the relation between linear velocity and
angular velocity in uniform motion in a circle. Suppose a
point P moves along the circumference of a circle at a
constant velocity v, describing an arc a in time t ; then a/t
is called the linear velocity and is represented by v. During
the same time t, the angle 0 is generated ; and 6/t is called
the angular velocity, and is represented by the Greek letter
o> (omega).
10 TRIGONOMETRY
Dividing Equation (1), Art. 17, by t gives
« = ^or^ = r«>
t t
i.e. the linear velocity is equal to r times the angular velocity.
While in general angular velocity may be expressed in
any units whatsoever, in the equation v = rco the angular
velocity must be expressed in radians per unit of time.
PROBLEM 1. If a point moves 26 feet in the arc of a
circle of radius 7 feet in 3 seconds, what is its angular
velocity ?
Solution. The linear velocity of the point is %£- feet per
second.
Substituting in equation (1) we have
¥ = 7 a,
.-. w = |f = 1.238 radians per second.
PROBLEM 2. A flywheel makes 200 revolutions per
minute. Show that its angular velocity is 72,000° per
minute or — — radians per second.
If the radius of the flywheel is 3 feet, show that the
velocity of a point on the rim is 42.84 miles per hour.
19. EXAMPLES
1. Construct the following angles : 30°, 45°, 60°, 135°,
300°, - 60°, - 90°, - 390°, - 420°.
2. Construct approximately the following angles : 2
radians, 3|- radians, — ^ radian, — 4 radians, 9 radians.
3. Construct the following angles :
7T 7T 7T 5 7T 5 TT
2' ~3' 4'1 ~T~' T'
4. Eeduce the following angles to radian or circular
measure : 10°, 30°, 45°, 60°, 135°, 225°, - 270°, - 12°,
- 18°, 24° 15', - 612° 19' 25".
RECTANGULAR COORDINATES AND ANGLES 11
5. Reduce to degree measure : 2 radians, 5 radians,
— 3 radians, ^ radian, — ^ radian, b radians.
6. Reduce to degrees :
TT 3_7r _5jr 3.1416 2 TT + 1 2
3' 4 ' " 3 ' 7T "' TT' 2 ' 7T + 3*
7. If an arc of 30 ft. subtends an angle of 4 radians,
find the radius of the circle.
8. In a circle whose radius is 5, the length of an intercepted
arc is 12. Find the angle (a) in radians, (6) in degrees.
9. If two angles of a plane triangle are respectively equal
to 1 radian and ^ radian, express the third angle in degrees.
10. In a circle whose radius is 12 ft., find the length of
the arc intercepted by a central angle of 16°.
11. Find the angle between the tangents to a circle at
two points whose distance apart measured on the arc of the
circle is 378 ft., the radius of the circle being 900 ft.
12. An automobile whose wheels are 34 inches in
diameter travels at the rate of 25 miles per hour. How
many revolutions per minute does a wheel make ? What is
its angular velocity in radians per second ?
13. Assuming the earth's orbit to be a circle of radius
92,000,000 miles, what is the velocity of the earth in its
path in miles per second ?
14. The rotor of a steam turbine is two feet in diameter
and makes 25,000 revolutions per minute. The blades of
the turbine, situated on the circumference of the rotor,
have one-half the velocity of the steam which drives them.
What is the velocity of the steam in feet per second ?
15. A belt travels around two pulleys whose diameters
are 3 feet and 10 inches respectively. The larger pulley
makes 80 revolutions per minute. Find the angular
velocity of the smaller pulley in radians per second, also
the speed of the belt in feet per minute.
CHAPTER II
TRIGONOMETRIC FUNCTIONS
20. The trigonometric functions, upon which trigonom-
etry is based, are functions of an angle.
These functions are the sine, cosine, tangent, cotangent,
secant, cosecant, versed sine, and coversed sine of an angle.
For any angle a they are written sin a, cos a, tan a, cot a,
sec a, esc a, vers a, and covers a.
21. Definitions of the trigonometric functions. Let OX
and OP be the initial and terminal lines respectively of any
angle a.
Y
0
AX
X'A
X A
X'
Let P be any point on the terminal line,
OP or r the distance from the origin to the point P,
OA or x the abscissa of the point P, and
AP or y the ordinate of the point P.
12
TRIGONOMETRIC ^UNCTIONS 13
It should be noted that OP, OA, and AP are directed
lines and hence
r = OP and not PO
x = OA and not AO
y = AP and not PA.
Then the trigonometric functions are defined as follows :
sin a = - =
ordinate
r distance
r distance
Y ordinate
tan a = ^ = — — : —
x abscissa
= * = abscissa
y ordinate
= /: = distance
x abscissa
A* distance
esc a = - = —
/ ordinate
vers a = 1 — cos a
covers a = 1 — sin a
It will be observed that each of the first six functions is
defined as a ratio between two line segments. These are
the fundamental trigonometric functions, and the ratios de-
fining them are called the trigonometric ratios.
The first six functions are called trigonometric ratios.
The expression trigonometric functions is more general and
embraces the versed sine, coversed sine, and the trigono-
metric ratios. It is evident, from the definitions, that the
values of the trigonometric functions are abstract numbers.
22. Values of the trigonometric functions of 30°, 45°, and
606. A few concrete illustrations serve to show the nature
of the trigonometric functions, to fix ideas, and to prepare
the way for more general considerations.
14
TRIGONOMETRY
Functions of 30°. Let OPF be an equilateral triangle
having its sides equal to 2 units. Place the triangle with
a vertex at the origin so that OX bisects the angle POP.
Then, by geometry, the angle A OP =30°, the ordinate
AP= 1, and the abscissa OA = -\/3.
Hence, applying definitions,
sin30° = i=.500
P
cos 30° = =.866
X'
o
V3 jA X
\j
F
V3 3
cot 30° = — = V
1.732
vers30°=l-iV3=.134 sec 30° =-^z = |V3 = 1.155 •
covers 30°= 1 - - = i = .500 esc 30° = | = 2.000
2i 2i J-
PROBLEM. Find the values of the trigonometric functions
of 30°, as above, taking OP= 1.
Functions of 45°. Let OAP be
a right-angled isosceles triangle
having its sides OA and AP each
equal to 1. Then Z. AOP = 45° and X '
the distance OP = V2.
Hence, applying definitions,
sin 45° = — = - V2
V2 2
cos 45° = — = lV2
V2 2
1
1 A X
sec 45°=-^ =
esc 45° = ^— =
vers45° = l-i-
cot45° = =
covers 45°
1-V2
TRIGONOMETRIC FUNCTIONS
15
PROBLEM. Find the values
of the trigonometric functions
of 45°, taking OA = 3.
Functions of 60°. Let OPFy
an equilateral triangle having
each side equal to 2, be placed
as in the figure. Then the
abscissa and ordinate of P
are 1 and V5, respectively.
Hence, applying definitions,
sin 60° =
1 A F X
cos 60° = -
2
V3
2~2
cot 60° = -L = | V3 covers 60° = 1 - - V3
PROBLEM. Find the values of the functions of 60°, as
above, taking OP= 4 a.
The values of the sines and cosines of 30°, 45°, and 60°
are used frequently and should be memorized. The follow-
ing table may be found helpful :
sin 30° = i VI = cos 60°
2
23. Values of the trigonometric functions of 120°, 135°, and
150°. By using the magnitudes of the figures of Art. 22
and properly placing them with respect to the axes, the
16
TRIGONOMETRY
values of the trigonometric functions of various angles may
be obtained.
Functions of 120°. From the figure and definitions it is
evident that
V3
X' A -1
120
sin 120° = -^
-:
2 2
cos 120° = — = -1
covers 120° = 1 —
V3
cotl20° = -=i = --V3
= - sec 120° = — =-2
2 — 1
esc 120°=— 7==? A/3
Functions of 135°. From the figure and definitions it is
evident that
V2 2" P
1 1 V2
:=l = _lV2 i V
V2 2
±
1 X' A -1
_^ ~
"l""""1 Y
0 X.
sec 135° = = - V2
esc 135° = -^?= V2
covers 135° = l-
TRIGONOMETRIC FUNCTIONS
IT
Functions of 150°. From the figure and definitions it is
evident that
sin 150° = -
cos 150° =
tan 150° =
cot 150° =
sec 150° =
-V3
-V3:
-V3
X A -Y3
3V
V3
?V3
3
J50"
csc 150° = =
covers 150° = l- =
w w
PROBLEM. Find the values of the trigonometric functions
of 210°, 225°, 240°, 300°, 315°, and 330°.
24. Signs of the trigonometric ratios. The signs of the
trigonometric ratios of any angle depend upon the signs
of the ordinate, the abscissa, and the distance of any point
on its terminal line. As the terminal line passes from
one quadrant to another, there is always a change of
sign in either the abscissa or the ordinate of any point
on that line, but the distance remains positive. When a
coordinate changes its sign, every trigonometric ratio de-
pendent upon it must also change its sign.
The following table is constructed by taking account of
the signs of the abscissa x and of the ordinate y, and re-
membering that the distance r is always positive. It gives
the sign of each trigonometric ratio of an angle terminating
in any quadrant.
18
TRIGONOMETRY
1st Quadrant
2d Quadrant
3d Quadrant
4th Quadrant
-t-
+ .
—
—
+ -"*•
+ ~
+
+
cos a
± = +
T=
— = —
jH
tan a
H
±_
3=4-
T=
cot a
|^
+ ~
3= +
4- _
sec a
i^
+ _
±_
1^
pep /y
+.
+ .
+
+
+-
Hh
—
—
A2 A
25. Theorem. For every given angle there is one and only
one value of each trigonometric function.
The theorem is demonstrated
for the sine of an angle. The
same method is applicable to
each of the remaining functions.
- Let a be any angle. Eefer-
ring to Art. 21, it is clear that
if it be possible to obtain two
or more values for sin a they
must be obtained by taking dif-
ferent points on the terminal line.
Let Pj and P2 be any two points on the terminal line.
Then by definition
Xf
0
But since the right triangles OA^ and OA2P3 are similar,
TRIGONOMETRIC FUNCTIONS
19
and have their corresponding sides in the same direction, it
follows that
OP, OP2
Hence the value of sin a is independent of the position of
the point chosen on the terminal line, but depends solely
upon the position of the terminal line, i.e. upon the angle.
The above theorem may be stated as follows : The trigo-
nometric functions are single-valued functions of the angle.
26. Theorem. Every given value of a trigonometric func-
tion determines an unlimited or infinite number of positive,
and negative angles, among which there are always two positive
angles less than 360°.
The theorem is demonstrated for a given tangent. A
similar method is applicable to the remaining functions.
Let tan a = — — where m
n
and n are positive numbers.
Then
-f- m — m
tan a —
— n
Locate the point Pl5 whose
abscissa is — n, and whose
ordinate is m. This point
and the origin determine the terminal line of the angle a1}
whose tangent is — — •
n
Likewise the point P2 is located by using — m and n as
ordinate and abscissa respectively. Drawing the terminal
line OP2, a second angle a2 is found, which also has the
given tangent.
The angles a, and <*2> are evidently less than 360°, and
have the given tangent — — • There is an unlimited num-
ber of positive and negative angles coterminal with a, and
«2, all of which have the given tangent. Hence the theorem.
20
TRIGONOMETRY
27. EXAMPLES
In what quadrants does a terminate when
1. sin a = — i 3. tan a = 5. 5. sin a = £.
L <j
2. cos a = i. 4. cot a = — 8. "^ 6. cos a = — 4.
o 5
7. sin a is positive and cos « is negative.
8. tan ct is positive and cos a is negative.
9. cosec a is negative and cos a is negative.
10. tan a is negative and sin a is positive.
11. cos a is negative and sin « is negative.
Give the signs of the trigonometric functions of the fol-
lowing angles :
12. 750°.
i*=.
U14. 560°.
^ 15. 5|7T.
16. -15°.
17. - — •
18. -470°.
10 77r
"T
Find the negative angles, numerically less than 360°, that
are coterminal with the following angles :
20. ±^
3
21. *£.
22. 300°.
23.
24. 5.
25. -495°
26. Construct the positive angles, less than 360°, for which
the sine is equal to J- , and find the values of the other func-
tions of both angles.
SOLUTION. Determine the points
whose ordinates are 2 and whose
distances are 5, as follows :
With 0 as center and a radius 5,
describe a circle. Through a point
on the Y"-axis, 2 units above the
origin, draw a line parallel to the
X-axis, intersecting the circle in
the two required points PI and P2.
Draw the terminal lines OPi and
OP», giving the angles «i and #2, f°*
which we have
TRIGONOMETRIC FUNCTIONS 21
2 2
sin a,i = - and sin «2 = -
5 5
-V2T
COS «i = - COS «2 = -
tan 0! = - =
\/21 21 -V21 21
5 5
CSC «i = - CSC «2 = ~
27. Construct the positive angles, less than 360°, for
which the cosine is f, and find the values of the other func-
tions of both angles.
Find the values of the functions of all angles less than
360° determined by
^ 28. tana = -f ^32. sin « = -•}.
29. cot a = If 33. sin<* = ^.
30. esc a = 3J* 34. cos a = A.
-31. cota = -5.
35. Given tan a = - 4, find the value of sm a cos a
36. Given sec « = 6; find the value of
cot a
sin2 a + cos2 a
cos a
37. Given sin a = .3, find the value of tan a sec a cos a.
38. Given esc a = 8 and tan ft = 3, find the value of
sin a cos ft -f cos a sin ft.
39. Given tan« = i and cos/3 = — ^-VS, a terminating
in the first quadrant and /3 in the second, show that the angle
between the terminal lines of a and (3 is a right angle.
CHAPTER III
RIGHT TRIANGLES
\
28. In every triangle there are six elements or parts.
These are the three sides and the three angles.
When three elements are given, one of which is a side,
the other three elements can be determined.
The solution of a triangle is the process of determining
the unknown parts from the given parts.
In the present chapter it will be shown how the trigono-
metric functions can be employed to solve the right triangle.
29. Applications of definitions of the trigonometric func-
tions to the right triangle. Let ABC be any right tri-
angle. Place the triangle in the first quadrant with the
vertex of one acute angle coinciding with the origin, and
one side, not the hypotenuse, coinciding with the X-axis, as
in the figure.
Y
B
X'
Then, by definitions,
sin a = - =
side opposite
hypotenuse
side adjacent
cos a = - =
c hypotenuse
22
RIGHT TRIANGLES 23
= g = side opposite
b side adjacent
6 side adjacent
cot a = - = —
a side opposite
==c= hypotenuse
6 side adjacent
a side opposite
It is seen that the functions of any acute angle of a right
triangle are expressed in terms of the side adjacent, the
side opposite, and the hypotenuse, without reference to the
coordinate axes.
It follows that
sin/3 = - cot0 = -
C 0
cos/? = - sec/3 = -
c a
30. Trigonometric tables. In the preceding chapter the
trigonometric functions of 30°, 45°, and 60° were calculated.
By processes too complicated to introduce here, tables have
been computed, giving the values of the trigonometric func-
tions of acute angles. These tables generally contain two
parts. In one part the values of the functions, called
natural functions, are given ; in the other part the logarithms
of the trigonometric functions, called logarithmic functions,
are given.
When an angle is given its trigonometric functions can
be taken from the tables, and vice versa. The functions of
known angles thus become known numbers and can be used
in problems of computation.
Approximate values of the trigonometric functions can be
obtained by graphical methods.
24 TRIGONOMETRY
PROBLEM. Measure the distance, abscissas and ordinates
of the points a, 6, c, • • • , j. From these measurements
compute to two figures the sine, cosine, and tangent of 0°,
10°, 20°, • . • , 90°. By arranging the results in tabular
form, a two place table is constructed.
70C
60<
,IOe
This table may be used to solve Examples 1 to 6, Art. 36.
On account of its extreme simplicity the use of this table
allows the attention to be focused upon the fundamental
processes involved in the solution of triangles.
lno
RIGHT TRIANGLES 25
A
31. Formulas used in the solution of right triangles. A
right triangle can always be solved when, in addition to
the right angle y, two independent parts are given. The
formulas usually employed are :
c a2 + 62 = c2
cos a = - cos 8 = -
c c
tona=- tan/? = -
o a
When the computations are made without logarithms,
the formulas involving the cotangent, secant, and cosecant
may sometimes be used advantageously.
32. Selection of formulas.
(a) If an angle and a side are given, it is always possible
to find the unknown parts directly from the given parts
without the use of the formula a2 -f- 62 = c2. To find the
unknown side, select that formula which contains the given
parts and the desired unknown side. The unknown angle
can always be found from a. -f /? = 90°.
(6) If two sides are given, the third side would naturally
be found by the use of a2 + b2 = c2, but in practice it is
generally preferable first to compute an angle by the use of
a formula involving the given sides and an angle. To find
the third side, select a formula containing one of the given
sides, the angle already computed, and the required side.
33. Check formulas. In all computations it is necessary
constantly to guard against numerical errors. However
carefully the computations are made, errors may still occur,
and therefore computed parts should be checked by means
of check formulas. Any formula which was not used in
the solution of the triangle may be used for this purpose.
For the right triangle the formula
may conveniently be used, and in general it is sufficient.
26 TRIGONOMETRY
34. Suggestions on solving a triangle. Make a careful
free-hand construction of the required triangle, and write
down an estimate of the values of the unknown parts.
Large errors will be detected readily, without the use of
check formulas, when the computed parts are compared
with the estimates.
Before entering the tables, and before making any com-
putations, select all the formulas to be used, solve the
formulas for the required parts, and make an outline in
which a place is provided for every number to be used in
the computation. This will often lessen the actual work,
for frequently several required numbers are found on the
same page of the table.
The arrangement of the work is of considerable impor-
tance in every extended computation.
35. Illustrative examples.
1. In a right triangle, given b = 14, a = 35°, to find a, c,
and ft. ^
SOLUTION. Approximate construction. S^<\
Estimate a = 9, c = 17. %,''' \a
By natural functions
a-- cosa = -
0 C
— 7) tan flf ' r — ft — Q0° ct.
6=14
Check
c2 = a2 + &2
a2= 96.10
62 = 196.0
cos a
a = 14 x. 7002 c = 14 -*- .8192 j8 = 56°
a = 9.803 c = 17.09
By logarithms
292.1
c2 = (17.09)2 = 292.1
a = 6tana c = — 0 = 90° - «
b
a
log 6
log tan a
log a
a
14 log b
35° log cos a
logo
c
c
6
c-b
c+ b
RIGHT TRIANGLES
Check
6)
log (c - 6)
log (c + 6)
2 log a
log a
Filling in the above outline, the completed work appears as follows ;
b 0 nno
a = b tan a
b
a
log 6
log tan a
log a
a
14
35°
1.14613
9.84523
-10
0.99136
9.8030
c =
cos a
log b
1.14613
0= 55°
log cos a
9.91336 - 10
logc
1.23277
c
17.091
Check
ai - C2 _ 52 _ (C _ 5) (C + 5^
c
b
c-b
c-f &
17.091
14.
log(c-&)
log(c + &)
0.49010
1.49263
3.091
31.091
2 log a
log a
1.98273
0.99136
Since the check formula gives the same value for log a as that found
in the solution, the computation is in all probability correct.
2. In a right triangle, given c = 6.275, /3 = 18°47f, to find
a, &, and a.
SOLUTION. Approximate construction.
Estimated a = 5, b = 2.
By natural functions Check
a = c cos j8 b = c sin 0 c2 = a2 + 62
a = 6.275 x .9468 b = 6.275 x .3220 a2 = 35.30
a = 5.941 b = 2.021 62 = 4.084
a = 90° - ft 39.384
a = 71° 13' c2 = (6.275)2 = 39.38
28
TRIGONOMETRY
By logarithms
a = c cos /3
b = csin/3
a = 90° - /a
c
ft
logc
log cos /3
log a
a
6.275
18° 47'
0.79761
9.97623 - 10
logc
log sin /3
log 6
6
Check
2 = (c - 6)
0.79761
9.50784
a = 71° 13'
-10
0.30545
2.0205
(c+6)
0.77384
5.9407
. c 6.275
6 2.0205
log (c - 6)
log (c + 6)
0.62885
0.91884
c - & 4.2545
c + 6 8.2955
2 log a
log a
1.54769
0.77384
3. Given a = .064873, b = .12574, to
find a, /?, and c.
SOLUTION. Approximate construction.
Estimate a — 30°, /3 - 60°, c = .15.
sin a
a
.064873
6
.12574
log a
log 6
8.81206 - 10
9.09948 - 10
log tan a
a
9.71258 - 10
27° 17' 23"
/3 62° 42' 37"
c
.14149
b
c-6
.12574
.01575
c + 6
.26723
log a
log sin- a
logc
c
8.81206 - 10
9.66133 - 10
9.15073 - 10
.14149
Check
log (c - 6)
log (c + 6)
2 log a
log a
8.19728-10
9.42689 - 10
17.62417 - 20
8.81208 - 10
RIGHT TRIANGLES
36. EXAMPLES
Solve the following right triangles, y being the right
angle. It is recommended that the first ten problems be
solved by the use of a three-place table of natural func-
tions, or by means of a slide rule.
1. a = 6 +A. c = .0091
a = 20° a = .0029
2. c = 2.5 7. 6=371
a = 35° a = 43°
3. 6 = .84 8. c = 7.72
ft = 75° 6 = 6.87
4. a = 25 9. a=18°
6 = 60 c = .0938
5. c = 82 10. /?=49°30'
a = 37 c = 12.47
The following problems should be solved by the use of
a four-place * or a five-place table.
11. a = 1870 16. c = 12.145
a = 19° 55' a = 9.321
<**-12. c=.3194 17. 6 = 78.545
a = 25° 41' « = 31° 41' 6"
13. 6 = .9292 18. 6=3.4572
/3 = 32° 43' ft = 57° 57' 57"
14. a = .00006 19. c = 20.082
6 = .000019 6 = 16.174
15. c = 1200.7 20. a = 78° 0' 3"
a = 885.6 a = 271.82
* Whenever sides are given to four significant figures and angles to
minutes.
30 TRIGONOMETRY
a = 5987.2 j6. /? = 83° 15' 6"
ft= 88° 53' 2" c = 7000
22. c = . 09008 27. a =66° 6' 18"
a = .07654 c = 8070.6
23. a = 46° 39' 50" 28. a = 978.45
a = 26.434 6 = 1067.2
24. a = 30.008 -^29. a = 5280
b = 29.924 6 = 5608
25. a = 111.45 /A30- « = 17° 26' 34"
b = 121.69 c = 46.474
31. Solve the isosceles triangle, one of the equal sides
being 690.13, and one of the base angles being 15° 20' 25".
32. Solve the isosceles triangle whose altitude is 606.6,
one of the equal sides being 955.7.
33. Solve the isosceles triangle whose base is 2558, and
whose vertical angle is 104° 0' 46".
34. Solve the isosceles triangle whose base is 161.4, and
whose altitude is 204.4.
35. Find the length of a side of a regular octagon in-
scribed in a circle whose radius is 49.
37. Oblique triangles. When any three independent parts
of an oblique triangle are given it can be solved by means
of right triangles. The oblique triangle may be divided
into two right triangles by drawing a perpendicular from
one vertex to the opposite side, or the opposite side pro-
duced.
When one of the given parts is an angle, the perpendicu-
lar must be selected so that one of the resulting right
triangles will contain two of the given parts.
In case the three sides are given, a second part of one of
RIGHT TRIANGLES 31
the right triangles can be obtained by equating the expres-
sions for the length of the perpendicular obtained from each
right triangle.
Thus a2 - a2 = 62 - (c - a)2,
from which a? = —
2c
38. APPLICATIONS
1. To find the distance from B to (7, two points on op-
posite sides of a river, a line BA, 200 feet long, was laid off
at right angles to the line BC, and the angle BAC was
measured and found to equal 55° 29'. What was the re-
quired distance ?
2. A railway is inclined 4° 23 '20" to the horizontal.
How many feet does it rise per mile, measured along the
horizontal ?
3. From the top of a tower 120 feet high the angle of
depression of an object in the horizontal plane of the base
of the tower is 24° 27'. How far is the object from the foot
of the tower ? How far from the observer ?
Given any point A and a second point B at a greater elevation
than A.
The angle of elevation of B from A is the
angle that the line AB makes with its orthogonal
projection upon the horizontal plane through A.
_ In the figure it is the angle a.
A The angle of depression of A from B is the
angle that the line BA makes with its orthogonal projection upon the
horizontal plane through B. In the figure it is the angle /3. It is clear
that a = 0.
4. Find the length of one side of a regular pentagon
inscribed in a circle whose radius is 18.24 feet.
5. Find the perimeter of a regular polygon of n sides
inscribed in a circle whose radius is r.
32 TRIGONOMETRY
6. Find the perimeter of a regular pentagon circum-
scribed about a circle whose radius is 18.24 feet.
7. Find the perimeter of a regular polygon of n sides
circumscribed about a circle whose radius is r.
