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., OSMAJJIA VNIVERSITY LIBRARY
Gall No. Sfllf* *tf{) 77/^?cession No. / ^
Title
This book should be retrned on or before tKe date-
last marked below.
PLANE
TRIGONOME'ln I
BY
ARNOLD DRSSDEN, PH.D.
Assistant Professor of Mathematics in the University
of Wisconsin
NEW YORK
JOHN WILEY & SONS, INC.
CHAPMAN & HALL, LIMITED
1921
COPYRIGHT, 1921
BY
ARNOLD DRESDEN
TECHNICAL COMPOSITION CO.
CAMBRIDGE], MASS., IT. 8. A.
PREFACE
WHILE the importance of the function concept for elementary
mathematics has become recognized by many writers of college
algebra texts and of " unified freshman mathematics " books, it
has received little recognition from writers on elementary trigo-
nometry. To emphasize this importance has been the leading
motive in writing the present book. A somewhat detailed study of
the graphs of the trigonometric functions (Chapter V) and of
the inverse functions (Chapter VIII) has been introduced for
this purpose. Much more could and should be done in this di-
rection; perhaps the present effort may suffice as a first step.
The opportunity afforded by the writing of a new text has been
used to make some changes in the presentation of the traditional
material. Circular measurement of angles is introduced in the
first chapter so as to be available for use throughout the course.
The fundamental theorems on projections are presented early and
are used subsequently so that the student may be familiar with
them when they are applied in a general proof of the addition
theorems, based on a method quite generally followed by conti-
nental writers. Recognizing the value of the " solution of tri-
angles/' a good deal of space has been devoted to this subject,
and an attempt has been made to develop it in such a manner
that the student can appreciate the reasons for the different
methods that are discussed.
On the question of " applied problems/' I have taken a definite
position. I do not think it feasible to introduce into an ele-
mentary text technical material from the applied sciences, impor-
tant though such material may be. Without such material, how-
ever, applications cannot well be anything ^but problems which
use the language of the applied sciences without really belonging
to them. An elementary text can render useful service, even to
applied science, by stressing the fundamental concepts of trigonom-
etry and by setting problems which connect with the student's
actual experience and which suggest ways in which these concepts.
iii
iv PREFACE
may be applied, leaving actual applications to the fields to which
they belong.
It has not seemed desirable to add to the number of tables of
logarithms already available. The elementary treatment of log-
arithms in Chapter III and the problems scattered throughout
the book call for the use of a set of five-place tables, of which
there are many excellent ones in existence.
No attempt at logical completeness has been made, but rather
has it been my aim to adapt the treatment to the stage of logical
development which may be expected of students who begin the
study of trigonometry. I am aware of the fact that a fuller dis-
cussion might be made in several instances and I shall be happy
if the treatment as given should a ouse the critical powers of some
students and develop in them a desire for more penetrating analysis.
The material as here presented was used originally in mimeo-
graphed form by a few classes in the University of Wisconsin.
I am under a debt of gratitude to the Department of Mathematics
for allowing the material to be thus tried out. And it is with
special pleasure that I recognize my indebtedness to colleagues in
that department and to Professor T. M. Simpson, now of the
University of Florida, to some for suggestions and criticisms, to
others for assistance in the reading of the proofs. If I add to this
my appreciation of the courtesy shown by the publishers of the
book, I am ready to rest my case with the jury consisting of the
teachers and students of trigonometry.
ARNOLD DRESDEN
UNIVERSITY OF WISCONSIN
March. 1921
CONTENTS
CHAPTER I
POSITIVE AND NEGATIVE LINES AND ANGLES. COORDINATES.