8. Find the length of a chord subtending a central angle
of 63° 14' 20" in a circle whose radius is 124.93 feet.
9. A straight road, PR, makes an angle of 19° 27' 30"
with another straight road, PS. Having given PR = 640
feet, find US, the perpendicular distance from R to PS.
/"* I
ACUAt a certain point the angle of elevation of a moun-
tain is 34° 28'. At a second point 500 feet farther away,
the angle of elevation is 31° 12'. Find the height of the
mountain above the table-land.
11. One side of the square base of a right pyramid is 15
inches, and the altitude is 20 inches. Find the slant height
and the edge. Find the inclination of a face to the base of
the pyramid.
12. How far can you see from an elevation of 2000 feet,
assuming the earth to be a sphere with a radius of 3960
miles ?
13. Two forces of 95.75 pounds and 120.25 pounds, at
right angles to each other, act at a point ; find the magnitude
of their resultant and the angle it makes with the greater
force.
14. Find the velocity of a point, whose latitude is
44° 30' 20", due to the rotation of the earth. Assume the
radius of the earth to be 3960 miles.
15. Find the length of a belt running around two pulleys
whose radii are 12 inches and 4 inches, respectively, the
distance between the centers of the pulleys being 6 feet.
16. A diagonal of a cube and a diagonal of a face of a
cube intersect at a vertex. Find the angle between them.
RIGHT TRIANGLES 33
17. A force of 2000 pounds applied at the origin makes
an angle of 33° 25' with the positive X-axis. Find its
components along the X and Y axes respectively.
18. A force of 185 pounds applied at the origin makes an
angle of 82° 12' with the positive X-axis. Another force of
327 pounds applied at the origin makes an angle of 11° 32'
with the positive X-axis. Find the X and Y components
of each of these forces. Add the X-components and also
the y-components and then find the resultant in magnitude
and direction.
19. A cylindrical tank, whose axis is horizontal, is 8 feet
in diameter and 12 feet long. The tank is partly filled with
water, so that the depth of the water at the deepest point is
3 feet. How many gallons of water are in the tank, there
being 7-J- gallons in a cubic foot ?
20. A man who can paddle his canoe 5 miles per hour in
still water paddles at his usual rate directly across a river
one-half mile wide. If the river 'flows 4 miles per hour,
where will the canoe land and what is its speed in the water ?
* ' •*=
•
•-A— &-o
CHAPTER IV
O-^v
VARIATIONS OF THE TRIGONOMETRIC FUNCTIONS
REDUCTION OF FUNCTIONS OF /i90°±a
39. In the study of the variations of the trigonometric
functions, their values at 0°, 90°, ISO0, 270°, and 360° are of
special importance.
Values of functions of 0°. The terminal line of 0° coin-
cides with the positive X-axis. For a point on this line, at
a distance r from the origin, we have
Y
x=r,
y=(
)
Hence
sin
0° =
0
= 0
cotO°
_r*_
00
P
r
0~
X' 0
X
cos
0° =
r
= 1
secO°
_ r _
1
r
r
tan
0° =
0
= 0
cscO°
_ r _
00
Y
r
0
Values of functions of 90°. The terminal line of 90° coin-
cides with the positive F-axis, therefore
X-
= U, y = r.
Y
Hence
P
sin 90° = - = 1
cot90° = - = 0
r
r
cos 90° = - = 0
r X'
sec 90° =. - = oo
0
X
r
0
tan90° = - = oo
csc90° = - = l
i
0
r
Y
* For the
interpretation of - see any College
Algebra.
34
VARIATIONS OF TRIGONOMETRIC FUNCTIONS 35
Values of functions of 180°. The terminal line of 180° co
incides with the negative X-axis,
therefore
x= — r,
Hence
sin 180°= -- =0
r
cos 180° = — = - 1
X'
tan 180° = — = 0
— r
cot 180° = — - = oo
sec 180° = — = -1
— r
esc 180°= =00
Values of functions of 270°. The terminal line of 270°
coincides with the negative Y-axis, therefore
x = 0, y = — r
Hence
sin 270° =nr= -1 cot 270° =—=0
r —r
cos 270°= - =0
r
sec 270°= £ =00
tan 270° =— =00 esc 270°=— =-1
0 — r
The values of the functions of 360° are identical with the
values of the functions of 0°, since these angles are
coterminal.
40. Variation of the functions. It has been shown that
the trigonometric functions are functions of an angle. As
the angle varies, the trigonometric functions depending upon
the angle also vary.
36
TRIGONOMETRY
Let the line OP, of fixed
length r, revolve about 0
from the initial position OX.
Then the angle XOP or a
increases from 0° to 360°.
The variations of the trigo-
nometric functions can be
traced by observing the
changes in the abscissa and
ordinate of the point P.
a. As the angle a increases from 0° to 90°,
y increases from 0 to r,
and x decreases from r to 0.
Hence
sin a or - is positive and increases from 0 to 1
fm
cos a or - is positive and decreases from 1 to 0
tan a or - is positive and increases from 0 to oo
cot a or - is positive and decreases from co to 0
y
T
sec a or - is positive and increases from 1 to oo
esc a or - is positive and decreases from oo to 1
y
As the angle a increases through 90°, x passes through
zero, changing from a positive number to a negative num-
ber. Then, immediately before a becomes 90°, cos a or -
r
is a very small positive number ; while immediately after
a has passed 90°, cos a is a very small negative number.
This may be expressed by saying that cos a passes through
zero and changes sign as a passes through 90°.
VARIATIONS OF TRIGONOMETRIC FUNCTIONS 37
Likewise, immediately before a becomes 90°, tan a or £
x
is a very large positive number ; while immediately after
a has passed 90°, tan a is a very large negative number.
It has been seen that tan 90° = oo . Hence tan a passes
through oo and changes sign as a passes through 90°.
Hence we may say that tan 90° = ± GO , choosing the posi-
tive sign when associating 90° with the first quadrant and
the negative sign when associating 90° with the second
quadrant.
Similarly, whenever a trigonometric function passes throiigh
oo it changes sign.
b. As the angle a increases from 90° to 180°,
y decreases from r to 0,
and x decreases from 0 to — r.
Hence
sin a or ^ is positive and decreases from 1 to 0
r
cos a or - is negative and decreases from 0 to — 1
tan a or ^ is negative and increases from — oo to 0
x
cot a or - is negative and decreases from 0 to —GO
y
sec a or - is negative and increases from — oo to — 1
x
esc a or — is positive and increases from 1 to + oo
y
c. As the angle a increases through 180°, y passes through
zero, changing from a positive number to a negative number.
As the angle a increases from 180° to 270°,
y decreases from 0 to — r,
and a increases from — r to 0.
Hence
sin a or ^ is negative and decreases from 0 to — 1
T
38 TRIGONOMETRY
cos a or - is negative and increases from — 1 to 0
T
tan a or - is positive and increases from 0 to -f oo
cot a or - is positive and decreases from + oo to 0
sec a or - is negative and decreases from — 1 to — oo
x
esc a or - is negative and increases from — oo to — 1
d. As the angle a increases through 270°, x passes through
zero, changing from a negative number to a positive
number.
As the angle a increases from 270° to 360°,
y increases from — r to 0,
and x increases from 0 to r.
Hence
sin a or - is negative and increases from — 1 to 0
flf*
cos a or - is positive and increases from 0 to 1
tan « or - is negative and increases from — oo to 0
cot a or - is negative and decreases from 0 to — oo
sec a or - is positive and decreases from + oo to 1
7*
esc a or - is negative and decreases from — 1 to — oo
The above results are presented in tabular form.
VARIATIONS OF TRIGONOMETRIC FUNCTIONS 39
(
1st QUADRANT |
P 9C
2d QUADRANT
0 18
8d QUADRANT
3° 27
4th QUADRANT
0° 360°
sin «=-
+ 0 inc. -f 1
+ 1 dec. + 0
- 0 dec. - 1
- 1 inc. - 0
X
cos a=-
r
+ 1 dec. + 0
-Odec. - 1
- 1 inc. - 0
+ 0 inc. + 1
V
tan a=-
X
-f 0 inc. +00
—oo inc. — 0
+ 0 inc. +00
—ao inc. — 0
cot«=*
+ 00 dec. + 0
— 0 dec. —oo
+00 dec. + 0
- 0 dec. -oo
sec «=-
X
+ 1 inc. +00
—oo inc. — 1
— 1 dec. —oo
+ 00 dec. + 1
esc a— -
y
+ oo dec. + 1
+ 1 inc. +00
— oo inc. — 1
— 1 dec. —oo
It is thus seen that the sine and cosine can never be
greater than -+- 1 nor less than — 1, while the secant and
cosecant have no values between -f 1 and —1, but have
values ranging from +- 1 to +00 and from — 1 to — oo .
The tangent and cotangent may
have any value from +- oo to — oo .
41, Graphical representation. A
graphical representation of the trigo-
nometric functions is effected by first
locating points using the different
values of the angle as abscissas and
the corresponding function-values as
ordinates, and then drawing a smooth
curve through these points taken in
the order of increasing angles.
The values of the functions of
the angles previously calculated are
sufficient to determine an approxi-
mate graph. For greater accuracy
the values of the functions may be
taken from the table of natural
functions.
a
since
a
tan «
0
0
0
0
7T
1
7T
1
6~
2
6
gV3
it
-A/~
IT
1
4
2
4
7T
3"
|V3
7T
3
V3'
7T
1
IT
00
2
2
27T
1 _
2*
V3
3
2
3
37T
1 -
Sir
-1
4
2
4
67T
1
STT
V3
6
2
6
3
7T
0
IT
0
?7T
1
77T
1
6
2
"6~
gV3
etc.
etc.
40
TRIGONOMETRY
Sine Curve y = sin x
Tangent Curve y—tanx
The sine and tangent curves illustrate the truth of the
theorems of Arts. 25 and 26.
o
Thus for any given value of the angle, as — ^, there is but
4
one value of each function, the curves showing the sine and
tangent to be 1 V2 and — 1 respectively.
But for any given value of a trigonometric function, as
sin a, = ^, there are an unlimited number of angles, as
, Z, 2J», 2f», etc. .
Also for tan a = V^we see that a may have the values
I HTT, 21 TT, 31 TT, etc.
42. Periodicity of the trigonometric functions. From the
sine curve it is readily seen that the sine is 0 when the
VARIATIONS OF TRIGONOMETRIC FUNCTIONS 41
angle is 0, and that it increases to 1 as the angle increases to
?-. At this point the sine begins to decrease and has the
2
value 0 when the angle is TT, and finally reaches the value
o
— 1 when the angle is — - ; then the sine again begins to in-
2i
crease and has the value 0 when the angle has the value 27r.
When the angle increases from 2 TT to 4 TT, the sine repeats
its values of the interval from 0 to 2?r. If the angle were
increased indefinitely, the sine would repeat its values for
each interval of 2 ?r. For this reason the sine is called a
periodic function, and 2 TT is its period.
A study of the tangent curve shows that the tangent has
the same values between 0 and TT that it has between ?r and
27T or between 2?r and 3?r. Hence the tangent is also a
periodic function having the period- TT.
43. EXAMPLES
1. Plot y = cos x and give the period of the cosine.
2. Plot y = cot x and give the period of the cotangent.
3. Plot y = sec x and give the period of the secant.
4. Plot y = esc x and give the period of the cosecant.
5. Plot y = sin x + cos x.
6. Plot y = x + sin x.
TRIGONOMETRY
REDUCTION OF FUNCTIONS OF n 90° ± a
44. The formulas to be developed in this section enable
us to express any function of any angle in terms of a func-
tion of an angle differing from the given angle by any
multiple of 90°. They may be used to express any func-
tion of any angle in terms of a function of an angle less
than 90° or less than 45°.
45. Functions of — a in terms of functions of a(n = 0).
Let XOP be any positive angle and XOP' a numerically
equal negative angle.
Then taking OP' = OP we have, for each figure,
xf = x, y' = - y, r' = r,
where #, #, r and #', y', r' are associated with P and Pf
respectively.
Then in each quadrant
sn — a = ~f = _ — _ sn a
xf x
cos (—«)= — = - = cos a
VARIATIONS OF TRIGONOMETRIC FUNCTIONS 43
tan (— a) = ^- = — ^ = — tan a
x x
Xf X
cot (—«) = — = = — cot a
y' -y
r' r
sec (— a) — — — - — sec a
x' x
esc (— a) = — = -^- = — esc a
2/' -2/
46. Functions of 90° + a in terms of functions of a(n = I),
Let XOP be any positive angle a and XOP' the angle
90° + a.
Then taking OP' = OP we have, for each figure,
where a?, ?/, r and #', #', r' are associated with P and Pr
respectively.
X'
x
Similar figures can be constructed for the other quadrants.
Then in each quadrant
sin (90° + «) = £' = - = cos a
-
r'
= - snce
tan (90° + a) = = — = - cot «
v -y
44
TRIGONOMETRY
cot (90° + «)=- = — ^ = - tan a
y' »
sec (90° + «)== — = — = - esc «
a; — y
esc (90° + «)= — = - = sec a
'
47. Functions of 90° — a in terms of functions of a (n = 1).
Let XOP be any positive angle and XOP' the angle
90° - a.
Then taking OP' = OP we have, for each figure,
x =
y
x, rf = r.
Y
Y
D'
7
y P
&&
•>. .• v
T>
.X' 0
x' x x X' J
aj^V
/ Y
y\/<
^
•$*
p' ry
/
Y
\
/
D '
Y
Similar figures can be constructed for the other quadrants.
Then in each quadrant
sin (90° - a) = £ = - = cos a
cos (90° -«) = - = % = sin a
r' r
tan(90°-<x) = ^ = -
x' y
cot (90° - a) = - = ^ = tan a
y %
sec (90° - a) = — = - = esc a
x' y
esc (90° — a) = - = - = sec a
y1 ^
VARIATIONS OF TRIGONOMETRIC FUNCTIONS 45
48. Functions of 180° — a in terms of functions of a(n=2).
Let XOP be any positive angle and XOP' the angle
180° - a.
Then taking OP' = OP we have, for each figure,
»' = -», y'=y, r' = r.
P'Y
P
Y
«IY
r
\i
Bg.
2/
\
fij
a?' X5
"^ a?
X •** vj
« x X' 2/'
^S-J^ x
P
p
r
|
Y
Y
Similar figures can be constructed for the other quadrants.
Then in each quadrant
sin (180° - «) = ^ = ^ = sin «
T T
cos (180° - a) = -,= — = - cosa
tan (180° -«) = ^ = -^- = - tana
x' — x
cot (180° - «)= - = -^^ = - cot a
2/' y
sec (180° -«) = - = -^- = - sec a
x' — x
esc (180° - a) = - = -= csca
y1 y
49. Laws of reduction. The method of the last four
articles may be applied to any angle of the form n 90° ± a
to obtain formulas of reduction for all positive and negative
integral values of w, a being any angle. ' A complete inves-
tigation would show the following laws to be true :
I
46 TRIGONOMETRY
When n is even, any function of n 90° ± a is numerically
equal to the same function of a. ; when n is odd, any function
ofn 90° ± a is numerically equal to the cofunction of a.
If the function of n 90° ± a is positive, a being considered
acute, the members of the equation have like signs ; if the
function of n 90° ± a. is negative, a being considered acute,
the members of the equation have unlike signs. The results
thus obtained are valid for all positive and negative values
of a.
50. EXAMPLES
Express as functions of a positive angle less than 90° :
1. sin 130°.
SUGGESTION. 130° = 90° + 40°, or 130° = 180° - 60°.
2. cos 170°. 5. cos (-20°).
3. tan 110°. 6. tan (-80°).
4. cot 160°. 7. sin (-120°).
Express as functions of 0 :
8. sin (810° -0). 12. tan (0 - 180°).
9. tan(360°-0). 13. sec (- 180° - 6).
10. cot (270° + 0). 14. esc (- 630° + 0).
11. sin (9 - 90°). 15. cos (990° - 0).
Express each of the trigonometric functions of the follow-
ing angles as functions of a without using the laws of
reduction.
16. 180° + a. 18. 270° + a.
17. 270° -a. 19. 360° -a.
CHAPTER V
FUNDAMENTAL RELATIONS. LINE VALUES
51. In the present chapter eight fundamental relations
between the trigonometric functions are developed. It is
then shown that each trigonometric function may be repre-
sented geometrically by a single line, giving the so-called
line values. This gives a second view of the trigonometric
functions. The line values offer a simple method of demon-
strating the fundamental relations, and of developing the
properties already derived by means of the trigonometric
ratios. In fact the line values might serve as the funda-
mental definitions of the trigonometric functions and thus
trigonometry could be based upon these values. Inciden-
tally the line values suggest the origin of some of the terms
used to designate the several trigonometric functions.
. FUNDAMENTAL RELATIONS
52. Certain relations of fundamental importance exist
between the trigonometric functions of an angle. These
will now be developed.
From definitions,
sin a = -,
r
csca=-. .-. sina = — - — . (1)
y csc a
x
cos a = -.
seca=-» .-. cosa=— - — (2)
x sec a
47
48 TRIGONOMETRY
tan a = ^ ,
x
cot a = -. .-. tan a =
cot a
sin a = -j
cos a
tan a = . .-. = tan a. (4)
x cos a
cota=-, and - cot a. (5)
sin a
From geometry, 2/2 + x2 = r2.
Dividing successively by r2, #2, and yz,
2? + 12 = 1, .-. sin2 a + cos2 a = 1. (6)
•£ + 1=^, ... tan2 a + 1 = sec2 a. (7)
a2 x2
^ + 1=^, /. cot2 a + 1 = esc2 a. (8)
These eight formulas are called the fundamental relations.
They are used frequently, and should be memorized. In
doing this, it is well to have the method of derivation
clearly in mind.
53. The use of exponents. In affecting trigonometric
functions with exponents, the exponent is usually placed
immediately after the function. Thus, in the formulas of
the preceding article, sin2 a is identical in meaning with
(sin a)2, and tan2 a is identical' with (tan a)2.
An exception to the above is made for the exponent — 1,
in which case it is always necessary to make use of paren-
FUNDAMENTAL RELATIONS 49
theses. Thus, (cos a)"1 is used to express , while
cos a
cos"1 a has an entirely different meaning, as will be ex-
plained later.
54. Trigonometric identities. An identical equation, or
an identity, is an equation which is satisfied for all values
of the unknown quantities. The eight fundamental rela-
tions are trigonometric identities. There are many other
identities depending upon these.
The truth of an identity can be established in two ways :
First. Begin with one of the fundamental relations, and
produce the given identity by means of the fundamental
relations and algebraic principles.
Second. Begin with the given identity and transform it
to one of the fundamental relations, or reduce one member
of the equation to the other by means of the fundamental
relations and algebraic principles.
The choice of the trigonometric transformations neces-
sary to effect the reductions is often suggested by the
functions involved in the given problem. When no trans-
formation is thus suggested, the problem may generally be
simplified by expressing each of the functions in terms
of sines and cosines, and making use of the relation
sin2 a 4- cos2 a = 1.
Avoid the use of radicals whenever possible.
55. Trigonometric equations. A conditional equation is
an equation which is not satisfied for all values of the
unknown quantity, but is satisfied only for particular values.
(a) Trigonometric equations involving different trigono-
metric functions of the same angle may often be solved by
simplifying the equation by the use of the fundamental
relations. Thus sin a _ COs a
may be written in a = 1 or tan a = 1
cos a
/.a = 45° or 225°.
50 TRIGONOMETRY
(5) Some equations may conveniently be solved by trans-
posing all of the terms to the first member of the equation,
factoring, and then placing each factor equal to zero. Thus
2 sin2 a + V3 cos a = 2 V3 sin a cos a -f sin a
may be written
sin a (2 sin a — 1) -f- V3 cos a (1 — 2 sin a) = 0
or
(sin a — V3 cos a) (2 sin a — 1) = 0.
This equation is satisfied if either
sin a — V3 cos a = 0 or 2 sin a — 1 = 0
whence tan«=V3 or sina = i.
/.a = 60°, 240° or a =30°, 150°.
(c) When no other method suggests itself, the equation
may be transformed, by the use of the fundamental rela-
tions, into an equation containing only one function. The
equation thus obtained may be solved algebraically for the
function involved, from which the values of the angle may
be obtained. Thus
2
10 cos3 a tan a — 9 cos2 a 10 sin a + 9 = 0
esc a
10 sin «(1 - sin2 a)- 9(1 - sin2 a)— 12 sin « + 9 = 0
10 sin a — 10 sin3 a + 9 sin2 a — 12 sin a = 0
.-. sin a = 0 or 10 sin2 a — 9 sin a -f 2 = 0
from which sin a = 0, f , or •£•.
.-. a = 0°, 23° 35', 30°, 150°, 156° 25', or 180°.
A trigonometric equation is considered completely solved
when every positive angle less than 360° which satisfies it
has been determined. All other angles coterminal with
these angles also satisfy the equation.
In the solution of trigonometric equations extraneous
roots may occur. These may be detected by substitution in
the original equation.
FUNDAMENTAL RELATIONS 51
56. EXAMPLES
Prove the following trigonometric identities :
1. sin 6 = cos 0 tan 0. cot A
3. cos 6 =
2. tan 0 = sin 0 sec 0. esc 0
4. (1 — sin #)(! + sin x) = cos2 x.
5. (sec x — tan #)(sec x + tan a?) = 1.
6. (sin x -f- cos #)2 = 2 sin a; cos cc -}- 1.
7.
cot a;
8. cos2 a tan2 « + sin2 a. cot2 « = 1.
9. Given cos a. = -| , a being in the fourth quadrant, find
the values of the1 remaining trigonometric functions.
SOLUTION. Since sin2 a + cos2 a = 1,
sin2 a = 1 - cos2 a =1 - |f = |f.
.*. sin a = — \/|f = — $\/l4.
sin a
Also tan a =
cot a =
esc a =
sin a — fv/14 2\/14
Compare with the method of examples 26 and 27, Art. 27.
10. Given tan y — 4, y being in the third quadrant, find
the values of the remaining trigonometric functions.
SOLUTION. Since sec2 y = I -f tan2 y,
sec y = — Vl -f 16 = — \/17.
Also cot y = = - ;
tan y 4
esc2 y = 1 + cot2 y = 1 + ^ = £f ;
esc y = —
52 I/ TRIGONOMETRY
sin 3, = -!- = — 1— = -AVl7;
cos y = — = — I— = - TV VI7.
secy _vi7
11. Given sin 0 = -|, 0 terminating in the second quadrant,
find the values of the remaining trigonometric functions.
12. Given cot 0 = T7^, 0 terminating in the first quadrant,
find the values of the remaining trigonometric functions.
13. Given sec 0 = — 2, 0 being in the third quadrant,
find the values of the remaining trigonometric functions.
14. Given tan a = — -£, find the remaining functions of-
the angles less than 360° which satisfy the equation.
Solve the following trigonometric equations for angles
less than 90°:
15. 5 sin x + 8 = 3(4 — sin x).
16. sinw — cos w = 0.
17. 2 cos2 x — 3 cos x 4- 1 = 0.
18. (tan 6 - l)(tan 0 - V3) = 0.
19. Solve sin x = cot x for sin x.
Some of the following examples are identities and others
are equations. Establish the identities and solve the equa-
tions for angles less than 180°.
20. sin4 x = 1 — 2 cos2 x + cos4 x.
21. sin4 x = 2 — 6 cos2 x + cos4 x.
22. (V§ + l)L+ = sin
tan 6
23 2Bin! + costf VS
tan $ cot $ 2
24. tan u + cot u = sec u esc u.
25. 2 sin2 y esc ?/ + 3 esc y = 1.
FUNDAMENTAL RELATIONS 53
Prove the following trigonometric identities :
26. tan* 0 + cot2 0 = sec2 0 esc2 6-2. .
27. sec2 ft + cos2 (3 = tan2 ft sin2 ft-}- 2.
28. esc2 y — sec2 y = cos2 y esc2 y — sin2 y sec2 y.
rtrt . sin a 4- tan a
29. sin a tan ce = — — •
cot « + csc <*
30. cot2 a — cos2 a = cot2 a cos2 a.
31.
sin 0 cos 0
32. (sin 0 4- cos 0)2 4- (sin 0 - cos 0)2 = 2.
33. sin2 x 4- vers2 a? = 2(1 — cos x).
34. sec2 a csc2 a _(l~tan2a)2= 4
tan2 a
Solve the following trigonometric equations for angles
less than 180°:
35. 2 cos2 a — 3 sin a = 0.
36. sec2 a 4- cot2 a = 4±,
37. 1 4- tan2 a; — 4 cos2 a; = 0.
38. tan x 4- cot x = 2.
39. 2 sin2 x 4- 3 cos x == 0.
40. V3 cos x 4- sin # = V2.
41. csc2 a: — 4 sin2 a; = 0.
42. cot aj 4- csc2 x = 3.
LINE VALUES
57. Representation of the trigonometric functions by lines.
The trigonometric functions of an angle have been studied
solely from the standpoint of ratios ; but each function can
also be represented in magnitude and sign by a single line.
54
TRIGONOMETRY
Let LOP be any angle. Take OP=l. Draw the ordi-
nate and abscissa of the point P. About 0 as the center
construct a circle with OP as a radius. Draw the tangents
to this circle at L and F, the beginning of the first and second
quadrants respectively, and produce them to intersect the
terminal line, or the terminal line produced, in M and G.
Then
y AP AP
sin « = - = — — = — — •
OP
AP
x OA OA KA
cos a = - = = — — = OA
r OP 1
x OA OL
= x=OA = FG==FG
OP
x OA OL
FUNDAMENTAL RELATIONS 55
or
Thus the trigonometric functions are represented by the
segments AP, OA, LM, FG, OM and OG. It is evident
that these segments represent the trigonometric functions
in magnitude. They also represent the trigonometric func-
tions in sign. In accordance with the conventions of Arts.
6 and 11, the segments AP and LM are positive when drawn
upward from the X-axis and negative when drawn down-
ward ; the segments OA and FG are positive when drawn
from the F-axis to the right and negative when drawn to
the left; the segments OM and OG are positive when they
coincide with the terminal line of the angle and negative
when they coincide with the terminal line produced.
It is evident that the sign of each trigonometric function
is determined by the segment representing the function,
since in each quadrant the segment is positive whenever the
ratio defining the function is positive, and negative whenever
the ratio is negative. Thus in each quadrant LM has the
same sign as the ratio ?, and similarly for the other func-
x
tions. Therefore the segments represent the trigonometric
functions in sign as well as in magnitude.
The segments which represent the trigonometric functions
are called the line values of the functions.
PROBLEM. Represent vers a and covers a by line values.
58. Variations of the trigonometric functions as shown by
line values. The line values of the trigonometric functions
give a simple method of tracing the variations in the func-
tions as the angle varies from 0° to 360°. Thus as the point
P, in the figures of Art. 57, describes the circle of radius
unity, the changes in the segments AP, OA, LM, etc.,
represent the variations in the sine, cosine, tangent, etc.,
respectively.
56
TRIGONOMETRY
Numerous figures, with three or four values of the angle in
each quadrant, serve to suggest these variations.
59. Fundamental relations by line values. By the use of
line values the fundamental relations of Art. 52 may be
simply obtained and easily memorized.
F« cot a -,
Thus
From the triangle OAP,
AP2 +OA2=OP2 or sin2 a + cos2a = 1.
From the triangle OLM,
or tan2 a + 1 = sec2 a.
or cot2a+l = csc2a.
From the triangle OGF,
From definitions,
AP sin a
tan a = =
OA cos a
OA cos a
cot a
AP sin a
sec« = -— =
OA cos a
esca-O?.-1
AP sin a
or
or
or
or
tana =
cot a =
seca =
csca =
sin a
cos a
cos a
sin a
cos a
1
sin a
FUNDAMENTAL RELATIONS 57
co«_= = __ or
AP LM tan a tan a
The process of derivation of these formulas applies
equally well to an angle of the third or fourth quadrant ;
hence the formulas are true for all values of a.
To memorize these formulas most easily, form a clear
mental picture of the figure for the first quadrant and read
each formula directly from the figure, applying the ratio
definitions and the Pythagorean theorem.
60. EXAMPLES
1. Construct the line values of an angle terminating in
the third quadrant.
2. Construct the line values of an angle terminating in
the fourth quadrant.
3. Obtain the fundamental relations for an angle in the
third quadrant, by the use of line values.
4. Obtain the fundamental relations for an angle termi-
nating in the fourth quadrant, by the use of line values.
5. Deduce the relations between the functions of 90° + x
and the functions of x by means of line values.
SUGGESTION. Construct two figures, one for the angle 90° + x and
the other for the angle x.
6. Deduce the relations between the functions of 180° — x
and the functions of x by means of line values.
7. Deduce the relations between the functions of 90° — x
and the functions of 270° + x by means of line values.
By means of line values, trace the variations in the follow-
ing functions as a varies from 0° to 360° :
8. sin a. 11. cot «.
9. cos «. 12. sec a.
10. tan a. 13. esc a.
58 TRIGONOMETRY
Prove the following trigonometric identities :
14. = sin x cos x.
cot x + tan x
15. cos2 a (1 + cot2 a) = cot2 a.
16. sin 0 cot2 0 = esc 0 - sin 0.
17. tan a — tan « sin2 a = sin a cos a.
18. cos a tan2 a = sec a — cos a.
jo cos « sin « _ sin3 a -f- cos3 a
tan « cot a sin « cos a
on sin3 a + cos3 « .<
20. — = 1 — sin a cos a.
sin a , cos a
Oi sin x . tan x . sec x 2 cot a? 4- 1
1
22.
23.
04
cosa; cotx csca; cot2 re
1 — cos x _ sec a; — 1
1 -|- cos a? sec x + 1
1 — cos x sin2 a;
1 +• cos x (1 + cos a:)
1 — cos # (1 — cos #2
1 -j- cos a? sin'" a;
25. 2 sin ?/ cos2 y + (2 cos2 ?/ — 1) sin y = 3 sin y — 4 sin3 y.
26. cos4 a — sin4 a = cos2 a — sin2 a.
27. sec2 a + cosec2 a = sec2 a esc2 a.
no tan a; -f- tan y _ sec x -+- sec y
\ sec a; — sec ?/ tan a; — tan y
29. tan4 w — cot4 u = (tan2 u + cot2 w)(sec2 w — esc2 w).
on tan a; -f tan ?/
30. - — -i- - ? = tan x tan y.
cot x + cot y
01 tana? — tan y
31. - - 2 == — tan x tan w.
cot x — cot ?/
FUNDAMENTAL RELATIONS 59
on tan x -f cot y tan x
64. = — — •
cot x -f- tan y • tan y
33. Express sin2 a; — cos2 a; in terms of tan x.
34. Express vers x — covers x in terms of sin x.
35. Express cot2 a? -f- csc2# in terms of cos x.
•yp-pQ /V»
36. Express — — in terms of tan x.
covers x
37. Express tan2 x + sec2 x in terms of cot x.
38. Express '(4 sin3 0 - 3 sin 0)(2 cos2 0-1)
-f- (3 cos 0 — 4 cos3 6) (2 sin 6 cos 0) in
terms of sin 0.
39. Given sin 0 = a, find the remaining functions of 0.
40. Given cot 0 = 6, find the remaining functions of 0.
41. Express each trigonometric function of 0 in terms of
cos 0.
42. Express each trigonometric function of 0 in terms of
tan0.
43. Simplify (a + 6) sin 30P - 6 cos 60° + a tan 180°.
44. Simplify I sin (270° - a;) + m cos (180° + a?)
+ w sin (90° - a;).
45. Simplify a sin 135° +(a — 6) cos 225° + & cos 315°.
46. Simplify tan (-120°) + cot 150°-tan 210°+ cot 240°.
Find the positive values of 0, less than 180°, that satisfy
the following equations :
47. sin0 cos 0 = 0.
48. sin 0 + cos 0 = 0.
49. sin0(2sin0-l)(2cos0-l) = 0.
50. sin 0 + cos 0 cot 0 = 2.
60 TRIGONOMETRY
51. sin0cos0 = |.
52. 2 sin 0 cos <9 + sin 0-2 cos0 = l.
53. 2 cos 0 + sec 6 = 3.
54. sec (9 + tan 0 = 2.
55. 2 sin 0 + 5 cos 9 = 2.
56. 1 + sin2 0 = 3 sin 0 cos 0.
57. tan4 0-4 tan2 0-5 = 0.
58. What negative angles, numerically less than 180°,
satisfy the equation sin2 x -f- sin x = cos2 x — 1 ?
59. Given 9 cos2 w + 9 sin ^ = 11, find tan u.
60. Given tan2 x — 5 sec a; + 7 = 0, find sin x.
Find the positive values of 0, less than 360°, that satisfy
the following equations :
61. sin 0 = - cos 285°.
62. tan 0 = cot (-144°).
63. cos ( - 0) = sin 190°.
64. sin0 = — sin 50°.
65. If sin 122° = k, prove that tan 32° = ~V/J
66. If cot 255° = a, prove that cos 345° =
67. If cos(- 100°)= k, prove that tan 80° = -
k
Solve the following equations for any trigonometric func-
tion of a.
68. 2 sin a + esc a = 1. 71. tan2 a -f 4 sec a = 5.
69. esc a cot a = j. 72. tan a + cot a = 2.
70. 2 sin a + cos2 a = 1.
73. Given sin ^ = k sin v and tan u = Z tan v, find sin w
and sinv.
CHAPTER VI
FUNCTIONS OF THE SUM OF TWO ANGLES
DOUBLE ANGLES. HALF ANGLES
61. We have thus far considered the properties and rela-
tions of the functions of a single angle. We have also
shown that the functions of the angles n 90 ± a depend
upon the functions of the angle a.
In the present chapter we consider the relations between
the functions of the sum of two independent angles and the
functions of the separate angles, and develops some related
formulas.
-* 62. The sine of the sum of two acute angles expressed in
terms of the sines and cosines of the angles.
B X
Let a and /? be any given acute angles.
Construct the angle (a -f- /?), Art. 12.
Then (a + ft} may be acute as in Fig. 1, or obtuse as in
Fig. 2.
From any point P, in the terminal line of the angle
(a 4- /?), draw PA perpendicular to OX, and PQ perpendicu-
lar to OS. Through Q draw RQT parallel to OX, and QB
perpendicular to OX.
61
62 TRIGONOMETRY
Then the angle TQP is equal to 90° + a.
By definition
OP OP OP OP
BQ.OQ.RP QP
OQ' OP QP' OP
and =s
therefore sin (a + P) = sin a cos p 4- cos a sin p. (1)
PROBLEM 1. Show that formula (1) is true when either
angle is 90°.
PROBLEM 2. Show that formula (1) is true when each
angle is 90°.
63. The cosine of the sum of two acute angles expressed in
terms of the sines and cosines of the angles. ' Referring to the
figures of the previous article, we have by definition
OB OQ QR QP
OQ' OP QP' OP
But = cos «, = cos ft
=cos (90° +«) = - sin a, and = sinft
therefore cos (a -f p) =cos a cos p — sin a sin p. *" (1)
PROBLEM 1. Show that the formula (1) is true when
either angle is 90°.
PROBLEM 2. Show that formula (1) is true when each
angle is 90°.
64. The formulas developed in the last two articles are
of the utmost importance, since many other formulas are
FUNCTIONS OF THE SUM OF TWO ANGLES 63
derived from them. They form the basis of the present
chapter. In developing these formulas a and ft were con-
sidered positive acute angles, but the formulas are true for
all values of a and ft} as may be shown either geometrically
or analytically.
It is shown geometrically by constructing figures in
which a and (3 are of any magnitude, and following the
method of proof given above for acute angles.
We present the analytic demonstration as the more satis-
factory.
65. To prove that
sin (« -|- ft) = sin a cos ft 4- cos a sin ft (1)
and cos (a -f- ft) = cos a cos ft — sin a sin ft (2)
are true for all values of a and ft.
First. To show that a can be replaced by a', where a' =
a + 90°.
Let a' = a -f 90° where a is acute.
As a varies from 0° to 90°, a' varies from 90° to 180°.
Also sin a' = sin (a + 90°) = cos a, (3)
cos a' = cos (a + 90°) = — sin a. (4)
Now sin («' + /?)= sin (90° + a + £)= cos(a -f £).
Since a and /? are both acute, cos (a + ft) may be ex-
panded by (2), hence
sin (a' + ft) = cos (a + ft) = cos a cos ft — sin a sin ft.
Therefore, by (3) and (4),
sin (a' + ft) = sin a' cos ft + cos «' sin ft. (5)
Similarly,
cos(«' -f ft) = cos (90° + a + ft) = - sin (« -f /8)
= — sin a cos /? — cos a sin /?.
64 TRIGONOMETRY
Therefore, by (3) and (4),
cos («' + /?) = cos a' cos ft — sin a' sin j3. (6)
Formulas (5) and (6) show that one angle, a, in (1) and
(2) may be extended to 180°.
Second. To show that a can be replaced by a" where a" =
a + 18CP-
Let a" = a + 180°.
As a varies from 0° to 180°, a!' varies from 180° to 360°.
Also sin a" = sin (a + 180°) = — sin a, (7)
cos a" = cos (a + 180°)= — cos a. (8)
Now sin (a" -f 0)= sin (180° + a + /?)= - sin (a + 0).
Since a is less than 180° and ft acute, sin (« 4- /?) may be
expanded by (1), as extended in the first part of the proof,
hence
sin (a" + /?) = — sin (a + ft) = — sin a cos ft — cos a sin ft.
Therefore, by (7) and (8),
sin (a" + ft) = sin a!' cos ft -f cos a" sin ft. (9)
Similarly,
cos (a" 4- /?)= cos (180° 4- a + 0) = - cos (a + 0)
= — cos a cos /? + sin a sin /3.
Therefore, by (7) and (8),
cos (a" -f ft} = cos a" cos 0 — sin a" sin /?. (10)
Formulas (9) and (10) show that one angle, a, of (1) and
(2) may be extended to 360°.
Third. In a similar manner it may be shown that (1) and
(2) are true when ft varies from 0° to 360°, a having any
value from 0° to 360°, from which it easily follows that (1)
and (2) are true for all positive values of a and ft.
FUNCTIONS OF THE SUM OF TWO ANGLES 65
Fourth. Formulas (1) and (2) may be shown to be true
when either a or ft or both are negative, by the use of the
substitutions
a' = a-n 360°
hence (1) and (2) are true for all positive and negative
values of a and (3.
66. To find the tangent of the sum of any two given angles
in terms of the tangents of the given angles.
Let a and ft be the given angles.
Then tanQ + 8) = sin (" + ffl = sina cos ft + cos « sin ^
cos (a + ft) cos a cos ft — sin a sin ft
Dividing numerator and denominator of the last fraction
by cos a cos ft and simplifying, we have
1 — tan a tan p
67. To find the cotangent of the sum of any two given
angles in terms of the cotangents of the given angles.
Let a and ft be the given angles.
Then cot (« + /?) = cos <« + ^ = cos « cos /3 - sin « sin/?.
sm (a -h ft) sm a cos ft + cos a sm ft
Dividing numerator and denominator of the last fraction
by sin a sin ft and simplifying we have
cot p+ cot a
68. Addition formulas. Collecting the formulas for the
sum of two angles for convenience of reference, we have
sin (a -h ft) = sin a cos ft + cos a sin ft (1)
cos (a + ft) = cos a cos ft — sin a sin /3 (2)
66 TRIGONOMETRY
69. To find the sine, cosine, tangent, and cotangent of the
difference of two given angles.
Since the formulas of Art. 68 are true for all values of ot
and ft they are true when — ft is substituted for ft.
Hence
sin (a — ft) = sin a cos (— ft) + cos a sin (— ft)
cos (a — ft)= cos a cos (— ft) — sin a sin (— ft)
tan (ct K\ - tang +
~^-
cot(-0)+cota
But remembering that
sin (— ft) = — sin ft, cos (— ft) = cos ft
tan (- ft) = - tan ft, and cot (-£) = - cot ft
these formulas become U
sin (a — ft) = sin a cos ft — cos « sin /? (1) Ir.
cos (a— ft)= cos a cos £ + sin a sin ft (2)
tan (a - ft) = an<*~ — ?L/L (3)
1 4- tan
' cot/3-cot«
70. EXERCISES
1. Find sin 75°.
SOLUTION. 75° = sin (45° + 30°) = sin 46° cos 30° + cos 45° sin 30°
FUNCTIONS OF THE SUM OF TWO ANGLES 67
2. Find cos 75°. 7. Find sin (90° - £).
3. Find sin 15°. 8. Find tan (90° + a).
4. Find cos 15°. 9. Find cos (180° - a).
5. Find tan 75°. 10. Find sec 15°.
6. Find cot 15°. 11. Find esc (90° + a).
71. Double angles. To find the sine, cosine, tangent, and
cotangent of twice a given angle in terms of the functions of
the given angle.
From the formulas of Art. 68, letting /3 = a, we have,
after reduction,
sin 2 a = 2 sin a cos a (1)
cos 2 a . = cos2 a — sin2 a (2)
= l-2sin2a (3)
= 2 cos2 a - 1 (4)
rt 2 tan a /K-,
tan 2 a = — (5)
1 - tan2 a
cot2a = cot2a-1. (6)
2 cot a
To express clearly the real significance of these formulas,
we may state them from two points of view.
If a be the angle under consideration, the formula
sin 2 a = 2 sin a cos a
may be stated : The sine of twice an angle is equal to twice
the sine of the angle times the cosine of the angle.
If 2 a be the angle under consideration, the same formula
may be stated : TJie sine of an angle is equal to twice the
sine of half the angle times the cosine of half the angle.
It then follows that
sin a = 2 sin J a cos | a. (7)
Similarly, if a be the. angle under consideration, the
formula cos 2 a. = cos2 a — sin2 a
68 TRIGONOMETRY
may be stated : The cosine of twice an angle is equal to the
square of the cosine of the angle minus the square of the sine
of the angle.
If 2 a be the angle under consideration, the same formula
may be stated : The cosine of an angle is equal to the square
of the cosine of half the angle minus the square of the sine of
half the angle.
It then follows that
cos a = cos2 1 a — sin2 ^ a. (8)
Similarly,
from cos 2 a = 1 — 2 sin2 « we have
cosa = l -2sin2ia, (9)
from cos 2 a = 2 cos2 a — 1 we have
l, (10)
from tan 2 a = 2 tan " we have
1 — tan2 a
2 tan i a
from cot 2 a = cot2 a ~ l we have
2 cot a
cota^****-1. (12)
2 cot |a
72. Half angles. To find the sine, cosine, tangent, and co-
tangent of one half a given angle in terms of functions of the
given angle.
From formula (9), Art. 71, we have
cos «= 1 — 2 sin2 ~j
FUNCTIONS OF THE SUM OF TWO ANGLES 69
From formula (10), Art. 71, we have
cos a = 2 cos2 - — 1,
2
or COS^ = ±A/- "~ (2)
& A
From (1) and (2),
^^ (3)
2 ^ 1 + cos a
and cot
2
In each, of these formulas the positive or negative sign
is chosen to agree with the sign of the function of ^ , depend-
2i
ing on the quadrant in which - lies.
2
These formulas may be considered from two view points.
For example, formula (1) may be stated : The sine of half
an angle is equal to the square root of the fraction whose
numerator is one minus the cosine of the angle, and whose
denominator is two; or, Tlie sine of an angle is equal to the
square root of the fraction whose numerator is one minus the
cosine of twice the angle, and whose denominator is two.
73. To find the sum and difference of the sines of any two
angles, also the sum and difference of the cosines of any two
angles.
From the formulas of Arts. 68 and 69 we have, by ad-
dition and subtraction,
sin (a + ft) + sin (a — ft) = 2 sin a cos (3
sin (a -f- ft) — sin (a — (3) = 2 cos a sin ft
cos (a + ft) + cos (a — ft) = 2 cos a cos ft
cos (a + ft) — cos (a — ft) = — 2 sin a sin ft.
70 TRIGONOMETRY
Let a -J- ft = y and a — ft =• d. Solving for a and /?, we
have a = £(y 4- 3) and ft = 1 (y — d). Substituting these
values in the above equation, it follows that
sin Y + sin d = 2 sin £ (y + d) cos 1 (Y - 3), (1)
sin Y- sin 3 = 2 cos 1 (y + 3) sin | (>- 3), (2)
cos Y 4- cos 3 = 2 cos J (Y 4- 3) cos J (? — 3), (3)
cos Y — cos d = — 2 sin $(i + d) sin -J (<y — 3). (4)
Equations (1), (2), (3), and (4) may be read from two
view points.
Eegarding y and 3 as the given angles, (1) may be
stated : The sum of the sines of two angles is equal to twice
the product of the sine of half the sum of the given angles into
the cosine of half the difference of the given angles.
Thus, sin 6 x -f- sin 4 x = 2 sin 5 x cos x.
Eegarding i(y + 3) and -J(y — 3) as the given angles, it is
clear that their sum is y and their difference 3. Then, by
reading the second member first, equation (1) may be stated :
Twice the sine of any angle times the cosine of any other angle
is equal to the sine of the sum of the angles plus the sine of the
difference of the angles.
Thus, 2 sin 20° cos 5° = sin 25° + sin 15°.
74. Equations and identities. The formulas developed in
the present chapter are true for all values of the angles
involved ; hence they are trigonometric identities. By the
use of these identities many others may be established.
The remarks of Art. 54, concerning the use of the funda-
mental relations in establishing identities, apply here.
The identities of the present chapter are also useful in
solving trigonometric equations. By their aid an equation
involving functions of multiple angles may be transformed
into an equation containing functions of a single angle
(see Ex. 33, Art. 75). This transformed equation can then
be solved as indicated in Art. 55. Frequently equations
FUNCTIONS OF THE SUM OF TWO ANGLES 71
may be much simplified by reducing sums or differences of
sines and cosines to products by the relations of Art. 73
(see Ex. 41, Art. 75).
75. EXAMPLES
1. Find sin 221°, cos 221°, tan 22J°, and cot 22£°.
2. Given cosa = J; find sin 2 a, cos 2 a, tan2«, and
cot 2 a.
3. Given tan a = 3 ; find sin 1 a, cos % a, tan % a, and
cot \ a.
Prove the following identities :
4. (cos a + sin a) (cos a — sin a) = cos 2 «.
5. (sin a + cos a)2 = 1 -f sin 2 a.
6. sin 3 a = 3 sin a — 4 sin3 a.
SUGGESTION : sin 3 a = sin (2 a + a).
7. cos3a = 4cos3a — 3 cos a.
8. tan3« = 3tan"-tan3".
l-3tan2a
1 + COS X
, f. sin x
10. cot i x =
1 — cos x
- - 1 + tan i x _ 1 4-sin^
1 — tan i x ~ cos x
- ft 1 — cos 2 a
12. tan2a=-
1 + cos 2 «
13. sec2«= sec tt —
2 — sec2«
14. 4 sin2 1 a cos2 i a = 1 — cos2 «.
15. sin4« + sin2 « = 2sin 3acos a. (See Art. 73.)
16. cos 6 a — cos 2a= — 2 sin 4 a sin 2 a.
72 TRIGONOMETRY
17. sin (45° + «) — sin (45° - a) = V2 sin a.
18. sin (30° + a) + sin (30° - a) = cos a.
19. Express cos 4 a sin 3 a as the difference of two sines.
20. Express cos 5 a cos a as the sum of two cosines.
Prove the following identities :
21. cos2 a — cos2/? = — sin (a + ft) sin (a — ft).
sin 2 a + sin a 3 a
00
22.
- --^
cos 2 a -j- cos a 2
cos /? — cos a
24.
1 + tan2! 0
25. sin |- a; + 2 sin2 ^a? cos J # cot # = sin x cos -|-a/*.
26. 2 tani»csca; = l4-tan2^aj.
27. cos (a -f- ft) cos /? — cos ( «+ y) cos y =
sin (a -f y) sin y — sin (a -f- /?) sin ft.
28. cos2a = 8sin4-i-a-8sin2£a + l.
29. cos 2 a - cos a = 2 (4 sin4! a - 3 sin2! a).
30. (1 — cos 2 a) cot2! a = cos 2 a + 4 cos « + 3.
31. 2sin2|-0cot0 — csc20 + csc0 = cot2 0 + sin0.
32. Given tan a; = a tan 1 a? ; find tan ^ a; in terms of a.
Solve the following equations for all positive values of a
less than 180°:
33. sin2a-4cos2ia-sina + 3 = 0.
SOLUTION. 2 sin a cos a - 4 /1 + cos<*\ - sin a + 3 = 0,
or 2 sin a cos a — 2 cos a — sin « + 1 = 0,
or (2 cos a — 1) (sin a — 1) =0,
cos a = ^ or sin a = 1.
and a = 60° or 90°.
FUNCTIONS OF THE SUM OF TWO ANGLES 73
34. sin 2 a — cos « = 0.
35. sin 2 « + sin « = 0.
36.
37.
38. cos 3 a + 2 cos 2 « — 8 cos2a + 2 sin2 a 4- 5 cos a = 0.
39. cos 2 a H- 2 sin2! a — 1 = 0.
40 sin2a + cos2a + sin a = l.
41. sin 5 a — sin3cc + sin a = 0.
SOLUTION. 2 sin 3 a cos 2 a — sin 3 a = 0. Art. 73.
or sin3a(2cos2a-l) = 0.
sin 3 a = 0, or cos 2a = i.
3 a = 0°, 180°, 360°, and 2 a = 60°, 300°.
a = 0°, 60°, 120°, 30, 150°.
42. cos 3 a + sin 2 a — cos a = 0.
43. sin 3 a -+- sin 2 « -|- sin a = 0.
44. cos 3 a — sin 2 a + cos a = 0.
45. sin 3 a + cos 3 a — sin a — cos a '= 0.