RADIAN MEASUREMENT
Art. Page
1. Directed magnitudes 1
2. Points on a line. Directed Jines 1
3, 4. Points in a plane 2
5. A fundamental theorem 3
6, 7. Projections of line segments 4
8, 9. Directed angles 5
10, 11. Radian measure 6
CHAPTER II
THE TRIGONOMETRIC RATIOS. SIMPLE IDENTITIES
12, 13. Standard position 9
14-17. Trigonometric ratios 9
18, 19. Reduction to the first quadrant 11
20. Ratios of acute angles * 13
21, 22. Ratios of 30, 60, and 45 13
23. Ratios of 90, 180, 270, and 360 15
24, 25. Trigonometric functions 16
26. Periodicity 17
27, 28. Relations between the trigonometric functions 18
CHAPTER III
LOGARITHMS
29. Theory of exponents 20
30. The use of exponents in calculation 21
31, 32. Definition of logarithms 21
33, 34. Fundamental theorems on logarithms 22
35. Common logarithms 24
36, 37. Use of a table of logarithms 26
38, 39. Calculation by means of logarithms 28
v
i CONTENTS
CHAPTER IV
SOLUTION OF RIGHT TRIANGLES. APPLICATIONS
Art. Page
40, 41. The right triangle 32
42. Accuracy of the calculation. Checking the results 33
43, 44. Isosceles triangles 34
45-47. Projection 36
48, 49. Applications 38
CHAPTER V
THE GRAPHS OF THE TRIGONOMETRIC FUNCTIONS
50. Graphs of sin and cos 41
51, 52. Examples of graphs 42
53, 54. Operations on graphs 45
55, 56. Applications 47
57, 58. Graphs of tan 6 and cot 48
59. Graphs of tan (ad + b) and cot (a0 + b) 50
60, 61. Applications 51
62, 63. Graphs of sec and cosec 52
CHAPTER VI
THE ADDITION FORMULAE
64. A special case 55
65-67. Addition and subtraction formulae for the sine and the cosine . 55
68, 69. Addition formulae for the tangent and the cotangent 58
70, 71. Double angle and half -angle formulae 59
72, 73. Factorization formulae 61
74. Summary 63
75. Exercises on Chapter VI 63
CHAPTER VII
THE SOLUTION OF TRIANGLES
76. The Law of Sines. The area of a triangle 65
77, 78. Two angles and one side 66
79, 80. Two sides and an angle opposite one of them 68
81, 82. The Law of Cosines 73
83. Summary and critique 76
84. The Law of Tangents 76
85, 86. Mollweide's [equations 78
87, 88. Two sides and the included angle 79
89. The half-angle formulae for the angles of a triangle 80
CONTENTS VU
Art. Page
90, 91. Threesides 81
92. Inscribed and circumscribed circles. Area 82
93. Summary 84
94, 95. Exercises on Chapter VII 86
CHAPTER VIII
INVERSE TRIGONOMETRIC FUNCTIONS. TRIGONOMETRIC
EQUATIONS
96. Inverse functions 90
97, 98. The graphs of -a pair of inverse functions 92
99. The inverse sine function 94
100. The other inverse trigonometric functions 95
101. Exercises 97
102. Relations between multiple-valued and single-valued inverse
functions 98
103-106. Trigonometric equations 98
LIST OP ANSWERS TO THE EXERCISES 105
INDEX 109
PLANE TRIGONOMETRY
CHAPTER I
POSITIVE AND NEGATIVE LINES AND ANGLES. CO-
ORDINATES. RADIAN MEASUREMENT
1. Directed magnitudes. The use of positive and negative
numbers is a familiar method of indicating temperatures either
above or below a certain fixed temperature, which is taken as the
zero of the scale. In the same way, it is convenient to use posi-
tive and negative numbers to designate the measures of other
magnitudes, the values of which may fall on either side of a
certain fixed value, taken as a point of departure. For example,
northern latitude may be designated as positive, southern lati-
tude as negative; eastern longitude as positive, western longi-
tude as negative; a credit balance as positive, a debit balance as
negative; altitude above sea level as positive, altitude below sea
level as negative, etc.
2. Points on a line. Directed lines. A simple graphical rep-
resentation of such magnitudes is obtained by means of a straight
line, called the axis, upon which are indicated a fixed point,
called the origin, a unit distance and a positive direction.
O (Origin) 1 Positive Direction
, 1 1 ->-
K Unit H Axis
FIG. 1
For every number, positive or negative, there is a single cor-
responding point, P, on this line. P is the point whose distance
from the origin, 0, measured in terms of the unit distance, is, in
magnitude and direction, equal to the given number. Conversely,
to every point, P, on the line there corresponds a real number,
viz., the measurement of the distance OP in terms of the unit
distance, prefixed by a plus or minus sign. With the conventions
1
2 POINTS IN A PLANE
of Figure 1, positive numbers will correspond to points to the
right of 0, negative numbers to points to the left of 0. The
number which is in this way associated with a point P is called
the coordinate of P.
3. Points in a plane. A simple extension of this method
enables us to designate the position of a point in a plane by means
of a pair of numbers. We take two mutually perpendicular
lines, called the axes of coordinates, one horizontal and one verti-
cal. Their point of intersection, called the origin of coordinates,
is used as origin on each axis, as explained in the preceding section.
Moreover, a unit distance and a positive direction are specified
on each axis. The horizontal axis is called the axis of abscissae
or the X-axis; the vertical line the axis of ordinates or the Y-axis.