46. sin a; + sin 2 # + sin 3 x = 1 + cos x -f- cos 2 #.
47. ^-^=20032*.
1 + tan x
Prove the following identities :
48. tan (30° + a) tan (30° - <*) = 2 cos 2 a ~ 1.
' 2 cos 2 a + 1
49. tan a -J- cot a = 2 esc 2 a.
50. sin 80° = sin 40° + sin 20°.
51. 1 + sintt-cosft^^
1 + sin a + cos a 2
52. cos2 a - sin (30° + a) sin (30° - a) = f .
cos « -h cos
CHAPTER VII
INVERSE FUNCTIONS
76. Inverse trigonometric functions are closely related to
the trigonometric functions previously studied.
After introducing the fundamental idea of an inverse
function, it will be shown that they lead to new relations,
closely allied to the relations already developed.
77. Fundamental idea of an inverse function. From the
equation sin a = u, it is evident that a is an angle whose
sine is it. The statement a is an
angle whose sine is u is abbreviated
into
7 a, = sin"1 u.
The equation u = sin a expresses
u in terms of a.
The equation a = sin"1 u expresses a in terms of u.
We thus have two methods of expressing the relation
between an angle and its sine.
The symbol sin"1?* is the inverse sine of u. It may be
read the inverse sine of u, the anti sine of u, arc sine u, or an
angle whose sine is u. The last form should be used until
the conception of an inverse function is perfectly clear.
Corresponding to each direct function there is an inverse
function. Thus,
a = sin"1 u corresponds to sin a = u,
a = cos""1 u corresponds to cos a = u,
a — tan"1 u corresponds to tan a = u, etc.
CAUTION. Since sin"1 u is an inverse function, the — \
74
INVERSE FUNCTIONS 75
cannot be an exponent. It is. simply part of a symbol for an
inverse function. When a function is affected by — 1 as an
exponent, it must be written with parentheses, thus (sin a;)
-i
78. Multiple values of an inverse function. It has been
shown that the trigonometric functions are single-valued
functions. Thus, if a is given, there is only one value for
sin a. See Arts. 25 and 41.
On the contrary, the inverse functions are multiple-valued
functions. Thus, if u is given there are an infinite number
of values for sin"1^. See Arts. 26 and 41. A few ex-
amples will illustrate this property of inverse functions,
and show that all the values of any given inverse function
may be combined in a single expression.
To find all the values of tan"1
V3
tan-1-lr= 30°
V3
= 180° + 30°
= 360° + 30°
- _ 180° -f 30°, or - 150°
= _ 360° -|- 30°, or — 330°
= _5400^30°, or -510°
All these angles may be expressed by n 180° + 30°, n
being any positive or negative integer. Hence
_1 i
VB=
In general, if a is an angle whose tangent is u, it may be
shown that tan-i u = n>n + Q..
76 TRIGONOMETRY
To find all the values o/cos"1^.
cos-1^ 60°
= -60°
= 360° + 60°
= 360° - 60°
= 720° + 60°
= 720° - 60°
. = _ 360° + 60°
= -360° -60°
= _ 720° + 60°
= _ 720° - 60°
All these angles may be expressed by n 360° ± 60°, n
being any positive or negative integer. Hence
o
In general, if a is an angle whose cosine is u, it may be
shown that cos-i u = 2 nir ± a..
To find all the values o/sin-1!.
sin-1i= 30°
= 180° - 30°
= 360° + 30°
= 540° - 30°
= 720° + 30°
= 900° - 30°
- 180° - 30°
- 360° + 30°
- 540° - 30°
: - 720° + 30°
INVERSE FUNCTIONS 77
All these angles may be expressed by n 180° + ( - 1)"30°,
n being any positive or negative integer. Hence
In general, if a is an angle whose sine is w, it may be
own that sin-' U = /nr +(-!)" a.
In a similar manner it may be shown that
sec"1 u = 2 nir ± a,
csc~lu = nir + (— l)n a.
79. Principal values. The smallest numerical value of an
inverse function is called its principal value, preference be-
ing given to positive angles in case of ambiguity.
The principal values of the inverse sine and the inverse
cosecant lie between — — and - ; of the inverse cosine and
2 2
the inverse secant, between 0 and ?r ; of the inverse tangent
and the inverse cotangent, between — - and - •
The principal values of an inverse function are sometimes
distinguished from the general values by the use of a capital
letter.
Thus Sin-1-^-,
while sin-1 1 = mr + ( - l)w- •
6
80. To interpret sin sin~lu and sin~l sin a.
The expression sin sin"1!* is read: the sine of the angle
whose sine is u. This sine is evidently u, hence
sin sin'1 u = u.
78 TRIGONOMETRY
The expression sin"1 sin a is read : the angle whose sine
is the sine of a. This angle is evidently a, hence
sin"1 sin a == a,
or more generally,
sin"1 sin a = nir -\- (— l)n«.
Similar relations exist between any direct function and
the corresponding inverse function.
Thus cos cos"1 u = u ;
cos"1 cos a = a, or cos"1 cos a = 2 n-n ± a ;
tan tan"1 u = u ;
tan"1 tan a = a, or tan"1 tan « = nir + a, etc.
81. Application of the fundamental relations to angles ex-
pressed as inverse functions. The fundamental relations,
being true for all angles, must necessarily be true when
the angles are expressed as inverse functions.
Thus, letting a = tan"1 u in the identity
sin2 a + cos2 a = 1,
we have (sin tan"1 u)2 -f- (cos tan"1 u)2 = 1.
Similarly (sin esc"1 u)2 -f- (cos esc"1 u)2 = 1 ;
(tan sec"1 u)2 + 1 = (sec sec"1 u)2 j
sin cos"1 u = • •
esc cos"1 u
By expressing the angle of the fundamental relations as
an inverse function, we may develop relations between the
inverse functions.
82. Given an angle, expressed as an inverse function of u,
to find the value of any function of the angle in terms of u.
By the application of one or more of the fundamental
relations, it is always possible to solve the stated problem.
INVERSE FUNCTIONS
79
Several illustrations are given below. The method em-
ployed can be readily applied to the other functions.
1. To find the value of tan cos"1 u in terms of u.
If tan cos"1 u is expressed in terms of the cosine of
cos"1 u, the problem is solved, since cos cos"1 u = u.
This may be done as follows :
tan cos"1 u
sin cos"1 u _ ± Vl — (cos cos"1 u)2 __ ± Vl — u2
COS COS"1 U U U
This result may be obtained geometrically. Since u is
given, it is evident that cos"1^ represents, among others,
two positive angles, a± and «o, each less than 360°.
Let us assume u positive and let us construct these angles
defined by cos"1 u.
Then from the figure and the definition of the tangent,
and
Since the tangent of any angle Co-
terminal with «! is -t- — — , and the tangent of any
angle coterminal with 0% is — — — , and since cos"1^
u
represents all angles coterminal with either ai or #2, we have
tan cos"1 u =
± Vl - u*
2. To find the value of sec cot 1 u in terms of u.
To solve this problem it is only necessary to express
sec cot"1 u in terms of cot cot"1 u.
80
TRIGONOMETRY
Thus
sec coir1 u = ± Vl 4- (tan cot"1 u)2 = -v/1 4-
(cot cot"1 w)2
This result may be obtained geometrically. Construct
the angles given by cot"1 u. Let us assume in this problem
that 16 is negative and hence that — u is positive. If the
cotangent of an
^angle is nega-
tive the angle
must terminate
in either the
second or fourth
quadrant. Since
OA and OB are
the terminal lines of otj and «2 respectively, and since the
terminal line is always positive, we have
OA = OB =
u —u
or by considerations similar to those in the previous ex-
ample, we have
+ Vl 4- u2 ± Vl 4- w2
sec cot"1 16=-
±16 U
3. To find the value of sin cos"1 u in terms ofu.
We have sin cos"1 u = ± Vl —(cos cos"1 16)2 = ± Vl — i*2.
4. To find the value of cot vers"1 16 in terms of u.
We have
1-u
± Vl — (1 — vers vers"1 w)2
1-u
±^/2u — u*
INVERSE FUNCTIONS 81
83. Some inverse functions expressed in terms of other in-
verse functions.
1. To express cos"1 u in terms of an inverse tangent.
From tan cos"1 u = — , (Art. 82, prob. 1) by
u
taking the inverse tangent of each member (Art. 80), there
results ± yi _ ^2
cos"1 u = tan"1 — •
u
2. To express the cot"1 u in terms of an inverse secant.
From sec cot"1 u = ± + M> (Art. 82, prob. 2) there
u
results
3. Similarly from sin cos"1 u = ± Vl — w2 there results
cos"1 u = sin"1 ( ± Vl — w2).
4. Also from cot vers"1 M = ~ there results
± V2 u - u*
1-u
vers"1 u = cot"1
± V2 u - u2
By the method exemplified in Arts. 82 and 83 it is possi-
ble to express any inverse function in terms of any other
inverse function.
In applying the above formulas care must be exercised
in selecting the angles, since each inverse function repre-
sents an infinite number of angles and one member of
the equation may represent angles not represented by the
other. For example, in problem 1, if u be positive the
cos"1 u represents angles terminating in the first and fourth
quadrants ; but tan"1 -5^- — -^- represents angles terminat-
ing in the second and third quadrants as well as angles
terminating in the first and fourth quadrants.
82 TRIGONOMETRY
84. Some relations between inverse functions derived from
the formulas for double angles, half angles, and the addition
formulas.
The general method applicable to this class of problems
will be illustrated by a few examples.
1. To express cos (2 sec"1 u) in terms of u.
cos (2 sec"1 u) = 2 (cos sec"1 u)z - 1 Art. 71, Eq. 4.
1=1-1.
(sec sec"1 u)2 u2
From this relation it follows that,
(2 \
1 ).
u2 )
2. To express tan (1 cos"1 u) in terms of u.
lY^^L
COS
Art. 72, Eq. 3.
From this relation it follows that,
l , *. ,, _i_ /I — if
2
3. To express sin (sin"1 u + cos"1 v) in terms of u and v.
sin (sin"1 u + cos"1 v) = sin sin"1 u • cos cos"1 v
+ cos sin"1 u • sin cos"1 v
Art. 62, Eq. 1.
From this relation it follows that,
sin"1 u + cos"1 v = sin"1 (uv ± Vl — w2 Vl — v2).
85. EXAMPLES
Find the value of each of the following :
1. sin-^VS. 3. tan"1!.
2. cos-^-iVS). 4. tan cot"1 4.
5. sin cot"1 4.
INVERSE FUNCTIONS 83
Express the following in terms of u and v:
6. cos cot"1 w. 17. cos (2 cos'1 w).
7. sec cot"1 u. 18. cos (2 skr1 u).
8. esc cot"1 u. 19. cos (2 tan"1 u).
9. cos sin"1 u. 20. sin (sin"1 u -f sin"1 v).
10. cos tan"1 u. 21. cos (sin"1 w 4- sin"1 v).
11. cossec^w. 22. tan (tan"1 u + cot"1 v).
12. cos esc"1 u. 23. tan (sec"1 u + sec"1 v).
13. sin (2 cos"1 w). 24. cos (see"1 w + osc"1 v).
14. tan (2 tan-1 w). 25. sin (£ cos-1 w)«
15. tan (2 sec-1 w). 26. cos (J cos'1 u).
16. tan (2 cos-1 w). 27. sin (i sec'1 w).
28. sin (| tan-1 w).
Find x in terms of a.
29. tan"1 a; = cot"1 a.
30. sin -1 x = tan"1 a.
31. cos"1 x = 2 sin""1 a.
32. cos"1 # = sin"1 a + tan"1 a.
33. sin"1 x = -J- sec"1 a.
Find a; in terms of a and &.
34. tan"1 x = sin"1 a + sin"1 6.
35. cos"1 x = sec"1 a — sec"1 b.
36. sin"1 x = 2 cos"1 a -f- £ cos"1 6.
CHAPTER VIII
OBLIQUE TRIANGLE
86. In the present chapter we develop the formulas by
means of which a triangle may be completely solved when
any three independent parts are given.
87. Law of sines. In a plane triangle any two sides are to
each other as the sines of the opposite angles.
c c
Let a, b, c be the sides of a triangle and a, ft, y the angles
opposite these sides, respectively.
From the vertex of y draw h perpendicular to the side c,
or c produced.
Then, for each figure, sin a = - ,
b
and
siny3=-<
a
Art. 21
Art. 29
Dividing the first equation by the second, we have
a _ sin a
b sin/?
In a similar manner, dropping a perpendicular from the
vertex of a, it is seen that
c sny
The last two equations may be written
a b c
sin a sin p sin -y
84
OBLIQUE TRIANGLE 85
88. Law of tangents. The tangent of half the difference of
two angles is to the tangent of half their sum, as the difference
of the corresponding opposite sides is to their sum.
From the law of sines we have
a _ sin a
b sin/3
By division and composition, this becomes
a — b _ sin a — sin ft
a + b sin a + sin/?'
which reduces to
a- 6 2cos -(« + sina— ff)
a + 6 2 sin 1 (a + J3) cos | (a - ft
= cot i (a + J3) tan *- (« - J3)
tana + a +
Singly tanHP-V) ^_g, (2)
tani(p+Y> 6+*
and tanj^^Lo) = c--a> (3)
tan i (-y + a) c + a
89. Cyclic interchange of letters. Each formula pertaining
to the oblique triangle gives rise to two other formulas of
the same type by a cyclic interchange of
letters. A cyclic interchange of letters
may be accomplished by arranging the
letters around the circumference of a circle
as in the figure, and then replacing each
letter by the next in order as indicated by
the arrows. Thus by this cyclic interchange of letters
formula (1) of Art. 88 gives rise to formula (2) ; likewise
formula (2) gives rise to formula (3).
90. Law of cosines. The square of any side of a plane tri-
angle is equal to the sum of the squares of the other sides
minus twice the product of those sides into the cosine of the
included angle.
86
TRIGONOMETRY
c D
B
DA B
For each figure
AB = AD + DB, or DB = AB- AD. Art. 2
But AB = c, ^4Z> = 6 cos a, hence DB = c — b cos a.
Also D(7 = b sin a.
From the right triangle CDB we have
BC2 = DC2 -\- DB2
Substituting values, this equation becomes
a2 = (6 sin a)2 + (c — 6 cos a)2
= 62sin2a + c2 — 2 fcccos a + 62cos2a
= 62(sin2 a + cos2 a) + c2 — 2 6c cos a.
Hence a2 = 62 + c2 - 2 be cos a.
Similarly 62 = c2 + a2
and c == fl ~f~ o •
91. To find the sine of half an angle of a plane triangle in
terms of the sides of the triangle.
From equation (1), Art. 72,
— cos «
whence
From the cosine law
. cos a. =
52 + c2 _
2 be
(1)
(2)
OBLIQUE TRIANGLE 87
•
From equations (1) and (2)
2 be
2 be
(a — 6
2 be
Let a + b -f c = 2 s. (4)
Subtracting 2 a, 26, and 2 c from each member of (4) we
have respectively
-a + 6 + c = 2(S-a)
a -f- 6 — c = 2 (s — c).
Then equation (3) becomes
a
Similarly sin •£ = -j-
ca
-82.
^/in ter
and 6in? = + J(* -")(*-*).
2 * a0
In these formulas the positive sign is given to the radical,
since it is known that half an angle of any plane triangle is
less than 90°. The same applies to the corresponding for-
mulas for the cosine and the tangent of half an angle.
To find the cosine of half an angle of a plane triangle
'n terms of the sides of the triangle.
88
TRIGONOMETRY
From equation (2), Art. 72, we have
2cos2- = l + cosa.
2
Then
1 . fr2 + c2 - a2
l+___
(b + c)2 - a2
2 be
26c
and
Hence
Similarly
and
93. To find the tangent of half an angle of a plane tri-
s angle in terms of the sides of the triangle.
Since
Similarly
and
sin?
tan"- 2-
(1)
(2)
(3}
HUI
A , a
tan a — + •% fe
_ £)($ _ c)
2 "" \
5(5 - a)
tan P — 4- A /(5
- c)(s — a)
2 *
s(s - b)
,nY- ^fe
-a)(s-b)t
OBLIQUE TRIANGLE
Formula (1) may be written
89
Letting
Similarly
and
tan" 1 ,/(«-«)(*-&)(* -c>-
(4)
(5)
(6)
(7)
Ln2 s-a\
r_:,_.J(«-«)(*-W*-c)
* *
tana- r
2 s -a
tanp- r
2~«-6'
tan'' r
2 s-o
94. To find the area of a plane triangle in terms of two
sides and the included angle.
Let A be the area and h the altitude. Then in each figure
A = ^ ch,
and h = b sin a.
Therefore A = \ be sin a.
Similarly A = J ca sin 0,
and A = \ab sin -y.
95. To find the area of a plane triangle in terms of a side
and two adjacent angles.
From Art. 94, A = % be sin a.
90 TRIGONOMETRY
From the sine law , _ csin/?
siny
. _ c2 sin a sin J3
2 siny
Then since a + ft + y = 180°,
M _c2 sin a sin p
~2sin(a + p)'
96. To find the area of a plane triangle in terms of the
three sides.
From Art. 94.
A = ±bc sin a.
Since sin a = 2 sin - cos - .
2 2'
A — be sin - cos - •
2 2
Substituting the values of sin ^ and cos ^ as found in
Arts. 91 and 92, we have, after reduction,
-a)(S-b)(s-c).
97. Formulas for solving an oblique triangle. The formu-
las developed in the present chapter are sufficient to solve a
plane triangle when three independent parts are given.
The law of sines
a b c
sin a sin ft sin y
is used when two of the given parts are an angle and the
opposite side.
The law of tangents
tan i (a- ft) = ^| tan 1 (a + 0)
a + 6
is used when two sides and the included angle are given.
OBLIQUE TRIANGLE 91
If a, 6, and y are the given parts, \ (a + ft) is obtained from
the relation a + /? + y = 180°. The formula then gives the
value of -J (a — /3), which, united with the value of % (a
gives a and /?.
TTie ta;s of half angles
s (s — a) s — a
are used when the three sides are given. The last formula
is the most accurate, since the tangent varies more rapidly
than either the sine or the cosine. The formula involving
r is advantageous when all the angles are to be computed.
TJie law of cosines
may be used to determine the third side when two sides
and an angle are given. It may also be used to determine
an angle when the three sides are given.
This formula is used with natural functions, not being
adapted to logarithmic computation.
98. Check formulas. Any formula which was not used
in the solution of the triangle may be used as a check
formula.
The relation a + (3 + y = 180° cannot be used as a check
when a problem has been solved by the law of tangents,
since the law of tangents involves this relation.
When two equations from the sine law have been used to
find two elements of a triangle, the third equation from the
sine law cannot be used as a check, since the first two
equations involve the third.
92
TRIGONOMETRY
99. Illustrative problems.
1. Given c = 127.32
a = 71° 58' 22"
J3 =52° 19' 40"
SOLUTION. Construction and estimates.
A
A\
7 V
to find a
b
y.
a = 140
6 = 120
7 = 60°
••A
Outline
c sin a tan * fa ^ a
C%ecfc
_ 5
ian 2 ^« p;
sm 7 a
+ 6
c
a
ft
a-6
a + 6
y
log a — b )
colog (a + 6)
log sin a
log tan £ (a + /3)
logc
colog sin 7
log tan £ (a — /3)
log a
a
!>+«
,
~
sin 7
r, more compactly,
log sin /3
logc
colog sin 7
log&
6
log sin a
. w> j colog sin 7
^ [ log sin /3
log a
log&
a
6
OBLIQUE TRIANGLE
93
Filling in the above outline, the completed work appears as follows
c sin a
sin 7
c
a
V
127.32 c, ,
71° 58' 22"
52° 19' 40" tan*Ca /3) a ~ & tan *fa + /3)
124° 18' 2" a + 6
55° 41' 58" a-6
24.57
268.55
62° 9' 1"
log sin a
logc
colog sin 7
log a
a
9.97814-10 a + b
2.10490 ?(« + £)
0.08297 log (a -6)
1.39041
7.57098
0.27708
2.16601 colog (a + 6)
146.56 log tan £ (a + /3)
3 sin/* logtanKa-0)
9.23847-10
9° 49' 28"
62° 9' 1"
71° 58' 29"
52° 19' 33"
Shl7 Ha -I-/9)
log sin ft
logc
colog sin 7
log 6
6
Or, in the
9.89846-10 ' r'a
2.10490 ,3
0.08297
2.08633
121.99
more compact form,
(log sin a 9.97814-10
f logc 2.10490
"S)j colog sin 7 0.08297
-2 1 log sin ft 9.89846
log a 2.16601
log b 2.08633
a 146.56
b 121.99
2. Given
a = 1674.3
c = 1021.7
£ = 28° 44' 39"
to find a
7
b.
Estimates
a = 120°
7 = 30°
6 = 900.
94
TRIGONOMETRY
a -f c
T _ c sin /3
sin 7
a
c
/s
a — c
a + c
a + 7
log (a - c)
colog (a + c)
log tan £ (a + 7)
log tan H«-7)
1674.3
1021.7
28° 44' 39"
652.6
2696.0
151° 15' 21"
75° 37' 40"
logo
log sin /3
colog sin 7
log 6
3.00932
9.68205-10
0.27268
2.96405
920.66
Check
2.81465
6.56928-10
0.59135
b =
sin a
9.97528-10
43° 22' 12"
75° 37' 40"
118° 59' 52"
32° 15' 28"
log a
log sin 0
colog sin a
log 6
3.22384
9.68205-10
0.05817
2.96406
3. Given
a = 1.4932
b = 2.8711
= 1.9005
to find
a
Estimates
ft
a= 25°
/3 = 120°
y-
7 = 35°
r —
tin7 r
2 S - C
a
1.4932
6
2.8711
c
1.9005
2s
6.2648
s
3.1324
s — a
1.6392
s- 6
0.2613
* — c
1.2319
s
3.1324
log (s - a)
0.21463
log 0-6)
9.41714 - 10
log (s - c)
0.09058
colog s
9.50412 - 10
logr2
19.22647 - 20
logr
9.61324 - 10
log tan J
9.39861 - 10
log tan |
0.19610
log tan J
9.52266 - 10
•
2
14° 3' 26"
e
2
57° 31' 2"
2
2
18° 25' 34"
a
28° 6' 52"
ft
115° 2' 4"
7
36° 51' 8"
CAecfc : a + 0 + 7
180° 00' 4"
OBLIQUE TRIANGLE
95
4. Given
6 = .0060041
c = .0093284
= 44° 47' 58"
to find
Estimates
j8= 30°
a = 105°
a = .012.
Check
sin/3
6 sin 7
tin "" ("ft "v} — a c tan 1 (OL 1 V^
c
a + c
b
c
7
.0060041
.0093284
44° 47' 58"
a
c
a — c
a + c
a + 7
K« + 7)
.012574
.0093284
.0032456
.021902
153° 1'48"
76° 30' 54"
log&
colog c
log sin 7
log sin /3
a
a
7.77845 - 10
2.03019
9.84796 - 10
9.65660 - 10
26° 58' 12"
71° 46' 10"
108° 13' 50"
csina
log(a-c)
colog (a + c
log tan Ka + 7)
log tan £ (« - 7)
U«-7)
7
7.51129 - 10
1.65952
0.62015
9.79096 - 10
31° 42' 51
76° 30' 54"
108° 13' 45"
44° 48' 3"
sin 7
logc
colog sin 7
log sin a
log a
a
7.96981 - 10
0.15204
9.97764-10
8.09949-10
0.012574
100. The ambiguous case. When two sides and an angle
opposite one of them are given, the triangle may admit of
no solution, of one solution, or of two solutions.
Let a, b, a be given. The formula
sin/?;
determines
96
TRIGONOMETRY
If the calculated
sin ft is greater than 1,
there can be no solu-
tion.
a < b sin a
If the calculated
sin ft equals 1, ft =
90° and there is one
solution.
.5 1 a
a = 6 sin a
If the calculated
sin ft is less than 1,
two supplementary
values of ft are deter-
mined, giving two so-
lutions unless the
larger value of ft plus
a is equal to or greater
than 180°.
a > b sin a
a<b
Given
6 = 420
c = 389.73
y =53° 47' 20"
to find ft
a
a.
Estimates
/3 = 65°
a =60°
a = 390.
Ttoo solutions
/S' = 115°
a' = 10°
a' = 90.
a
OBLIQUE TRIANGLE
97
Ch'jck of 1st solution
sin/3 =
6 sin 7
tan \ (a - 7) =
- -
a + c
b
c
y
420
389.73
53° 47' 20"
a
c
a — c
a + c
ft + 7
*(« + 7)
440.61
389.73
50.88
830.34
119° 35' 51'
59° 47' 50"
log b
colog c
log sin 7
log sin p
ft
180° -p = p'
P + y
£' + 7
ft
ft'
2.62325
7.40924 - 10
9.90679 - 10
9.93928 - 10
60° 24' 9"
119° 35' 51"
114° 11 '29"
173° 22' 11"
65° 48' 31"
6° 37' 49"
log (a - c)
colog (a + c)
log tan £ (« + 7)
log tan £ (a - 7)
i(«-7)
H«+7)
ft
7
1.70655
7.08074- 10
0.23502
9.02231 - 10
6° 0'34"
59° 47' 50"
65° 48' 24"
53° 47' 16"
Check of 2d solution
tan \ (7 - a') =
c + a
tan | (7 +
c sin ft
c
a'
c — a'
c + a'
7 + ft'
* (y + «f)
389.73
55.771
333.96
445.50
60° 25' 9"
30° 12' 34"
sin 7
, c sin a'
sin 7
log sin a
r logo
II colog sin 7
log sin a'
log a
log a'
a
a'
9.96008 - 10
2.59076
0.09321
9.06244 - 10
log (c - a')
colog (c + a')
log tan i (7 + a')
log tan £ (7 - a')
H7-ft')
K7 + «0
7
ft'
2.52370
7.35115 -10
9.76509-10
9.63994 - 10
23° 34' 45"
30° 12' 34"
53° 47' 19"
6° 37' 49"
2.64405
1.74641
440.61
55.771
FIRST SOLUTION
p = 60° 24' 9"
ft = 65° 48' 31"
a = 440.61
SECOND SOLUTION
j8' = 119°35'51"
a*= 6° 37' 49"
a' = 55.771
98 TRIGONOMETRY
101. EXAMPLES
Solve the following triangles, using a three-place table or
a slide-rule.
1. a =26 2. a = 48
a = 53° ft — 61°
ft = 49° y = 69°
3. a = 73° 4. a = 69°
6 = 55 6 = 64
5. a = 54° 40' 6. a = 163
6 = 122 6 = 241
c = 110 y = 34° 20'
7. a = 42 8. 6 = 115
6 = 28 c = 96
y = 72° a = 110°
9. a = 51 10. a = 8.03
6 = 63 6 = 6.42
c = 70 c = 7.15
11. a = 20° 12. y = 43°50'
a = 15 a = .34
6 = 25 c = .30
Solve the following triangles :
^13. 6 = 63.67 14. 6' =20.007
ft = 100° 10' a = 40° 27' 30"
« = 40°0'10" y = 42° 30' 15"
15. a = 238.61 . NsJ6. a = 8.0038
6 = 216.77 6 = 4.6259
c = 98.435 c = 4.3167
17. 6 = .76328 ^4.8- 6 = 85.249
c = 2.4359 c = 105.63
y = 120° 46' 18" a = 50° 40' 24"
OBLIQUE TRIANGLE
99
19. a = 1.4562
c = .45296
/3 = 74° 19' 38"
21. a = 2.1469
6 = 3.2824
j^= 4.0026
23. 6 = .06532
c = .01846
y = 8° 0' 20"
25.; a = 764.38
« = 143° 18' 31"
13 = 13° 34' 26"
20. a =
6 =
c =
22. c
a
a
24. 6
c
26. a
6
c
83.831
56.479
74.025
7.2693
.54871
5° 41' 30"
10.246
18.075
33° 30' 5"
962.27
637.34
655.80
Find the areas of the following triangles
28. a =
c =
30. a =
6 =
\32. c =
a =
^54. 6 =
27. a = 15
6 = 20
c = 25
29. a = 20.46
6 = 19.72
c = 15.04
31. 6 = 3.46
c = 4.09
a = 56° 10'
33. a = 48
v = 43°
172
103
141
18.3
22.4
32°
435.3
289.6
31° 7'
10.34
83° 22'
60° 40'
35. The horizontal distance from a point on top of a
tower to a distant flagpole is 468 ft. The angle of elevation
of the top of the flagpole is 5° 8' 30.". The angle of depres-
sion of the foot of the pole is 15° 36'. Find the height of
the flagpole.
100 TRIGONOMETRY
36. A tower is situated on a hill which inclines at an
angle of 23° 19' 10" to the horizontal. The angle of eleva-
tion of the top of the tower, from a point on the hillside,
was measured and found to be 43° 39' 50". At a point 75.5
ft. farther down, the angle of elevation was found to be
39° 23' 20". How high is the tower ?
37. From a point 5 miles from one end of a lake and
3 miles from the other end, the lake subtends an angle of
47° 34' 30". Find the length of the lake.
38. Find the altitudes of a triangle whose sides are 125.4,
230.6, and 179.8.
39. A flagpole stands on the summit of a hill. The hill
inclines 32° 18' 20" to the horizontal. At a point 25.5 ft.
from the base of the flagpole, measured along the incline,
the angle subtended by the flagpole was found to be 41° 24'.
Find the height of the flagpole.
Two observers, A and B,
stationed 4000 ft. apart, at the same
instant observe the angles BAG and
CBA to an automobile traveling on
a straight road. Three minutes
later they measure the angles DAB and ABD to the second
position of the automobile.
If
Z CBA = 32° 8',
2. DAB = 40° 12',
what is the rate of the automobile ?
41. A man wishes to measure the length of a lake from
his position on a hill top 185 ft. above the level of the lake.
He finds the angles of depression of the ends of the lake to be
6° 18' and 2° 30'. The angle subtended by the lake, formed
OBLIQUE TRIANGLE
101
by the two lines of sight, is 66° 27'.
the lake.
Find the length of
(42) The angles of elevation of a cloud, directly above a
straight road, from two points of the road on opposite sides
of the cloud, are 78° 15' 20" and 59° 47' 40". Find the height
of the cloud, the distance between the two points of observa-
tion being 5000 ft.
43. Two observers at A and
B. whose longitudes are the
same, simultaneously observe
the moon and find the angles
ZAM&vdi Z'BMto be 35° 2' 20"
and 51° 17' 10" respectively.
The latitude of A (Greenwich)
is N, 51° 17' 15" and the lati-
tude of B (Cape of Good Hope) is S. 33° 45' 16". The moon
is in the plane determined by A, E} and B. Find EM, the
distance from the center of the earth to the moon, the
radius of the earth being 3960 miles.
44. Show that twice the radius, 2 Jf2, of a
circle circumscribing a triangle is. given by
the equations
2 £ _ a _ b _ c
sin a sin ft sin y
SUGGESTION. ZBAC=£DOC.
45. Find the radius of a circle in-
scribed in a triangle whose sides are
given.
SOLUTION. Representing the area of the
triangle by A, we have
A — \ ; ar + £ br + J cr = rs.
But A = Vs(s - a) («-&)(*- c). Art. 96.
Therefore r =J(* - «)(* ~ *»)(* - <0. See Art. 93.
\ s
MISCELLANEOUS EXERCISES
102. 1. An angle of 4 radians, having its vertex at the
center of a circle, intercepts an arc of 7 inches; find the
radius of the circle.
2. Express the following in degrees : \ radians, -^ radi-
o o
STT ,,
ans, -j- radians.
3. Express the following in radian measure: 40°, 55°,
38°, 52° 16'.
4. Eeduce f radians to degrees.
5. Find the number of degrees in a central angle which
intercepts an arc of 5 feet in a circle whose radius is 8 feet.
6. An arc of 12 inches subtends a central angle of 50° ;
find the radius of the circle.
7. The number of minutes in an angle is 1\ times the
number of degrees in its supplement ; find the number of
radians in the angle.
8. An angle exceeds another by ^ radians, and their sum
is 160° ; express each angle in radian measure.
9. The angles of a triangle are to each other as 2 : 3 : 4 ;
express each angle in radian measure.
10. The angles of a triangle are in arithmetical progres-
sion, and the mean angle is twice the smallest ; express each
angle in radian measure.
102
MISCELLANEOUS EXERCISES 103
11. The circumference of a circle is divided into 7 parts
in arithmetical progression, the greatest part being 10 times
the least ; express in radians the angle which each arc sub-
tends at the center.
12. An arc of 40° on a circle whose radius is 6 inches is
equal in length to an arc of 25° on another circle ; what is
the radius of the latter circle ?
Prove each of the following :
13. sin « cos a = sin3 a cos a -f cos3 a sin a.
14. cot a esc a = .
15.
sec a — cos a
1 + tan2 a _ sin2 a
1 + cot2 a cos2 a
1/S • a , SID. ft COS ft 0
16. sm£ + - ^ = — ___cos&
cot ft — 1 1 — tan ft
17. sin6 a -\- cos6 a = 1 — 3 sin2 « cos2 a.
18. cot a — tan cc == esc a sec a (1 — 2 sin2 a).
19 1 ~ tanfi _ Got ft -I
1 + tan £ ~~ cot/2-f l'
20. sec4 a — tan4 a = 2 sec2 a — 1.
21. tan 15° = 2 -V3.
22. sin 4 a; = 4 (cos3 cc sin a? — sin3 x cos #).
23. cos 4 # = 4 cos4 # 4- 4 sin4 x — 3.
24. sin 5 x = 16 sin5 x — 20 sin3 x -f- 5 sin a?.
25. cos 5 a? = 16 cos5 x — 20 cos3 x + 5 cos x.
26. sin (a + ft -f y) = sin a cos /? cos y
+ cos a sin /? cos y
+ cos a cos (3 sin y
— sin a sin /? sin y.
27 tan(a-f-#+ x _ tan ot+ tan ^ + tan y— tan «tan ft tan y
1— tan /3 tariy— tan y tan a— tan a tan ft'
104 TRIGONOMETRY
28. 2 sin (a - ft) cos a = sin (2 a - £) - sin/3.
29. cos2(*:=1
V2
31.
(* ) V2
32. 1
4 1 - tan $
33. tan a + tan ft = sinO + £)
cos a cos /3
34. cot« + cot^=s-H^±^l.
sin a sin /?
35. cos (a -f ft) cos (a — ft)= cos2 a — sin2)8.
I. COS0 =
1 + tan2 1 0
37.
cot 0 — tan ^
38. tan 0 + cot 0 = 2 cosec 2 0.
39. C0s" + sin'*
cos a — sin a
40.
cos (45° - 0)
41. sin (a + ft) cos («-/?) +cos (a+ft) sin (a-)8) = sin 2 a.
cos a -f- cos yS
MISCELLANEOUS EXERCISES 105
43. tan 2 « — tan a = tan a sec 2 a.
44. sin« = 2cos2K
cot^a
45. cos 3 a cos a + sin 3 a sin a = cos 2 a.
4,
[sin [£•-£] 4. cos [ ^-oTj = sin a + cos /?.
\ A ) \ * yj
47. 4 sin 0 sin (60° - 0) sin (60° + 0) = sin 3 0.
48.
2sin0 + sin20
Solve each of the following equations for all values of
the unknown quantity less than 360°.
49. tan 0= sin 0.
50. (3-4cos2a)cos2a=0.
51. sin a + cos a cot a = 2.
52. tan (45° + 0) = 3 tan (45° -0).
53. 5 sin 0 = tan 0.
54. 1 -f sin2 a = 3 sin a cos a.
55. 2 cos x + sec x = 3.
56. tan42/-4tan22/-h3 = 0.
57. sec J3 4- tan ft = 2.
58. 2 sin a + 5 cos a = 2.
59. sin 5 x -\- sin 3 x = 0.
60. cos 7 x — cos a? = 0.
61. tan60 = l.
62. tan 20 tan 0 = 1.
63. sin40 + sin20-f cos0 = 0.
106 TRIGONOMETRY
64. sin20-2cos0 + 2sin0-2 = 0.
65. sin20 + sin30 = 2sin0cos20 — J.
66. tan20cot0-tan20 + cot0-l = 0.
68. sec2 0 esc2 0 + 4 = 4 esc2 0 + sec2 0.
69. Given tan (3 = u, find sin 2 /?.
70. Given tan 0 = esc 2 0, find cos 0.
71. Given cos x = -|, find cos i #.
72. Given cos x = f , find tan i x and tan 2 a?.
73. Given tan 2x = m, find tan x.
74. If « + p + y = 180°, show that
tan a + tan /? + tan y = tan a tan /? tan y.
75. If a + ft + y = 180°, show that
cot ^ + cot f± + cot 2 = cot - cot £ cot 2 .
^ ^ 2 222
76. If a + £ + y = 180°, show that
sin a + sin ft + sin y = 4 cos £ cos ^ cos 2«
222
Prove the following :
77. tan-'i + tan-1!^.
^^S 4 54
78. tan"1 - + tan"1
6
79. tan-1 k + tan-1 1 =
1 — kl
80.
81.
MISCELLANEOUS EXERCISES 107
82. sin-1
83. cos-1
m + n * n
m _, / n
n
84. cot-1 fe2~2 - = 2csc-1fc.
Solve the following for y:
85.
86. sm-12/ + sin-12;y = -
2
87. cot-1 3 y •
88. tan'1 y = sin"1 a + cos"1 b.
89. Prove that in a plane triangle, right-angled at y,
90. In a right triangle, c being the hypothenuse, prove
that
. 91 c — b 91
l -
91. In an isosceles triangle in which a = 6, prove that
cos« = ^; versy = ^2.
92. In a triangle in which y = 60°, prove that
93. Show that the area of a regular polygon, inscribed in
• i u j* • . Tir2 . 2?r
a circle whose radius is r, is -^- sin --
108 TRIGONOMETRY
94. Show that the area of a regular polygon, circum-
scribed about a circle whose radius is r, is nr2 tan -.
95. Show that the area of a regular polygon of n sides is
-j- cot -i a being the length of a side.
96. Prove that the areas of an equilateral triangle and of
a regular hexagon, of equal perimeters, are to each other as
2:3.
97. A flagpole 125 ft. high, standing on a horizontal
plane, casts a shadow 250 ft. long ; find the altitude of the
sun.
98. From the top of a rock that rises vertically 325.6 ft.
out of the water, the angle of depression of a boat was found
to be 24° 35' ; find the distance of the beat from the rock.
99. A balloon is directly above a station A. From a
second station B, in the same horizontal plane with A, the
angle of elevation of the balloon is 60° 30'. If AB = 300 ft.,
find the height of the balloon.
— * 100. From a tower 64 ft. high the angles of depression
of two objects in the same horizontal line with the base of
the tower and on the same side of the tower, are measured
and found to be 28° 14' and 42° 47' respectively; find the
distance between the objects.
101. Find the area of a triangular field whose sides are
48 rods, 62 rods, and 74 rods.
*** 102. The earth subtends an angle of 17£" at the sun ; find
the distance of the sun from the earth, the radius of the
earth being 3960 mi.
103. A vertical tower makes an angle of 113° 12' with the
inclined plane on which it stands ; and at a distance of 89 ft.
8 in. from its base, measured down the plane, the angle sub-
tended by the tower is 23° 26'. Find the height of the tower.
MISCELLANEOUS EXERCISES 109
104. In a circle of radius 8, find the area of a sector with
an arc of 50°.
105. In a circle of radius 12, find the area of a segment
with an arc of 132°.
106. In a circle of radius r, find the area of a sector with
an arc of 1 radian.
107. The area of a square inscribed in a circle is 200 sq. ft. ;
find the area of an equilateral triangle inscribed in the same
circle.
'•* 108. Find the area of a triangle of which two sides are
6 + V5 and 6 - V5, the included angle being 32° 12'.
109. At what latitude is the radius of the circle of latitude
equal to 1 the radius of the earth ?
110. From the top of a lighthouse 85 feet high, standing
on a rock, the angle of depression of a ship was 3° 38', and
at the bottom of the lighthouse the angle of depression was
2° 43' ; find the horizontal distance of the vessel and the
height of the rock.
111. At a point directly south of a flagpole, in the hori-
zontal plane of its base, I observed its elevation, 45°; then
going east 200 ft. its elevation was 35°. Find the height
of the flagpole.
112. A castle and a monument stand on the same hori-
zontal plane. The angles of depression of the top and the
bottom of the monument viewed from the top of the castle
are 40° 32' 18" and 80° 17' 46", and the height of the castle
is 104f feet. Find the height of the monument.
113. At the distance a from the foot of a tower the angle
of elevation a of the top of the tower is the complement of
the angle of elevation of a flagstaff on top of the tower;
show that the length of the staff is 2 a cot 2 a.
110 TRIGONOMETRY
114. A flagstaff a feet high is on a tower 3 a feet high ;
the observer's eye is on a level with the top of the staff,
and the staff and tower subtend equal angles. How far is
the observer from the top of. the staff?
115. Two towers on a horizontal plane are 120 feet apart.
A person standing successively at their bases observes that
the angular elevation of one is double that of the other;
but, when he is halfway between them, the elevations are
complementary. Find the heights of the towers.
116. An observer sailing north sees two lighthouses 8
miles apart, in a line due west ; after an hour's sailing one
lighthouse bears S. W., and the other S. S.W. Find the
ship's rate.
117. From the top of a house 42 ft. high, the angle of
elevation of the top of a pole is 14° 26' 9"; at the bottom of
the house it is 23° 21' 33" ; find the height of the pole.
118. From the top of a hill I observe that the angles of
depression of two successive milestones in the horizontal
plain below, in a straight line before me, are 14° 20' 24"
and 5° 31' 14". Find the height of the hill.
119. Along the bank of a river is measured a line 500
ft. in length ; the angles between this line and the lines
of sight from its extremities to an object on the opposite
bank are 53° 1' 7" and 79° 44' 55". Find the breadth of
the river.
120. A cape bears N. by E. (1 of 90° E. of K), as seen
from a ship. The ship sails N. W. 30 miles and then the
cape bears E. How far is it from the second point of
observation ?
121. Find the height of a precipice, its angles of eleva-
tion at two stations in a horizontal line with its base being
39° 30' and 34° 15' and the distance between the stations
being 145 ft.
MISCELLANEOUS EXERCISES 111
122. From the summit of a hill, 360 ft. above a plain,
the angles of depression of the top and bottom of a tower
standing on the plain were 41° and 54° ; required the height
of the tower.
123. At one side of a canal is a flagstaff 21 feet high
fixed on the top of a wall 15 feet high; on the other side
of the canal, at a point on the ground directly opposite, the
flagstaff and the wall subtend equal angles. Find the width
of the canal.
124. From the top C of a cliff 600 feet high, the angle of
elevation of a balloon B was observed to be 47° 22', and the
angle of depression of its shadow S upon the sea was 61°
10' ; find the height of the balloon, the altitude of the sun
being 65° 31' and J5, S, C being in the same vertical plane
and the sun being behind the observer.
125. At each extremity of a base AB = 758 yards, the
angles between the other extremity and two objects C and
D were observed, viz. CAB = 103° 50' 41", DAB = 53° 17'
24", DBA = S5° 47' 30", and CBA = ±6° 13' 27"; find CD.
126. Show that if r be the radius of the earth, h the
height of the observer above the sea, and d the angle of
depression of the horizon ; then tan d = — ^— -
127. A privateer lies 12.75 miles S. W. of a harbor, and
a merchantman leaves the harbor in a direction E. by S., at
the rate of 10 miles an hour ; on what course and at what
rate must the privateer sail in order to overtake the mer-
chantman in \\ hours ?
128. From the top of a hill the angles of depression of
two objects in the plain at its base were observed to be 45°
and 30°, and the horizontal angle between them is also 30°;
find the height of the hill in terms of the distance a be-
tween the objects.
112 TRIGONOMETRY
V-^129. The topmast, 120 feet above the water line of a
man-of-war coming into port at the rate of 10 miles an hour,
was first seen on the horizon at 8.45 A.M. by a person swim-
ming near the water's edge; and at 10.06 A.M. she cast
anchor. Find an approximate value for the radius of the
earth.
130. A railway curve which is a circular quadrant has
telegraph poles at its extremities and at equal distance
along the arc, the whole number of poles being 10. A per-
son in one of the extreme radii produced and at a distance
of 300 feet from its extremity, sees the third and sixth
poles in line. Find the radius of the curve.
CHAPTER IX
DE MOIVRE'S THEOREM WITH APPLICATIONS
103. In the present chapter De Moivre's theorem is
introduced with some of its applications, including the
demonstration of the fundamental series by means of which
we may calculate the trigonometric tables.
A few preliminary considerations pertaining to the com-
plex number furnish the necessary basis.
104. Geometric representation of a complex number. In
algebra it is shown that every complex number can be re-
duced to the form a + b V— 1 or a + ib where a and b are
any real numbers, and i=V — 1.
Having given a complex number a + ib, we can construct
a point P whose coordinates are y
a and b. This point may be
considered the geometric repre-
sentation of the complex num-
ber. It is thus seen that to
every complex number there cor-
responds a point in the plane.
Conversely, to every point in
the plane there corresponds a
complex number, since the coor-
dinates of the point represent the two elements, a and 6, of
the complex number.
The line OP, instead of the point P, may be considered
the geometric representation of the complex number.
113
X'-
114 TRIGONOMETRY
From the figure it is evident that
a = r cos a,
and b = r sin a.
Therefore a + ib = r cos a + ir sin a.
Hence any complex number, a -f ib, can be reduced to the
r(cos a + t" sin a). (1)
The form (1) includes all real and all pure imaginary
numbers as special cases. Letting a = 0° or 180°, (1) reduces
to r or — r, which represents any real number since r repre-
sents any real positive number. Letting a = 90° or 270°,
(1) reduces to ± ri, any pure imaginary number.
The angle XOP or a is the argument of the complex num-
ber a -f ib. It is determined by the equation
tana=*.
a
The distance OP or r is the modulus of the complex num-
ber a + ib. It is determined by the equation
r— Va2 -f- 62.
105. To show that
[r (cos a + i sin «)]n = rn (cos n« + 1 sin wa),
91 6eift<7 a positive vnieger.
Squaring the quantity r (cos a -\-i sin a), we have
[r (cos a + i sin a)]2 = r 2 (cos2 « — sin2 a-\-2i sin a cos a)
= r2(cos 2 a + i sin 2 a) by Art. 71. (1)
Multiplying each member of (1) by r (cos a + i sin a) we
have
[r (cos a + 1 sin a)]3= r3[(cos 2 a cos « — sin 2 a sin a)
+ /(sin 2 a cos a -f- sin a cos 2 a)]
= j-3 (cos 3 a + i sin 3 a) by Art. 68. (2)
From (2) we have, similarly,
[r (cos a-\-i sin a)]4 = r4 (cos 4 a -f- i sin 4 a). (3)
Equations (1), (2), and (3) have the form
[r (cos « + i sin «)]tt = rn (cos na + i sin w«). (4)
DE MOIVRE'S THEOREM WITH APPLICATIONS 115
Multiplying both members of (4) by r (cos a + i sin a), we
have
[r (cos a+i sin a)]n+1 = rn+1[cos (w + 1) <*+i sin (n +!)«]. (5)
Hence, assuming the law expressed in (4) to be true,
equation (5) shows that it is true when n is increased by
unity. But the law is true for n = 4, by equation (3) ;
hence it is true for n = 5. Being true for n — 5, it must
also be true for n =± 6, etc. Hence equation (4) is true for
all positive integral values of n.
It can be shown that equation (4) is still true when n is a
negative integer, or a fraction.
From equation (4) it is seen that the ?ith power of a com-
plex number is a complex number having an argument n
times the argument of the given number and a modulus
equal to the nth power of the given modulus.
Letting r=l, equations (1), (2), (3), and (4) become
(cos a + i sin a)2 = cos 2 a + * sin 2 a (5)
(cos a + i sin a)3 = cos 3 a + i sin 3 a (6)
(cos a + i sin a)4 = cos 4 a + i sin 4 a (7)
( cos a + / sin a)n = cos /ia + / sin /ia. (8)
The last equation is known as De Moivre's theorem.
PROB. Show De Moivre's theorem is true when n = — 3.
106. Geometric Interpretation,
Since each of the complex num-
bers cos a + i sin a
cos 2 a + i sin 2 a
cos 3 a + i sin 3 a
cos 4 a + i sin 4 a
has a modulus equal to unity,
the lines representing these num-
bers terminate in points lying
on the circumference of a circle whose radius is unity.
The arguments of any two consecutive integral powers of
cos a + i sin a differ by a, hence the lines representing any
two consecutive powers differ in direction by a.
116 TRIGONOMETRY
107. Applications of De Moivre's theorem. De Moivre's
theorem may be used to find the various roots of unity, to
extract any root of a complex number, to obtain the sine
and cosine of any multiple of an angle, and to expand the
sine and cosine of an angle in a series of powers of the
angle.
108. To find the cube roots of unity.
If the cube roots of unity are real numbers, or complex
numbers, we may assume, by Art. 104, that
VI = r (cos a + i sin a). (1)
Then, cubing, 1 = r3 (cos 3 a 4- i sin 3 a). (2)
Also l = r3cos(3a-2n7r) + zV3sin(3a-2ri7r), (3)
n being an integer.
Equating the real and imaginary parts of equation (3), we
have
and ?*3 sin (3 a — 2 nir) = 0.
These equations of condition are satisfied when
3 a — 2 mr = 0 and r = 1,
whence a =2-^.
3
When n = 0, 1, 2, 3, 4, etc.
« - 0, £=, if, ^, S- etc., respectively. (4)
o o o • o
Every angle in this series is coterminal with either 0, -^,
o
4 7T
or — -; hence all the values of sin a and cos a are obtained
o
by using only the first three values of a. Substituting
these values of a in equation (1), and remembering that r=l,
we have
DE MOIVRE'S THEOREM WITH • APPLICATIONS 117
•^1=1 forn = 0,
•\/I = -i + »iV3 forn = l,
</l = - £ - 1 i V3 f or n = 2.
The three cube roots of unity
are represented geometrically by
P1? P2, and P3.
We can now write the three
cube roots of any real number
a, for letting al5 «2, and a3 be the cube roots of a, we have
Oi =</a- 1, 02 =%(- i + H V3), a3 =^a (- i - t£ V3),
where -\/a is the arithmetical cube root.
PROBLEM. Show that (— 1 -f i -J- V3)3 = 1, thus justifying
the assumption made in Eq. 1.
109. To find the fifth roots of unity.
Let -v/1 = r (cos a + i sin a). (1 )
Then, raising each member to the fifth power,
1 = r5 (cos 5 a + i sin 5 a). (2)
Also 1 = r5 cos (5 a — 2 mi) + it6 sin (5 a — 2 nw), (3)
n being an integer.
Equating the real and imaginary parts of equation (3), we
have r5 cos (5 a — 2 mi) = 1,
and r6 sin (5 a — 2 mi) = 0.
These equations of condition are satisfied when
5 a — 2 mr = 0, and r = 1.
Therefore « = — .
o
When w = 0, 1, 2, 3, 4, etc.
-jr, -jr, -jr> etc., respectively. (4)
118 TRIGONOMETRY
Substituting the values from (4) in equation (1), and
remembering that r = 1, we have
j->2 -v/1 = 1 for n = 0,
P», ^
il + isini? for n = 2,
o o
5 5
= cos — + isin— forn = 4.
5 5
110. To extract the square root of a + ib.
Let Va + ib = r (cos a + i sin a). (1)
Squaring, a + ib = r2 (cos 2 a + i sin 2 a). (2)
Also a + ib = ^[003(2 a — 2nir)+i sin (2 a — 2 rnr)], (3)
where n is any integer.
Equating the real and imaginary parts,
^008(2 a - 2tt7r)= a, (4)
r2sin(2a-2ri7r) = 6. (5)
Squaring and adding, we have
r4=a2 + &2, (6)
and r = Va2 + &2. (8)
Equation (8) gives the value of r in terms of the known
numbers a and 6.
From equations (4) and (7)
whence 2 a — 2 nir — cos'
the quadrant of 2 a — 2 n-ir being determined by (4) and (5).
DE MOIVRE'S THEOREM WITH APPLICATIONS 119
Then a = ±cos-1 a forn = 0, (9)
Va2 + b2
and « = TT + cos"1 — a for n = 1. (10)
The values of a for n = 2, 3, 4, etc., are coterminal with
the values of a in equations (9) or (10). Hence there are
only two values for sin a and cos a ; namely those for which
n = 0 and n = l.
Substituting these values of a in equation (1), we have,
finally,
Va + ib = \/a2 -f 62[~cos /£ cos-1 CT
and Va-M& = -\/a2 + 62|cos f «• + 4 cos'1 a
L V Va2 +
+ i sin f TT 4 i cos-1 - - - \\ . (12)
V Va2 + 6V J
These values of Va + 16, while more complicated than
Va H- ib itself, are nevertheless in the standard form of
complex numbers,
r (cos a + i sin a).
111. To extract the Mh root ofa + ib.
Let -fya + ib = r (cos a + i sin a). (1)
Then a + £6 = r* (cos fca + i sin fca) (2)
= r*[cos (Tea - 2 rnr) + i sin(fca - 2 WTT)], (3)
where n is an integer.
Equating the real and imaginary parts of equation (3), we
have r* cos (ka — 2 HIT) = a, (4)
and i* sin (&« — 2 WTT) = 6. (5)
120 TRIGONOMETRY
-Squaring and adding equations (4) and (5), we have
r2* = a2 + b2. (6)
Then r* = Va2 + 62, (7)
and r= ^a2 + 62. (8)
From equations (4) and (7), we have
cos (ka-
or fca — 2 nir = cos'1
the quadrant of ka — 2 mr being determined by the signs of
(4) and (5).
Therefore « = ?^Z[ + ! cos"1 a
k k
whence a =
«=^ + icos-1
k k
4?r 1
c^-r+'-cos-1
k k Va2 + b*
Substituting the values for r and a as found in equations
(8), (10), (11), (12), etc., in equation (1), we have the several
kih roots of a + ib.
When A; is an integer there are k roots. Consecutive
values of a, as given by (9), differ by — , hence all the
fc
values of a after the fcth are coterminal with one of the
first k values.
112. To express sin na and cos na in terms of sin a and
cosa.
We have
cos na + i sin na = (cos a + i sin «)n.
DE MOIVRE'S THEOREM WITH APPLICATIONS 121
Expanding the second member by the binomial theorem
and equating the real and imaginary parts, the problem is
solved.
Thus, for sin 4 a and cos 4 a we have
cos 4 a 4- i sin 4 a = (cos a -f i sin a)4 = cos4 a,
-h 4 i cos3 a sin a — 6 cos2 a sin2 a — 4 i cos a sin3 a + sin4 a.
Therefore sin 4 a = 4 cos3 a sin a — 4 cos a sin3 a.
and cos 4 a = cos4 a — 6 cos2 a sin2 a -f- sin4 a.
113. Comparison of the values of sin a, a, and tan a, a being
an acute angle. Let a be any acute angle expressed in ra-
dians. With the vertex 0 as
a center and any radius OB,
describe the arc EC. Draw AG
and BD perpendicular to OB,
and join B with O.
The area of the triangle OBC
is less than the area of the
sector OBC, and the sector OBC is less than the triangle
OBD.
But since
AC= OC sin a, BD = OB tan a, and arc BC = OB • a, Art. 17,
the area of the triangle OBC is equal to | • OB • OC sin a,
the area of the sector OBC is equal to ^ • OB • OB - a,
and the area of the triangle OBD is equal to |- • OB - OB tan «.
Hence
or sin a < a < tan a.
114. Value of for small values of a. In Art. 40 we
a
saw that sin a approaches 0 as a approaches 0. The value
of — therefore approaches . as a approaches 0.
a (J
122 TRIGONOMETRY
But from the previous article
sin a. < a < tan a.
Dividing by sin a, we have
a 1
sin a cos a
sin a
or 1 > > cos a.
a
sin a , . . sin a
Since lies between 1 and cos a, must approach
a a.
1 as a approaches 0, since cos a approaches 1.
Then for very small angles sin a may be replaced by «,
expressed in radians. The error thus introduced is so small
that it may be neglected in many problems. Thus, to five
decimal places,
sin 1° = 0.01745 1° = 0.01745 radians
sin 2° = 0.03490 2° = 0.03491 radians
sin 3° = 0.05234 3° = 0.05236 radians
sin 4° = 0.06976 4° = 0.06981 radians
115. To develop sin a and cos a in terms of a.
By De Moivre's theorem,
cos nO + i sin nO = (cos 0 -f i sin 0)n.
On expanding the second member by the binomial theorem,
we have
cos nO + * sin n& = cosn 0 -{-in cos""1 0 sin 6.
^j2 — -cosn~2 0 sin20 — ?! — j^ ' cosn~30 sin30
|n(n-l)(n~2)(n-8)cos,_4gsin4g | ... (1)
11
Equating the imaginary parts, we have '
sin n$ = n cos""1 0 sin 0 r^ cosw~3 0 sin3 0
tt(n-l)(n-2)(n-3)(tt-4)
+ _A & A A /cosn
l£
Let nO = a. Then equation (2) may be written
DE MOIVRE'S THEOREM WITH APPLICATIONS 123
sin a = ? cos"-1 0 sin 0 - ^ — ^ ^ cosw~3 0 sin3
0 |3
----
" "
Let a remain constant while n increases indefinitely.
Then 0 necessarily decreases indefinitely, since n0 = a, a
constant. By Art. 114, when 0 approaches 0, — ^— ap-
proaches 1, and cos 6 approaches 1. Making these substi-
tutions in equation (4), we have
a3 a5 a7
Equating the real parts of equation (1), we have
cos n6 = cos" 0 - n^~1) cos"-2 6 sin2 6
e
n(n- l)(n-2)(tt-3) **•-.•«* /KN
4. _1 - d\_^ — /i -- ? Cosn~4 6 sm4 ^ — ••-. (5)
By the same process as above, equation (5) becomes
The series for sin a, and cos a are convergent for all finite
values of a* They enable us to compute the sine and
cosine of any angle. It is then possible to construct a table
of natural functions, from which the logarithmic functions
may be obtained. In using these series a must, of course,
be expressed in radian measure.
* See any College Algebra on the convergency of series.
124 TRIGONOMETRY
116. EXAMPLES
1. Find the four fourth roots of unity by De Moivre's
theorem.
2. Find the six sixth roots of unity by De Moivre's
theorem.
3. Find the square root of 5 — 3 i.
SOLUTION. — Let V5 — 3 i = r (cos a + I'sin a). (1)
Then 5 - 3 1 = r2 (cos 2 a + i sin 2 «)
= r2[cos (2 a - 2 WTT) +isin(2a — 2nir)].
Equating the real and the imaginary parts,
r2 cos (2 a - 2 WTT) = 5, (2)
and r2 sin (2 a - 2 WTT) = - 3. (3)
Squaring and adding (2) and (3), we have
r* = 34, . •. r2 = V&, and r = ^34. (4)
Then cos (2 a — 2 nir) = »
V34
and 2 a - 2 nir = 329° 2', (5)
the quadrant being determined by (2) and (3).
When n = 0, a = 164° 31' ; (6)
» = !,« = 844° 31'. (7)
Substituting from (4) and (6) in (1), we have
\/5^3l = - 2.3271 + .6446 L
Substituting from (4) and (7) in (1), we have
V5^3l = 2.3271 -.6446i.
4. Find the square root of 3 -f 4 i.
5. Find the square root of — 3 — 4 i.
6. Find the square root of 1 + 2 1.
7. Find the square root of i.
8. Find the square root of — i.
9. Find the cube root of 2 — 3 i.
SOLUTION. — Let \/2 - 3 i - r (cos a + i sin a). (1)
Then 2 - 3 i =rs (cos 3 a + i sin 3 a)
(3a-2n7r)].
DE MOIVRE'S THEOREM WITH APPLICATIONS 125
Equating the real and imaginary parts,
r3 cos (3 a - 2 WTT) = 2 ; (2)
and r8 sin (3 a - 2 nir) = - 3. (3)
Squaring and adding (2) and (3), we have
i* = 13, .-. r8 = Vl3~ and r = \/13T (4)
2
Then cos (3 a - 2 WTT) =-— ,
and 3 a - 2 nir = 303° 41' ; (5)
the quadrant being determined by (2) and (3).
When n = 0, a = 101° 14' ; (6)
n = l, a = 221° 14'; (7)
n = 2, a = 341° 14'. (8)
Substituting from (4) and (6) in (1), we have
•v/2^3l = - .2987 + 1.5041 i.
Substituting from (4) and (7) in (1), we have
v 2-3i = - 1.1530 - 1.0107 *.
Substituting from (4) and (8) in (1), we have
•2/2 - 3 i = 1.4518 - .4933 i.
We have thus found the three cube roots of 2 — 3 i.
10. Find the cube root of 1 + i.
11. Find the cube root of — 1 +i.
12. Find the cube root of 2 -f 3 i.
13. Find the values of sin 3x and cos 3 a; in terms of
sin a? and cos a;.
14. Find the values of sin 5 a; and cos 5 a; in terms of
sin x and cos x.
15. Prove by De Moivre's theorem that
sin a = 2 sin ~ cos ~> also cos a = cos2 ~ — sin2 -•
2, L & &
I a . a\2
SUGGESTION, cos a + i sin a = I cos - + i sm - 1 . f
16. Show that cos a = cos3 ^ — 3 cos f sin2 1,
o o o
sin a = 3 cos2 ^ sin ^ — sin3 £
o o o
SPHERICAL TRIGONOMETRY
CHAPTER X
FUNDAMENTAL FORMULAS
117. The spherical triangle.* Spherical trigonometry
treats of the relations between the various parts of a
spherical triangle and of the methods of solving the spheri-
cal triangle.
The sides of a spherical triangle
are always arcs of great circles.
Having given a spherical triangle
ABC, situated upon a sphere S, a
triedral angle 0- ABC may be formed
by passing planes through 0, the
center of the sphere, and through
the sides of the triangle.
It is known from geometry that the arc AB and the
angle AOB contain the same number of degrees, and that
the angle CAB and the diedral angle C-AO-B contain the
same number of degrees.
The sides and angles of a spherical triangle may have
any values between 0° and 360°.
A triangle having one or more of its parts greater than
180° is called a general spherical triangle.
A triangle having each of its parts less than 180° is called
a spherical triangle.
We shall consider only those triangles whose parts are
each less than 180°.
* For a course on the right spherical triangle read Arts. 117 and 126
and from Art. 128 to end of Chapter XI.
127
128
TRIGONOMETRY
118. Law of sines. To find the relation between two sides
of a spherical triangle and the angles opposite.
/3 > 90°, 6 > 90°.
Given a spherical triangle and its accompanying triedral
angle ; through the vertex P pass planes perpendicular to
OR and OQ, intersecting in PF a line perpendicular
to the plane ORQ.
Then
sin ft
w
Also
sin a
-§-
*-»=&
therefore
Hence
Likewise
sin a
RP
sin b
sin a
QP
sin a
sin/3
sin 6
sn y sn c
Uniting these equations, we have
sing _ sin b _ sine
sin a sin p sin Y
FUNDAMENTAL FORMULAS
129
This demonstration applies to similar figures drawn for
all possible cases,* hence the theorem is always true.
To find the relation between the three
A
J
v
119. Law of cosines.
sides and an angle.
Given a spherical tri-
angle and its accompany-
ing triedral angle ; pass
a plane through the ver-
tex A perpendicular to
OA, intersecting the
planes of the triedral
angle in the lines AB,
AC, and BC.
b < 90°, c < 90° ; a < 180°, a < 180°.
Then AB = r tan c, OB = r sec c, AC = r tan b, OC = r sec b.
From the triangle OBC, by Art. 90,
B& = (r sec b)2 + (r sec c)2 — 2(r sec &) (r sec c) cos a. (1)
Likewise from the triangle ABC,
BC2 = (r tan b)2 + (r tan c)2 - 2 (r tan 6) (r tan c) cos a. (2)
Subtracting (2) from (1), we have
0 = r2 (sec2 6 - tan2 6) + r2 (sec2 c - tan2 c)
— 2 r2 sec b sec c cos a + 2 r2 tan 6 tan c cos <*,
which reduces to
0 == 1 — sec 6 sec c cos a + tan b tan c cos «,
or cos a = cos b cos c + sin b sin c cos a. (3)
Also cos b = cos c cos a -f- sin c sin a cos p, (4)
and cos c = cos a cos 6 + sin a sin 6 cos y. (5)
* The following seven cases can arise :
(1) 3 sides < 90°, 3 angles < 90° (4) 1 side < 90°, 2 angles < 90°
(2) 3 sides < 90°, 2 angles < 90° (5) 1 side < 90°, 1 angle < 90°
(8) 2 sides < 90°, 2 angles < 90° (6) 1 side < 90°, 0 angle < 90°
(7) 0 side < 90°, 0 angle <90°.
It is to be understood that all parts not mentioned are greater
than 90°.
130
TRIGONOMETRY
120- To extend the law of cosines.
In the derivation of the formula
cos a = cos b cos c -f- sin b sin c cos a,
b and c were less than 90°, while a and a were less than 180°.
To show that the formula is true in general, it is necessary
to consider two additional cases :
1st. Both b and c greater than 90°.
2d. Either b or c greater than 90°.
Since ft and y do not enter the formula, they may have
any value consistent with the above conditions.
First. Given the triangle ABC in
which b > 90° and c > 90°.
Extend the sides 6 and c of the tri-
angle ABC, forming the lune whose
angle is a. Then in the triangle A'BC,
the sides A'B and A'C are each less
than 90°, hence by Art. 119
cos a = cos (180° - b) cos (180° - c)
+ sin (180° - 6) sin (180° - c) cos «,
or cos a = cos b cos c + sin b sin c cos a.
Hence the law of cosines holds when
both b and c are greater than 90°.
Second. Given the triangle ABC}
in which b < 90° and c > 90°.
Extend the sides a and c of the
triangle ABC, forming the lune
whose angle is ft. Then in the tri-
angle AB'C, the sides AB' and AG
are each less than 90°, and the
angle B'AC is equal to 180° - a.
Then, by Art. 119,
cos (180° - a) = cos b cos (180° - c)
+ sin b sin (180° - c) cos (180° - a),
or cos a = cos b cos c -f sin b sin c cos a.
FUNDAMENTAL FORMULAS 131
Hence the law of cosines holds when either b or c is greater
than 90°. The law of cosines is therefore true in general.
121. To find the relation between one side and the three
angles.
Let a, b, and c be the sides of any
spherical triangle, and a', b'} c' the sides
of its polar triangle.
Applying the law of cosines to the
polar triangle, we have
cos a' = cos b' cos c' + sin b' sin c' cos a'.
But
a' = 180° - a, b'= 180° - /?, etc.
Therefore
cos (180° - a) = cos (180° - /?) cos (180° - y)
+ sin (180° - £) sin (180° - y) cos (180° - a),
or cos a = — cos /3 cos y + sin /? sin y cos a. (1)
Also cos (3 = — cos y cos a + sin y sin a cos 6, (2)
and cos y = — cos a cos ft + sin a sin /? cos c. (3)
122. The sine- cosine law. To find the relation between
three sides and two angles.
We have
cos a = cos 6 cos c + sin b sin c cos a, (1)
and cos b = cos c cos a -f sin c sin a cos /?. (2)
Eliminating cos a by substitution,
cos b = cos b cos2 c + sin b sin c cos c cos a + sin c sin a cos /?.
Transposing and factoring,
cos b (1 — cos2 c) = sin b sin c cos c cos a -f- sin a sin c cos /?.
Eeplacing (1 — cos2 c) by sin2 c, and dividing by sin c, we
have
cos b sin c = sin b cos c cos a + sin a cos ft.
Kearranging terms,
sin a cos /3 = cos 6 sin c — sin 6 cos c cos «. (3)
132 TRIGONOMETRY
Also sin 6 cos y = cos c sin a — sin. c cos a cos ft, (4)
and sin c cos a = cos a sin b — sin a cos 6 cos y. (5)
Interchanging ft and y and consequently b and c, we have
from (3)
sin a cos y = cos c sin b — sin c cos 6 cos a. (6)
Similarly from (4)
sin 6 cos a = cos a sin c — sin a cos c cos /?, (7)
and from (5)
sin c cos ft = cos 6 sin a — sin & cos a cos y. (8)
123. To find the relation between two sides and the three
angles.
Applying the sine-cosine law to the polar of the given
triangle, we have
sin a' cos ft' = cos bf sin c' — sin b! cos cr cos ex! .
But a'= 180° - a, 0' = 180°- 6, etc.
Then
sin (180° -a) cos (180°-&)=cos (180° -0) sin (180° -y)
-sin (180°- /?) cos (180°- y) cos (180°- a).
Therefore sin a cos 6 = cos ft sin y -f- sin ft cos y cos a. (1)
Also sin /? cos c = cos y sin a + sin y cos a cos &, (2)
sin y cos a = cos a sin /3 -f sin a cos /? cos c, (3)
sin a cos c = cos y sin ft + sin y cos /? cos a, (4)
sin ft cos a = cos a sin y 4- sin a cos y cos b, (5)
sin y cos b = cos /3 sin a + sin /? cos a cos c. (6)
124. To find the relation between two sides and two anglest
one of the angles being included between the given sides.
From Art. 122
sin a cos ft = cosb sin c — sin 6 cos c cos a.
Dividing this equation by
sin a sin ft = sin b sin a, Art. 118
FUNDAMENTAL FORMULAS 133
member by member, we have
, Q _ cot b sin c — cos c cos a
sin a
Therefore sin a cot (3 = cot b sin c — cos c cos a. , (1)
Similarly sin ft cot y = cot c sin a — cos a cos ft (2)
and sin y cot a = cot a sin b — cos 6 cos y. (3)
Interchanging a and ft and consequently a and 6, we
have from (1)
sin /? cot a = cot a sin c — cos c cos ft (4)
Similarly from (2)
sin y cot ft = cot b sin a — cos a cos y, (5)
and from (3)
sin a cot y = cot c sin 6 — cos b cos a. (6)
125. Formulas independent of the radius of the sphere. It
will be noticed that r, the radius of the sphere, does not
enter any of the formulas thus far developed ; hence they
are independent of the radius of the sphere, and may be
applied, without modification, to triangles on any sphere.
The fact is serviceable in problems of Astronomy and Ge-
odesy where the formulas are applied to triangles situated
upon the celestial and terrestrial spheres.
CHAPTER XI
SPHERICAL RIGHT TRIANGLE
126. The spherical right triangle is a spherical triangle
one of whose angles is a right angle. The other parts may
have any values between 0° and 180°.
In the work that follows the angle y will be taken as the
right angle.
127. Formulas for the solution of right triangles. The
formulas for the solution of any spherical right triangle are
obtained from the general formulas of Chapter X by letting
y = 90°. We thus have
from eq. (1) Art. 118 sin a = sin c sin a
from-eq. (1) Art. 118 sin b = sin c sin /3
from eq. (5) Art. 119 cos c = cos a cos b
from eq. (1) Art. 121 cos a = sin ft cos a
from eq. (2) Art. 121 cos ft = sin a cos b
from eq. (3) Art. 121 cos c = cot a cot ft
from eq. (2) Art. 124 cos ft = tan a cot c
from eq. (3) Art. 124 sin b = cot a tan a
from eq. (5) Art. 124 sin a =cot ft tan b
from eq. (6) Art. 124 cos a = tan b cot c.
128. Direct geometric derivation of formulas. Let a and b
be the sides of a given spherical right triangle, a and ft the
angles opposite, and c its hypotenuse.
Let 0-ABC be its accompanying triedral angle.
Through the vertex B pass a plane perpendicular to OA,
intersecting the planes of the triedral angle in AB} BC,
and CA.
134
SPHERICAL RIGHT TRIANGLE
135
Then Z BAO=a, Z BOC=a, Z AOC=b, and Z.AOB=c.
Also Z jB(L4, Z BOO, ZCAO,^BAO are each a right angle.
From the triangle ABC we have
OB
;sina? (1)
OB
AC
AC OA tan 6
cos a. = = -pr- = ,
AB AB tan c
OA
CB
CB OC tana
tana = __ = ^=___.
OC
By interchanging a and /? and consequently a and 6, or
by passing a plane through D JL to OB and proceeding as
above, we have
(2)
(3)
o
cos/? =
sin c
tana
- -- ,
tan c
sin a
Also from the figure.
OA
OA OC cos b
= OB=0^=seca
OC
Dividing (1) by (5) and reducing by (7) we have
,
COS 0
and by interchange of letters as before
(5)
(6)
(7)
(8)
sin (3
cos a
cos a
136
TRIGONOMETRY
Substituting the values of cos a and cos b from (8) and (9)
in (7), we have, after reduction,
cos c = cot a cot ft. (10)
In the demonstration of formulas (1) to (10) the parts
a, ft, a, b, and c were assumed less than 90°. To show that
these formulas are true in general, it is necessary to con-
sider two additional cases : viz. (1) when one side and the
hypotenuse are each greater than 90°, and (2) when the two
sides are each greater than 90°.
1. Wlien a > 90° and c > 90°.
Let ABC be the given
spherical right triangle.
Draw the lune BB'. Then
in the right triangle AB'C
a each part is less than 90°.
and formulas (1) to (10) are applicable.
sm(180°-a) = sin(180°-a)
' sin (180° - c)
sin a
From (1)
or
sin a =
sine
which shows that (1) holds when a and c are each greater
than 90°.
From (2) cos (180° - a) = tan b
or
cos a =
tan (180° -c)
tan b
tanc'
which shows that (2) also holds in this case.
Similarly it may be shown that formulas (3) to (10) hold
when a and c are each greater than 90°.
2. When a > 90° and b > 90°.
Let ABC be the given spherical
right triangle. Draw the lune (7(7.
Then in the right triangle ABC' c
each part is less than 90° and for-
mulas (1) to (10) are applicable.
SPHERICAL RIGHT TRIANGLE
137
From (1) sin (180° - a) = Bin (180° -a)
sine
or
sin a =
sin a
sin G
which shows that (1) holds when a and b are each greater
than 90°.
From (2) cos (180° - «) =
tan b
tan c
or
cos a =
tan c'
which shows that (2) also holds in this case.
Similarly it may be shown that formulas (3) to (10) hold
when a and 6 are each greater than 90°.
129. Sufficiency of formulas. It will be noticed that the
ten formulas of Arts. 127 and 128 contain all possible com-
binations of the five parts of a spherical right triangle, taken
three at a time ; hence they are sufficient to solve any spheri-
cal right triangle directly from two given parts.
130. Comparison of formulas of plane and spherical right
triangles. By rearranging the formulas of the previous
article, the analogy between the formulas of the plane and
the spherical right triangles is made apparent.
IN PLANE RIGHT TRIANGLES*
a
sin a = —
cos a = -
c
tan«=-
o
sin /?=*
C
cos/3 = -
c
tan/?=-
a
sin a = cos ft sin ft = cos a
c2 = a2 + b2
IN SPHERICAL RIGHT TRIANGLES
cos a =
sin a
sin c
tan 6
tanc
tana
sin b
sin ft =
cos ft =
sin b
sin c
tana
tanc
tan b
sin a
sn « =
1 = cot a cot ft.
*The above comparison is taken from Chauvenet's
Spherical Trigonometry."
cos b cos a
cos c = cos a cos b
cos c = cot a cot /?.
Plane and
138
TRIGONOMETRY
131. Napier's Rules. The ten formulas used in the solution
of spherical right triangles can all be expressed by means
of two rules, known as Napier's rules
of circular parts.
Napier's circular parts are the
sides a and 5, the complements of the
angles opposite or 90° — a, 90° — /?,
and the complement of the hypot-
enuse or 90° — c.
They are usually written
co a
a, b, GO a, co/3, coc.
It will be noticed that the right angle
is not one of the circular parts.
Let the five circular parts be placed
in the sectors of a circle in the order
in which they occur in the triangle.
Whenever any three parts are considered,
it is always possible to select one of
them in such a manner that the other two parts will either
be adjacent to this part, or opposite this part. The part
having the other two parts adjacent to it or opposite it is
called the middle part.
Thus let co a, 6, and a be the parts under consideration.
Then b is the middle part and co a and a are adjacent parts.
If co c, co /?, and b are the parts under consideration, b is
the middle part and co c and co y8 are opposite parts.
Napier's rules may now be stated as follows :
The sine of the middle part is equal to the product of the
cosines of the opposite parts.
The sine of the middle part is equal to the product of the
tangents of the adjacent parts.*
* To associate cosine with opposite and tangent with adjacent, it
may be noticed that the words cosine and opposite have the same
vowels ; likewise the words tangent and adjacent.
SPHERICAL RIGHT TRIANGLE 139
132. Theorem. In a spherical right triangle, a side and
the angle opposite terminate in the same quadrant.
From the equation
cos a = cos a sin ft
it is seen that cos a and cos a must always have the same
sign, since sin j3 is always positive. Hence a and a termi-
nate in the same quadrant.
133. Theorem. Of the three parts a, b, c, if any two ter-
minate in the same quadrant, the third terminates in the first
quadrant ; if any two terminate in different quadrants, the
third terminates in the second quadrant.
This follows directly from the equation
cos c = cos a cos b
by noticing that if any two of the quantities cos a, cos 6,
and cos c have like signs, the third is positive ; if any two
have unlike signs, the third is negative.
134. Two parts determine a triangle. In order to solve a
spherical right triangle two parts, in addition to the right
angle, must be given. Each of the required parts should be
obtained directly from the given parts.
Thus, given
•6 = 50°, c = 110°
we have cos a = — - ,
cos 50° '
cos a = tan 50° cot 110°,
using the formulas for spherical right triangles, or Napier's
Rules. If, in the solution of a problem, the sine of any
required part is found to be negative, no triangle is possible,
since no part of a spherical triangle can be greater than 180°.
Likewise if the logarithmic sine or cosine of any required
part is found to be greater than zero, the triangle is impossi-
ble, since no sine or cosine is numerically greater than unity.
140 TRIGONOMETRY
135. The quadrant of any required part. Since the parts
of a spherical triangle may have any value between 0° and
180°, it is always necessary to determine whether the required
parts are greater or less than 90°. This can be done by the
theorems of Arts. 132 and 133.
Thus, given
6 = 50°, c=110°,
we have a > 90°, a > 90°, and ft < 90°.
The quadrant in which any required part terminates may
also be determined from the formula used in calculating that
part, by observing the signs of the functions involved. But
when the unknown part is determined from its sine, the part
terminates in both quadrants, giving two solutions, unless
limited by the theorems of Arts. 132 and 133.
Thus, given
6 = 50°, c = 110°,
by writing the signs of each function above the function, we
have
cos 110°
cos a = — - ,
cos 50°
cos a = tan 50° cot 110°,
sin 110°
Then a > 90° and a > 90°, since their cosines are negative,
and (3 < 90° by Art. 132.
136. Check formula. The formula containing the three
computed parts may always be used as a check formula.
Thus, having given
6 = 50°, c = 110°,
the check formula is
cos a = cos a sin .
SPHERICAL RIGHT TRIANGLE
141
137. Solution of a right triangle.
Given b = 77° 35' 16" and a = 112° 19' 42"
tan a = ^=-^-
cota
6
a
log sin b
log cot a
log tan a
a
77° 35' 16"
112° 19' 42''
9.98973 - 10
9.61353 - 10
0.37620
67° 11' 30"
112° 48' 30"
cos a
+
tan b
log cos a
log tan b
log cot c
c
9.57969 - 10
0.65740
8.92229 - 10
85° 13' 13''
94° 46' 47"
cos /3 = cos & sin a
log cos 6
log sin a
log cos j3
§
9.33233 - 10
9.96615 - 10
9.29848 - 10
78° 31' 53"
Check
+
cos /3 = tan a cot c
log tan a
log cot c
log cos /3
0.37620
8.92229 - 10
9.29849 - 10
138. Two solutions or the ambiguous case. Whenever a
side and the angle opposite are given, there are two solutions.
Thus if a and a are given, the only formulas by which 6,
/3, and c can be determined are
sin 6 = tan a cot a,
sin/?.;
cos a
cos a3
sin a
Since the unknown parts are obtained from their sines,
each may have two values, giving two solutions, the
theorems of Arts. 132 and 133 not restricting the values of
the parts to one solution.
Having found the two values for each part, the theorems
142
TRIGONOMETRY
of Arts. 132 and 133 determine the values that belong to
each solution.
Thus, given a = 155° 27' 45" to find b
a = 100° 21' 50" c
ft
SOLUTION
sin b = tan a cot a
a
a
log tan a
log cot a
log sin b
b
155° 27' 45"
100° 21' 50"
9.65945 - 10
9.26217 - 10
8.92162 - 10
4° 47' 20" and
175° 12' 40"
stn
cos a
P — _
cos a
log cos a
log cos a
log sin /3
18
9.25503 - 10
9.95889 - 10
9.29614 - 10
11° 24' 22" and
168° 35' 38"
sin c =
sing
-4-
sina
log sin a
log sin a
log sin c
c
9.61835
9.99286
- 10
-10
9.62549
24° 58'
155° 1'
- 10
18" and
42"
Check
sin b = sin c sin /3
log sin c
log sin /3
log sin 6
9.62549 - 10
9.29614 - 10
8.92163 - 10
FIRST SOLUTION
&!= 4° 47' 20"
ft= 11° 24' 22"
C = 155° 1'42"
SECOND SOLUTION
62 = 175°12'40"
A = 168° 35' 38"
c2= 24° 58' 18"
SPHERICAL RIGHT TRIANGLE
143
139.
EXAMPLES
Solve the following spherical right triangles, right-
angled at y.
9. a = 132° 25'
a = 107° 30'
1. 6= 10° 32'
a= 12° 3'
2. a = 25° 18'
b= 32° 41'
3. c = 120°37'
ft = 9° 49'
4. c= 46° 40'
a= 20° 50'
5. a = 115° 6'
b = 123° 14'
6. a = 112° 43' 30"
c= 85° 10' 10"
7. a = 15° 18' 20"
c= 21° 30' 40"
8. 6 = 168° 13' 45"
c = 150° 9' 20"
140. Quadrantal triangles. A quadrantal triangle is a
triangle one of whose sides is 90°. Its polar triangle is
then a right triangle. The solution of a quadrantal tri-
angle is effected through the solution of its polar triangle.
141. Isosceles triangles. The solution of an isosceles
triangle is effected by solving the two equal right triangles
formed by dropping a perpendicular from the vertex to the
base.
10. c= 80° 3' 20"
0 = 135° 16' 30"
11. 6=171° 3' 15"
c= 12° 20' 30"
12. a= 35° 54' 20"
a= 47° 6' 10"
13. b= 15° 2' 30"
J3= 20° 11' 40"
14. «= 20° 26' 20"
£ = 84° 41' 40"
15. a= 25° 41' 30"
a= 34° 25' 40"
144 TRIGONOMETRY
•
142. EXAMPLES
Solve the following triangles,
1. a = 117° 54' 30" 3. £=153° 16'
b= 95° 42' 20" a= 19° 3'
c= 90° c= 90°
2. a = 69° 45' 4. a = 159°33'40"
(3= 94° 40' 6= 95° 18' 20"
c= 90° c= 90°
5. The base of an isosceles triangle is 51° 8'. The equal
angles are each 41° 57'. Find the equal sides and the
angle at the vertex.
6. The base angles of an isosceles triangle are each
100° 12' 30", the vertical angle is 50° 19' 40". Find the
equal sides and the base.
CHAPTER XII
OBLIQUE SPHERICAL TRIANGLE
143. In the present chapter the general formulas of
spherical trigonometry, already developed (Chap. X) are
transformed into standard formulas adapted to logarithmic
computation; and the problem of the solution of the spheri-
cal triangle is discussed.
GENEEAL SOLUTION
144. To find the angles when the three sides are given.
From Art. 119, we have
cos a = cos b cos c + sin b sin c cos a.
Therefore cos « = cos a.-cos b cos c. (1)
sin b sin c
But 2 sin2 i a = 1 — cos a
_ sin b sin c — cos a -f- cos b cos c
sin b sin c
cos a — cos (6 — c)
sin b sin c
Applying formula (4) of Art. 73, we have
sin b sin c
Letting 2s = a + b-\-c, we have
8in i a=
sin '6 sin c
145
146 TRIGONOMETRY
Similarly uniting (1) with
2 cos2 1 a = 1 4- cos a,
we have 2 cos2 4 « = sin 6 sin c + cos « -cos 6 cos c
sin b sin c
cos (b 4- c) — cos CT
sin 6 sin c
2 sin.±a + b c sin-— a
sin 6 sin c
Therefore cos^ a=Jsin * «h (*-«). (3)
* sin b sin c
Uniting (2) and (3),
ten 1 q = J8in 0 ~ *) Sin (' ~ C) = tan ** (4)
^ — —
sin 5 sin (s — a) sin (s — a)
Similarly tan * p = Jsin (J ~ c> f <« ~ ">= . ^n\. (5)
\ sin .« sin fx — h\ sin <^.c — A^ ^ '
and tan H = J8in (^ ~ ") 8in (
^ —
sin 5 sin (s — 6) sin (s — b)
tan
sin 5 sin (5 — (?) sin (5 — c)
where tan r = /sin (s - a) sin (s - fr) sin (3 - c) .
v sins
145. To find the sides when the three angles are given.
Following the method of the last article, the equation
cos a= —cos ft cos y 4- sin ft sin y cos a
gives
. - sin 6 sin Y
and cos J a = J«»(*-P) eo. (J-
^ sin p sin -y
and tan J a = J ~'°cts^Qcos (^~a) = tan /? cos (5 -a)
^- -
ctQ
cos(S-P)
OBLIQUE SPHERICAL TRIANGLE 147
where 2# = a
tan R
146. Delambre's or Gauss's formulas express relations be-
tween the six parts.
By Art. 68
sin i (a + (3) = sin i a cos i'/3 + cos ^ a sin i. /?.
Substituting for sin \ a, cos 1 a, sin 1 y8, and cos | /8 their
values in terms of the sides of the triangle, and simplifying,
we have
sin i (a 1 ff) = sin (s ~ 6) + sin (* ~ a^ A sln S Sln (.'
sin c ^ sin a si
2 sin c cos c
sin b
*r-
Then sin \ (a + p) = cos i (a -6) . CQs ^ ^ (1)
cos^c
Similarly sin £ (a - p) = sin^g~^ . cos i Y, (2)
sin^c
and cos J- (a + p) = . 8in J Y, (3)
COS C
and co8i(a-p) = .sin|Y. (4)
147. Napier's analogies express relations between five parts
of a triangle. They are easily obtained from Gauss's formulas.
Dividing (1) by (2), Art. 146,
8in|(a_p) tan|(a-6)
Dividing (3) by (4),
. tan c
cos|(a-p)
148 TRIGONOMETRY
Dividing (4) by (2),
cotjH
Dividing (3) by (1),
148. Formulas collected. The following formulas are suf-
ficient to solve a spherical triangle when any three parts
are given :
c 1nr
sin s sin (s — a) sin (s — a)
—)— = tan
sin !(<* + /?) _ tan^c
sin i (a — ft) tan 1 (a — 6)
cos -« + ) tan ± c
cos i (a — ./8) tan 1 (a + 6)
sin 1 (a + 6) _ cot ^- y
sin i (a— 6) tani(a — ^)
1-6) tanf
sin a _ sin b
sin a sin /?
VII
Formula I is used to determine an angle when the three
sides are given.
Formula II is used to determine a side when the three
angles are given.
Formulas III and IV are used when two angles and the
* These formulas are typical. Other formulas of the same type are
obtained by a cyclic change of letters.
OBLIQUE SPHERICAL TRIANGLE 149
included side are given. Formula III determines ^ (a — b)
and formula IV determines -j- (a -j- b), from which a and b
are obtained. Either formula may also be used to deter-
mine the side c when the other two sides and their opposite
angles are given.
Formulas V and VI are used when two sides and the in-
cluded angle are given. Formula V determines ^ (a — ft)
and formula VI determines J (a + (3), from which a and ft
are obtained. Either formula may also be used to deter-
mine the angle y when the other two angles and their op-
posite sides are given.
Formula VII is used when an angle and the side opposite
are among the given parts.
149. Whenever the formulas I to VI are employed in the
solution of a spherical triangle as indicated above, the
quadrant in which any part terminates may always be deter-
mined by noticing the signs of the functions involved.
But when the law of sines is employed, two values are
found for the required part. This leads to two solutions
unless limited to one solution by the following principles.
150. Theorem. Half the sum of any two angles is in the
same quadrant as half the sum of the sides opposite.
This follows from a consideration of the signs of the
functions involved in
cos \(a + ft) _ tan^- c
cos %(a — ft) tan |(a -f b) '
Since each part is less than 180°, tan 1 c and cos ^(« — ft)
are always positive. Hence cos|(« + /?) an(i tan i (a + 6)
must always have the same sign. Hence i(a + ft) and
|-(a + b) terminate in the same .quadrant.
151. Theorem. A side which differs more from 90° than an-
other side, terminates in the same quadrant as its opposite angle.
150
TRIGONOMETRY
We have from Art. 119
cos a — cos b cos c
cos a
sin b sin c
If a differs more from 90° than 6, cos a is numerically
greater than cos b. Cos a is also greater than cos b cos c,
since cos c is not greater than unity. Hence the numerator
of this fraction has the same sign as cos a. The denominator
being always positive, cos a and cos a have the same sign.
Hence a and a terminate in the same quadrant.
The negative of this theorem is not true.
Thus given, a = 165°, b = 120°, ft = 135°,
a terminates in the second quadrant, since a differs more
from 90° than 6.
152. Theorem. An angle which differs more from 90° than
another angle, terminates in the same quadrant as its opposite
side.
This follows from
sff cosy
sin ft sin y
by considerations similar to those of the previous article.
Thus given, a = 80°, y = 140°, and a = 120°,
c terminates in the second quadrant, since y differs more
from 90° than a.
153. Illustrative examples.
1. Given the three sides, a = 105° 27' 20", 6
c = 96° 53' 10", to find a, ft, and y.
83° 14' 40",
a
105° 27' 20"
b
83° 14' 40"
c
96° 53' 10"
2s
284° 94' 70"
s
142° 47' 35"
s— a
37° 20' 15"
s-b
59° 32 '55"
s — c
45° 54' 25"
Check-: s
142° 47' 35"
tan r = Jsin («-«) sin (s-&)sin Q-c)
* sins
tan \ a =
tanr
tan \ 8 =
sin (s — a)
tanr
sin (s — b)
tan r
sin (s — c)
OBLIQUE SPHERICAL TRIANGLE
151
log sin (s— a)
9.78284-10
log sin (s — 6)
9.93553-10
log sin (s— c)
9.85623-10
colog sin s
0.21846
2 log tan r
9.79306-10
log tan r
9.89653-10
log tan A a
0.11369
log tan A /3
9.96100-10
log tan A 7
0.04030
i«
52° 24' 55"
i£
42° 25' 51"
IT
47° 39' 17"
0E
104° 49' 50"
£
84° 51' 42"
T
95° 18' 34"
Check
^U2/ siiiK«-&)
a + b
a-b
*(«' + »)
*(*-&)
K«-0)
188° 42' 0
22° 12' 40'
94° 21' 0'
11° 6' 20'
9° 59' 4'
log sin £ (a + &)
log tan £(«-£)
colog sin A (a — 6)
log cot £ 7
log tan 1 7
9.99875 - 10
9.24563 - 10
0.71531
9.95969 - 10
0.04031
2. Given two sides and the included angle, a = 29° 18',
b = 37° 30', y = 51° 52', to find a, (3, and c.
*»>«=!!! J(,g+??*°K6-«)
b
37° 30'
a
29° 18'
y
51° 52'
b-a
8° 12'
b + a
66° 48'
4 (ft - «)
4° 6'
i (& + a)
33° 24'
17
25° 56'
log sin |(6 — a)
8.85429
-10
log cot A 7
0.31310
colog sin i(6 + a)
0.25926
log tan £ (£ — cc)
9.42665
- 10
1 03 - a)
14° 57' 14"
log cos A (& _ a)
log cot \ 7
colog cos \ (6 + a)
log tan A (|8 + «)
103 + 0)
a.
9.99889 - 10
0.31310
0.07839
0.39038
67° 51' 8"
14° 57' 14"
82° 48' 22"
52° 53' 54"
log tan \ (b - a)
log sin A (|8 + a)
colog sin A (^3 — «)
log tan A c
c
8.85540 - 10
9.96671 - 10
0.58831
9.41042 - 10
14° 25' 43"
28° 51' 26"
152
TRIGONOMETRY
Check
sin a _ sin 6 _ sin c
sin a sin j8 sin 7
log sin a
log sin a
9.68965
9.90177
-10
-10
log sin 6
log sin /3
9.78445-10
0.99656 - 10
log sin c
log sin 7
9.68361
9.89574
-10
-10
9.78788
-10
9.78789 - 10
9.78787
-10
3. Given two sides, and an angle opposite one of them,
a = 63° 24' 50", b = 17° 36' 40", a = 44° 48' 20", to find ft y,
and c.
sin a
taDic = *'"}(g + ^tanK«-6)
a
63° 24' 50"
log sin J (a -f
&)
9.71445 - 10
a
44° 48' 20"
log tan K« - /3)
9.57083 - 10
b
17° 36' 40"
colog sin \ (a —
ft)
0.62876
a + b
62° 25' 0"
log cot \ 7
9.91404 - 10
a-b
27° 11' 40"
i
7
50° 38' 0"
\ (a + 6)
31° 12' 30"
7
101° 16' 0'
£(a _ 5)
13° 35' 50"
log sin a
9.95147 - 10
log sin b
9.48081 - 10
log sin £ (a + £)
9.83375 - 10
colog sin a
0.15200
log tan £ (a —
&)
9.38359 - 10
log sin /3
9.58428 - 10
colog sin \ (a — /3)
0.45735
/3
22° 34' 44" .
log tan ,
c
9.67469 - 10
a — /3
40° 50' 6"
•
j C
25° 18' 20"
« + /3
85° 59' 34"
c
50° 36' 4(X
!(*-/»)
20° 25' 3"
i (a + /3)
42° 59' 47"
Cftecfc
*
sin ft _ sin c
sin /3 sin 7
log sin 6 9.48081 -
10 log sin c
9.88810 - 10
log sin £ 9.58428-
10 log sin 7
9.99155 - 10
9.89653 -
10
9.89655 - 10
OBLIQUE SPHERICAL TRIANGLE
153
154. Two solutions. There are two solutions, if any, when-
ever two sides and an angle opposite one of them, or two an-
gles and a side opposite one of them, are given, unless limited
to one solution by the principles of Arts. 150, 151, and 152.
Thus, having given £ = 45° 15' 12", b = 56° 49' 46", a =
68° 52' 48", to find a, y, and c.
sin 6
a
b
68° 52' 48"
56° 49' 46"
45° 15' 12"
a
a + P
a-p
ci + b
a-b
fo-&)
52° 19' 33"
97° 34' 45"
7° 4' 21"
48° 47' 22"
3° 32' 10"
125° 42' 34"
12° 3' 2"
62° 51' 17"
6° 1'31"
127° 40' 27"
172° 55' 39"
82° 25' 15"
86° 27' 50"
41° 12' 38"
log sin p
log sin a
colog sin b
. log sin a
a
9.85140
9.96980
0.07725
9.89845
52° 19' 33", or
127° 40' 27"
log sin £(« + £)
log tan $ (a - 6)
colog sin & (a — /9)
log tan £ c
I«
c
9.87639
9.02346
1.20984
9.99917
9.02346
0.18124
0.10969
52° 9' 35"
104° 19' 10''
9.20387
9° 5' 7"
18° 10' 14"
log sin £ (a + 6)
logtan£O-£)
colog sin } (a — ft)
log cot 1 7
i7
7
9.94932
8.79099
0.97895
9.94932
9.94238
0.97895
9.71926
62°21'1"
124° 42' 2"
0.87065
7° 40' 16"
15° 20' 32"
154
TRIGONOMETRY
Check
cosJ(a-/8) _ ;
logcos|(a + j3)
log tan %(a + 6)
cologcos£(ot — /3)
log tan \ c
9.81877
0.29012
0.00083
8.79013
0.29012
0.12361
0.10972
•&) tan^r
9.20386
a 4- B)
cos J(a - b)
FIRST SOLUTION
« = 52° 19' 33"
c = 104° 19' 10"
= 124°42' 2"
SECOND SOLUTION
a = 127° 40' 27"
c = 18° 10' 14"
y = 15° 20' 32"
155. Area of spherical triangle. Representing the area of
a sphere, S, by 720 spherical degrees, it is demonstrated in
geometry that the area of a spherical triangle, A, in terms
of spherical degrees, is equal to its spherical excess ; or
>f=(a + p+y — 180) spherical degrees
-'
log cos £( a + 6)
log tan £(« + £)
cologcos £(o — &)
log cot £ 7
9.65920
0.05762
0.00241
9.65920
0.20904
0.00241
9.71923
0.87065
Let A' and S' represent the area of the triangle and the
sphere respectively, in terms of the unit in which r is ex-
pressed. Since the ratio of the area of the spherical tri-
angle to the area of the sphere is independent of the units
used, we have A' __A
Therefore
But the area of the sphere expressed in terms of r is
4 ?rr2, therefore the area of the triangle is given by
A' =
180
156.
EXAMPLES
Solve the following spherical triangles and check the
results :
OBLIQUE SPHERICAL TRIANGLE
155
1.
6= 10° O'lO"
c = 114° 40' 40"
a= 92° 28' 20"
11. «= 61° 8'
J3= 59° 12'
y= 78° 25'
2.
b= 85° 4' 19"
c = 139° 58' 25"
«= 12° 20' 31"
12. a= 76° 43' 15"
b= 83° 35' 27"
c= 98° 26' 38"
3.
c= 82° 3' 4"
a= 70° 14' 12"
0 = 84° 20' 9"
13. « = 110°35'
^8 = 135° 42'
y = 146° 8'
4. a=, 95° 3' 30"
b = 128° 38' 50"
y = 170° 52' 20"
5. o= 29° 18'
6= 37° 30'
y = 51° 52'
6. a= 11° 21' 10"
y = 19° 0'20"
6= 66° 19' 30"
7. y = 179° 22' 11"
a = 148° 17' 17"
6= 25° 39' 34"
8. a = 108° 5' 18"
6 = 170° 30' 46"
c= 85° 50' 22"
9. a = 105° 27' 20"
6= 83° 14' 40"
c= 96° 53' 10"
10. a= 34° 19' 30"
6= 28° 37' 10"
c= 22° 44' 40"
14. y= 11° 34' 10"
6= 82° 56' 30"
c= 27° 9' 40"
15. a= 32° 4' 10"
/? = 128°56'20"
a= 39° 50' 30"
16. /?= 80° 40' 2"
c= 75° 54' 0"
6 = 100° 21' 28"
17. a= 21° I'lO"
y= 17° 22' 50"
a= 14° 13 '30"
18. )8= 77° 44' 55"
y= 92° 17 '24"
a= 26° 29' 39"
19. a = 114° 23' 9"
0= 88° 41' 11"
y= 79° 0' 4"
20. 0= 99° 4' 12"
y = 106° 0' 9"
a = 161° 2' 10"
156 TRIGONOMETRY
21. c = 100° 10' 40" 23. b= 42° 15' 20"
a= 65° 20' 30" c = 127° 3' 30"
y= 94° 30' 10" (3= 31° 44' 20"
22. a = 103° 19' 50" 24. y = 127° 4' 10"
/?= 92° 37' 30" a= 88° 12' 0"
y = 128° 54' 20" c = 141° 20' 30"
SOLUTION WHEN ONLY ONE PART IS REQUIRED
157. In many problems of astronomy and geodesy, it is re-
quired to find only one or two of the unknown parts of a spher-
ical triangle, the other unknown parts being of no importance
in the problem. It then becomes desirable to have a method
whereby the required parts can be computed without being
under the necessity of first computing any part not desired.
It has already been shown that any angle can be found
directly from three given sides (Art. 144), and that any
side can be found directly from three given angles (Art. 145).
It is evident that any part can be found from any three
given parts by the use of the general formulas containing
the required part and the three given parts. By the intro-
duction of auxiliary quantities, these formulas will now be
adapted to logarithmic computation.
158. Given two sides and the included angle, to find any
one of the remaining parts.
Let a, b, y be the given parts.
First. To find c.
The relation between a, 6, y, and c is (Art. 119),
cos c = cos a cos 6 -f sin a sin b cos y. (1)
To adapt this formula to logarithmic computation, let
m sin M= sin b cos y, (2)
and m cos M = cos b. (3)
OBLIQUE SPHERICAL TRIANGLE
157
Then eliminating b by uniting (1), (2), and (3), we have
cos c = m (cos a cos M+ sin a sin M ),
or cos c = m cos (a — M ).
From (2) and (3)
tan Jf = tan 6 cos y,
and from (3) and (4)
cos b cos (a — M")
cos c = »— = £t
cos
(4)
(5)
(6)
Equations (5) and (6) enable us to find c.
ILLUSTRATION. Given o= 75° 38' 20", & = 54° 54' 38", and
= 30° 17' 43".
tan M = tan & cos 7
a
75° 38' 20"
6
54° 54' 38"
7
30° 17' 43"
log tan 6
log cos 7
log tan Jf
M
0.15333
9.93623
0.08956
50° 51' 58"
a- M
24° 46' 22"
cos c = cos 5 cos (a -Jf)
cos Jf
log cos b
log cos (a — M)
colog cos M
log cos c
c
9.75956
9.95808
0.19987
9.91751
34° 12' 27"
Second. To find ft.
The relation between a, b, y, and ft is given by Art. 124,
(equation 5), from which we have
, Q _ cot b sin a — cos a cos y x^
siny
Multiplying numerator and denominator by sin b,
, ~_ cos b sin a — sin b cos a cos y
sin b siiiy
Again using equations (2) and (3) we have
m (cos M sin a — sin Jf cos a)
sin 6 sin y
(8)
, 0
cot « =
or
sin b sm y
(9)
(10)
158 TRIGONOMETRY
As before tan M = tan 6 cos y. (11)
From (2) and (10)
. (12)
.
sin M
Equations (11) and (12) enable us to find ft.
Third. To find y.
By interchanging a and 6 and consequently a and ft in
(11) and (12) we have, calling the auxiliary angle N9
tan JV= tan a cos y (13)
and cota = ™tysin(6-^) (14)
sm^"
to determine a.
159. Two parts required. It will be noticed that the same
auxiliary quantity M is used to find both c and ft. We thus
have a convenient method, much used in astronomy, for
finding a side and an angle when two sides and the included
angle are given.
For finding c and ft we have, collecting our formulas,
tan M = tan 6 cos y
cos M
cot/?=cotysin(q-Jf)t
sin M
Dividing equation (10) of Art. 158 by equation (4), we have
cot/?_tan (a — Jf)
cos c sin b sin y '
which serves as a check upon c and ft.
Similarly, for finding c and a, we have
OBLIQUE SPHERICAL TRIANGLE
tan N= tan a cos y
cos a cos (b — N}
cos c = 5s L
cos N
159
sin N
160.
PROBLEMS
An arc of lr on the earth's surface is equal to one English
geographical mile.
1. Find the distance between Boston, latitude 42° 21' N.,
longitude 71° 41' W., and San Francisco, latitude 37° 48' N.
and longitude 122° 28' W. P
SOLUTION. — Let APD be the me-
ridian of Greenwich from which longi-
tude is measured, ABCD the equator,
and P the north pole.
Let the positions of San Francisco
and Boston be represented by E and F
respectively. The desired distance is
EF.
Then
But
Letting
angle DPE = 122° 28',
angle DPF = 71° 41',
arc EB= 37° 48',
arc CF= 42° 21',
arc PB = arc PC = 90°.
PE=PB-EB = 52° 12',
PF = PC - FC = 47° 39',
Z FPE = /. DPE - Z DPF = 50° 47'.
Z FPE = 7, PE = a, PF= &,
we may find EF or c by the formulas
tan M = tan 6 cos 7, cos c = cos&cos(«
cos M
-M)
160
TRIGONOMETRY
_ cos b cos (a — M )
cos M
a
b
y
52° 12'
47° 39'
50° 47'
log cos b
log cos (a — M )
colog cos M
log cos c
c
9.82844 - 10
9.97951 - 10
0.08526
log tan b
log cos y
log tan M
M
a- M
0.04023
9.80089 - 10
9.89321 - 10
38° 33' 20", or
2313£ miles
9.84112 - 10
34° 44' 23"
17° 27' 37''
2. Find the distance between New York (lat. 40° 43' K,
long. 74° 0' W.) and San Francisco (lat. 37° 48' K, long.
122° 28' W.).
3. Find the distance between Calcutta (lat. 22° 33f N.,
long. 88° 19' E.) and Greenwich (lat. 51° 29' N.).
4. Find the distance between Baltimore (lat. 39° 17' N.,
long. 76° 37' W.) and Calcutta.
161. Given two angles and the included side, to find any
one of the rema ining parts.
Let a, ft c be the given parts.
First. To find y.
The relation between a, ft c, and y is, Art. 121, eq. (3),
cos y = — cos a cos ft + sin a sin ft cos c. - (1)
Let ra sin M = cos a, (2)
and m cos M = sin a cos c. (3)
Uniting (1), (2), and (3)
cos y = m (— sin M cos /? -f cos M sin /?),
or cos y = m sin (ft — M). (4)
From (2) and (3)
cot M = tan a cos c, (5)
OBLIQUE SPHERICAL TRIANGLE 161
and from (2) and (4)
cos a sin (8 —
sinM
Equations (5) and (6) enable us to find y.
(6)
Second. To find a.
The relation between a} (3, c, and a is given by Art. 124,
equation (4), from which
sin 8 cot a + cos c cos 8 ^
cot a = •— . ( 1 1
sine
, _ sin ft cos a + sin a cos c cos ft t ,^
sin a sin c
Uniting equations (2), (3), and (8),
cot a = — ^ " "" — ^— — ' , (9)
sin a sin c
sin a sin c
From (2), (3), and (10)
cot M = tan a cos c (11)
cot c cos (8 — Jf) /., 0\
and cota = ^ L, (12)
cos Jf
from which a is found.
To find b.
From (11) and (12) by interchanging a and /?, and conse-
quently a and 6, we have, calling the auxiliary quantity JV,
cot N= tan £ cos c (13)
and cot6 = - (14)
cos N
to determine &.
162. Given two sides and an angle opposite one of them, to
find any one of the remaining parts.
Let a, bj a be the given parts.
162 TRIGONOMETRY
First. To find c.
The relation between a, b} a and c is
cos a = cos b cos c + sin 6 sin c cos a. (1)
Let m sin Jf = sin b cos a, (2)
and m cos Jf = cos 6. (3)
Then cos a = m cos (c — M ). (4)
From equations (2), (3), and (4),
tan M = tan b cos a (5)
and cos (c - M) = Cos a cos ^ (6)
cos b
Equation (5) determines M and equation (6) determines
c — M. Adding these values, we have c.
In general there are 2 solutions for c. We may limit M
to positive values less than 180°. By equation (6) c — M
may have two values, numerically equal but opposite in
sign, giving two values for c unless the sum M -\- (c — M )
is greater than 180° or negative, in which case there is but
one solution.
Second. To find y.
The relation between a, 6, a, and y is
sin y cot a = cot a sin b — cos b cos y.
Multiplying by sin a and rearranging, we have
sin y cos a + cos y sin « cos b = sin a cot a sin 6. (7)
Let n cos JV= cos a, (8)
and n sin ^= sin a cos 6. (9)
From (7), (8), and (9) we have
n sin (y -f- JV) = sin a cot a sin 6. (10)
Then from (8), (9), and (10),
tan N== tan a cos 6, (11)
and sin (y -f JV) = sin N cot a tan 6. (12)
Equation (12) determines y + N and equation (11) deter-
mines H. Subtracting the second value from the first gives y.
OBLIQUE SPHERICAL TRIANGLE 163
In general there are two solutions, since y + N may have
two values.
Third. To find ft.
The angle ft is found from
. 0 sin 6 sin a ,* o\
sm ft = : , (lo)
sin a
which in general gives two values.
163. Given two angles and a side opposite one of them, to
find any one of the remaining parts.
Let a, ft, a be the given parts.
First. To find y.
The relation between a, ft, a, and y is
cos a = — cos ft cos y + sin ft sin y cos a. (1)
Let m sin M = cos ft, (2)
and m cos M = sin ft cos a. (3)
Then cos a = msin(y — M). (4)
From (2), (3), and (4)
cot M = tan ft cos a, (5)
-, . / •, f^ cos a sin J/" //»>.
and sm(y — M)=— — (6)
cosp
Equations (5) and (6) determine M and y — M, from
which y is found.
Two solutions may be possible, as in Art. 162.
Second. To find c.
The relation between a, ft, a, and c is
sin ft cot a = cot a sin c — cos c cos /?. (7)
Multiplying by sin a and transposing,
cos a sin c — sin a cos c cos /3 = sin /? sin a cot a. (8)
Let n cos JV= cos a, (9)
and w sin JV= sin a cos ft. (10)
164 TRIGONOMETRY
Then uniting (8), (9), and (10)
n sin (c — JV) = sin ft sin a cot a. (11)
From (9), (10), and (11)
tan N= tan a cos /?, (12)
and sin (c — N) = tan ft cot a sin N. (13)
Equations (12) and (13) determine N and c — JV, from
which c is found.
There may be two solutions.
Third. To find b.
We have sin b = sin ? sin *• (14)
Bin«
There are two values of b unless restricted to one solution
by the principles of Arts. 150, 151, and 152.
164. The general triangle. The parts of the general spheri-
cal triangle are not restricted to values less than 180°. It
can be shown that all the formulas developed for the oblique
spherical triangle are true for the general spherical triangle
if the double sign is introduced in the formulas of Arts.
144, 145, and 146.
ANSWERS
Art. 19 ; Page 10
5. 114° 35' 28", 286° 28' 40", etc.
6. 60°, 135°, - 300°, 57° 17' 44", 36° 28' 31", etc.
7. 7ift.
8. 2f radians, 137° .30' 34". 12. 247.16 E. P. M.
25.882.
9. 94° 3' 24". 13. 18.33 mi. per sec.
10. ifTrft. 14. 5236.
11. 2.7216 radians. 15. 9.6 TT
240 it ft. per min.
Art. 27 ; Page 20
1. 3d and 4th. 2. 1st and 4th. 3. 1st and 3d
7. 2d. 8. 3d. 20. -?^. 21. -^.
o o
28. sin «r = -g% V85, cos ^ = — ^ V85, cot a1 = — ^)
sec «! = — ^ V85, esc 06! = ^- V85,
sin «2 = — -^- V85, cos «2 = -g7- A/ 85, cot 02 = — -J-,
sec «2 = -J- V85, esc c^ = — 1 V85.
35. if
Art. 36 ; Page 29
1. 6 = 16.5 3. c-=.869 5. a = 27°
c = 17.5 a = .225 ft = 63°
ft = 70°. a = 15°. b = 72.6
2. a = 1.44 4. c = 65 6. a = 18£°
b = 2.05 a = 23° ft =± 71|°
0 = 55°. 0 = 67° & = . 00867
165
166
TRIGONOMETRY
7. a = 346 8. a = 27° 9. a = .029 10. a
c = 507 ft = 63° b = .089 a = 8.11
0 = 47° a = 3.50 0=72° 6 = 9.48
11. 6 = 5161
c = 5489
£=70° 5'
14. c=. 00006294
a = 72° 26'
= 17° 34'
12. a = .1384
6 = .2878
ft = 64° 19'
13. a = 1.446
c = 1.719
a = 57° 17'
15. b = 810.80 16. 17. etc.
a = 47° 31' 32" check your
= 42° 28' 28" results.
31. Base 1331.1, vertical angle 149* 19' 10".
32. Base angles 39° 23' 56", base 1477.0.
33. Equal sides 1622.9, base angles 37° 59' 37".
34. Equal sides 219.75, base angles 68° 27' 19".
35. 37.504.
Art. 38; Page 31
1.
290.83
ft.
7.
2nrtai
i ••
12.
54.775
mi.
2.
405.24
ft.
n
13.
153.72
Ibs.
3.
263.92
ft.
8.
130.99
ft.
38° 31'
46".
4.
289.93
21.442
ft.
ft.
9.
10.
226.11
2572.5
ft.
ft.
14.
739.38
hi.
mi.
per
5.
2 nr sii
180°
11.
21.360
22.638
in.
in.
15.
16 ft. £
l&
o Z
in.
n
6.
132.52
ft.
69° 26'
36".
16.
35° 16'.
Art. 50; Page 46
2. - cos 10°, - sin 80°. 5. cos 20°, sin 70°.
3. - cot 20°, - tan 70°. 6. - tan 80°, - cot 10°.
4. -.cot 20°, - tan 70°. 7. - sin 60°, - cos 30°.
8. cos B. 9. -tan B. 10. — tan0. 11. -cos
ANSWERS
16T
11.
12.
15.
18.
20.
21.
35.
37.
33.
39.
41.
Art. 56 ; Page 51
_, tan 0 = - £ V5, cot0 = -£V5,
sec 0 = — | V5, esc 0 = f .
sin 0 = ^VUfl, cos 0 = T|^ V149, tan 0 = J^,
sec 0 = I V149, esc 0 = TV VI49.
z = 30°. 16. « = 45°. 17. z=0°, 60°.
0 = 45°, 60°. 19. sina = ±^-i. + £V5.
Identity. 22. 0 = 120°. 24. Identity.
x = 60°, 120°. 23. 0 = 30°. 25. y = 30°, 150.
a = 30°, 150°. 36. a = 30°, 60°, 120°, 150°.
x = 45°, 135°. 38. x = 45°.
Art. 60; Page 57
tan2a-l. 34. sinzTVl-sin2*. 35. 1 + COSX
tan2# + l 1— cos2 a;
' ±vrro2'
_i_->/1 __ /«S 1 1
-, sec0
a
± Vl - a2
COS0
cot0
CSC0 =
, sec0
COS0'
43.
50.
55.
58.
59.
60.
?• 47. 0°, 90°. 48. 135°. 49. 0°, 30°, 60°, 150°.
2
30°, 150°. 51. 45°. 54. 36° 52'. (See tables.)
46° 24', 90°. 57. 65° 54', 114° 6'.
0°, _ 30°, - 150°.
tanu=±|V5, ±iV2. 61. 195°, 345°.
j, ±|V2. 62. 54°, 234°.
168
TRIGONOMETRY
Art. 70 ; Page 66
2. ^V2(V3 - 1). 3. iV2( V3 - 1). 4.
3+V3
1).
5.
1.
2.
3.
32.
34.
35.
1.
3.
5.
7.
10.
13.
14.
20.
3-V3
7. cos ft. 8. — cot a. 9. — cos a.
Art. 75 ; Page 71
* V2 - V2, fV2 + V2, V3 - 2V2, V3 + 2 V2.
± |V15, - |, ± | V15, ±
20
/O O£ /?^O O/W H fr
.raj;;- 400 65!;2
30°, 90°, 150°. 42. 0°, 30°, 90°, 150°.
0°, 120°. 43. 0°, 90°, 120°.
Art. 85 ; Page 82
60°, 120°, 420°, etc. 2. 150°, 210°, - 150°, etc.
45°, 225°, - 135°, etc. 4. A.
6.
u
± Vl 4- %2
± 2 ?^ Vl -
± Vw2 + 1
9. ±vr^:
u. i
u
15.
± 2 Vw2 - 1
— v*±v^/l — u2. 21. ± Vl— u2Vl— v2— uv.
ANSWERS
169
- V 2
' V 2u
2Q SoT TTTTOV T —
1 1
30 z ±a -
cot
31. 1
1 34. «.:
V
cot"1 a a
0 rt2
VI + a2
32 VT=rf-aV
.4 a .
\Xl-62_j_5Vl-a2
35. -i (1 + Va2— 1
a&
V62-l).
Art. 101 ; Page 98
1. 6 = 24.5
c = 31.8
y=78°.
2. a
&
c
= 50°
= 54.9
= 58.6
3. a = 61.4
c = 47.7
y = 48°.
4. /?
y
c
= 48° 20'
= 62° 40'
= 76.1
5. 0 = 68° 22'
y = 56° 58'
a = 107
6. «
c
= 40° 48'
= 104° 52'
= 141.
7. a = 69° 22'
0 = 38° 38'
c = 42.7
8. ft
y
c
= 38° 37'
= 31° 23'
= 173.
9. a = 44° 42'
£ = 60° 20'
y = 74° 54'
10. a
y
= 72° 21' •
= 49° 38'
= 58° V
11. /3 = 34° 44' or
y = 125° 16' or
c = 35.8 or
146° 16' 12. a
13° 44' ft
10.4 6
= 51° 44' or 128° 16'
= 84° 26' or 7° 54'
= .431 or .059
170 TRIGONOMETRY
Art. 101 ; Page 98
13. y = 39° 49' 50" 14. a = 13.081
a = 41.581 c = 13.620
c = 41.432 /? = 97°2'15"
15. a = 90° 20' 34" 16. a = 126° 59' 18"
ft = 65° 17' 34" ft = 27° 29' 38"
y= 24° 21' 50" y = 25°31'4"
17. a = 1.9555 18. y = 77° 22' 16"
a = 43° 36' 35" ft = 51° 57' 20"
ft = 15° 37' 7" a = 83.732
19. a = 87° 33' 58" 20. a = 78° 40' 32"
y = 18° 6' 24" ft = 41° 20' 47"
b = 1.4033 y = 59° 58' 41'^
21. a = 32° 24' 0" 22. Impossible.
ft = 55° 0' 28"
y = 92° 35' 32"
23. a = 142° 28' 9" or 21° 31' 11"
ft = 29° 31' 31" or 150° 28' 29"
a = .080746 or .04862
24. y = 76° 50' 20" or 103° 9' 40"
a =69° 39' 35" or 43° 20' 15"
a = 17.405 or 12.739
25. c = 502.28 26. a = 96° 9' 32"
b = 300.25 . ft = 41° 11' 10"
y = 23° 7' 3" y = 42° 39' 18"
27. .4 = 150. 30. ^ = 108.61
33. A = 368.91. 35. 172.8 ft.
36. 106.1ft. 37. 3.710 mi.
38. 97.14
124.59 39. 60.1ft.
178.64
40. 57.93 mi. per Lr. -41. 3888.0ft.
42. 6328.7 ft. 43. 239600 mi.
ANSWERS
171
Art. 102 ; Page 102
1.
4. 42°58/18//. 5. 35° 48' 35". 6. 13.754 in.
7. £. 8.
2. 60',72M35'. 3. , , ,
7T
9*
7T 7T 7T
6' 3' 2
55 7T 73 7T
144 144
9.
4 7T
9 ' 3' 9
12.
50.
51.
54.
56.
57.
59.
60.
61.
62.
63.
64.
66.
67.
68.
69.
73.
11 tE 10?r 16£ 22 TT 28 TT 34 TT 40 *•
' 77 ' 77 ' 77 ' 77 ' 77 ' 77 ' 77 '
9fin. 49. 0°, 180°.
30°, 150°, 210°, 330°, 45°, 135°, 225°, 315°.
30°, 150°. 52. tan-\(2 ± V3). 53. 0°, 180°, cos'1 f
45°, 135°, 225°, 215°, sin'1 Vf 55. 60°, 300°, 90°, 270°.
60°, 120°, 240°, 300°, 45°, 135°, 225°, 315°.
tan-1!. 58. 90°, 270°, sin-1 (-|i).
0°, 45°, 90°, 135°, 180°, 225a, 270°, 315°.
0°, 45°, 60°, 90°, 120°, 180°, 225°, 240°, 270°, 300°, 315°.
7i°, 37i°, 67$.°, 97i°, 127^°, 1571°, etc.
30°, 150°, 210°, 330°.
90°, 270°, 70°, 110°, 190°, 230°, 310°, 350°.
90°, 180°. 65. 210°, 330°.
45°, 215°, 671°, 1571°, 247f, 337^°.
60°, 120°, 240°, 300°, 18°, 54°, 90°, 126°, 162°, etc.
60°, 90°, 120°, 240°, 270°, 300°.
70. 0, ±V|. 71.
172
TRIGONOMETRY
99. 532. 100. 50.0. 101. 1478.5.
102. 93,470,000 mi. 103. 51.9 ft. 104. 27.925.
105. 112.36. 106. -• 107. 129.90 ft.
106. f
108. 8.2596. 109. 70° 31' 43".
111. 196. 112. 89.431.
115. 90 ft., 40 ft. 116. 13.66.
118. 820.54. 119. 535.4.
121. 567.3. 122. 132.6.
124. 641. 125. 962.605.
128. a. 129. 4009.
110. 5296 ft., 251 ft.
114. «V2.
117. 103.97.
120. 25.43.
123. 36.7 ft.
127. 16.33 N. 75° 36' E.
130. 1674.3.
Art. 116 ; Page 124
1. 1, i, — 1, — i.
2.1, i
4. ±(2 + 0-
6. ± (1.272 + .786 1).
7.
8. _
10. 1.0842 + .29051 i,
- .79370 + .79370 1,
- .29051 - 1.0842 1.
5. ±(l-2t).
Art. 139 ; Page 143
1. /3 = 78°9'22"
c = 10° 45' 55"
a = 2° 14' 5"
3. 6 = 8° 26' 14"
a = 120° 59' 19"
a = 95° 2' 10"
2? a= 41° 11' 53"
£ = 56° 19' 56"
c = 40° 27' 11"
4. ^ = 75° 21' 53"
6 = 44° 43' 49"
a = 14° 59' 33"
ANSWERS 173
5. a = 111° 23' 47'.' 6. ft =101° 38' 28"
£ = 120° 40' 56" a = 112° 13'. 48"
c = 76° 33' 24" b = 102° 35' 26"
7. £ = 46°r28" 8. a = 68° 42' 11"
b = 15° 18' 0" £ = 155°48'0"
a = 46° 2' 40" a = 27° 37' 26"
9. p = 153° 31' 29" or 26° 28' 31"
c = 50° 43' 22" or 129° 16' 38"
b = 159° 48' 44" or 20° 11' 16"
Art. 142; Page 144
1. a = 117° 45' 28"
2. y = 88° 23' 11"
0 = 96°27'1"
a = 69° 48' 42"
y = 93°0'61"
b = 94° 22' 46"
3. a = 8° 49' 46"
4. a =160° 13' 48"
y = 28°3'4"
0 = 105° 21' 16"
b = 10.6° 56' 53"
y = 104° 25' 45"
5. Sides, 32° 45' 6"
6. Sides, 112° 32' 20"
angles, 105° 49' 32"
base, 46° 15' 12"
Art.
156 ; Page 154
1. 0 = 11°0'47"
2. 0 = 14° 53' 47"
y=92°8'27"
y = 170° 26' 51"
a = 114° 42' 50"
a = 55° 56' 0"
3. « = 71°r23"
6. a = 24° 35' 10"
y = 84° 22' 25"
c = 43° 29' 48"
6 = 82°1'30"
0 = 154° 19' 20"
8. a = 133° 28' 34"
9. a = 104° 49' 50"
0 = 169° 38' 12"
0 = 84° 51' 42"
y = 132° 6' 14"
y = 95° 18' 24"
10.
a = 84° 57' 8"
0 = 57° 47' 44"
v = 43°4'36"
174 TRIGONOMETRY
14. a = 155° 14' 24" or 21° 52' 40"
£ = 25° 50' 58" or 154° 9' 2"
a = 107° 34' 50" or 58° 0' 44"
15. Two solutions. 24. No solution.
Art. 160 ; Page 159
2. 2229| miles. 3. 4291J miles.
^ -.. 1
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14 DAY USE
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