The position of an arbitrary point P in the plane is now deter-
mined by the horizontal distance from the F-axis to P, meas-
ured according to the unit and positive direction on the -XT-axis,
and by the vertical distance from the X-axis to P, measured in
accordance with the specifications on the F-axis. In this way
two real numbers are obtained, called respectively, the abscissa
or ^-coordinate (or simply the
TJ Y i "#") and the ordinate or
y-coordinate (or simply the
"y") of the point P. The
position of the point P is in-
dicated by means of these
a " 4 *x two numbers in parentheses,
the x-coordinate being placed
first, and the ^/-coordinate
in
IV second . Thus the points A , B
F 2 and C of Figure 2 are repre-
sented by the symbols (3,5)
(3, 2) and (2, -3) respectively. Conversely, for every pair
of real numbers, such as ( 2,4), a point is uniquely determined
of which these numbers are the x- and ^/-coordinates.
The four parts into which the two axes of coordinates divide
the plane are called the four quadrants; they are numbered as
indicated in Figure 2. We say that A lies in the first quadrant
(in I), B in the third quadrant (in III) and C in the fourth quad-
rant (in IV).
A FUNDAMENTAL THEOREM 3
4. Exercises.
1. Locate the points A (0,0), X(3,S), Y (-4,1), Z (0,3), F(-2,~2),
IP (5, -2), *7(-2,0), T(4,-3).
2. Determine the coordinates of the third vertex of an equilateral triangle
of which the origin and the point (6,0) are the other two vertices.
3. Determine the coordinates of the center of the circle which passes
through the points (0,0), (4,0), and (0,-4).
4. Show, by construction, that the points (0,2), (1,5), (3,11) and (2, 4)
lie on a straight line.
6. Show, by construction, that the points (1,5), (6,4), ( 1, 3) and
(6, 2) lie on the circumference of a circle of which the point (2,1) is the center.
6. Determine the coordinates of the center of the circle which passes
through the points A (5,0), B (0,-5) and C (-5,0).
7. Determine the codrdinates of the points midway between the points
A (2,7) and B (4,3); P (3,-4) and Q (7,4); X (-1,3) and Y (3,-l).
8. Find the distance from the origin of each of the following points:
A (3,-4), B (-5,4), C (-4, -6), D (5,5).
9. If r denote the distance of a point from the origin, determine both
coordinates for each of the following points:
A: r = 5, y = 3; B: r = 13, x = -12; C: r = 3, y = -4; D: r = 2, x = 1.
10. The point P lies on a circle about the origin as a center and with a
radius equal to 10; its ordinate is twice its abscissa, the two coordinates hav-
ing like signs. Determine the coordinates of P.
5. A fundamental theorem.
THEOREM I. If three points A 9 B and C be taken arbitrarily on a
directed straight line, the sum of the three segments AB, BC and
CA is equal to zero.
Proof. Reading from left to right, the three points must lie
in one of the following six orders: ABC' A C B^
* f y B,A',C\ B,C\A\
FIG. 3 C, A, B; C, B, A.
In the third case, illustrated in Figure 3, the segments BA, BC
and AC are positive segments, satisfying the relation BA+AC
- BC, or -BA + BC -AC = 0. But, -BA = AB and -AC
= CA. We conclude therefore
AB + BC + CA = 0.
In all the other cases the proof proceeds in entirely analogous
manner; the details are left to the reader.
PROJECTIONS OF LINE SEGMENTS
COROLLARY. If any number of points be taken arbitrarily on a
directed straight line, the sum of the segments from the first point
to the second point, from the second point to the third point, etc.,
and from the last point to the first point, is equal to zero.
Proof. Let the points be A, B, C, D, E and F. We know
then
AB + EC + CA = 0, CD + DE + EC = 0,
EF + FA + AE = 0, and CA + AE + EC = 0.
If we now add the first three of these equalities and subtract the
last one from their sum, we obtain
AB + EC + CD + DE + EF + FA = 0,
which proves the corollary.
6. Projections of line segments. In elementary geometry)
the projection of the segment AB of a line I upon a line m is de-
fined as the segment A\B\ of the line m determined by the feet
A\ and B\ of the perpendiculars dropped from A and B respec-
tively upon m. We prove now the following important theorem.
THEOREM II. The sum of the projections upon a directed line m of
the directed segments of a closed broken line is equal to zero.
D
FIG. 4
, and
Theorem I,
Proof. Let the brok-
en line be ABCDEFA
(see Fig. 4). The pro-
jections of the directed
^m segments A B, BC, CD,
DE, EF and FA up-
on the directed line
m are the directed
segments AiB ls Bid,
In virtue of the Corollary to
= 0;
hence the theorem is proved.
COROLLARY. The sum of the projections upon a directed line m
of the directed segments of an open broken line ABC . . . XP is equal
to the projection upon m of the directed line AP.
DIRECTED ANGLES 5
Proof. If the projections of A, 5, C, . . . X, P upon m be denoted
by Aij Bij Ci, . . . Xi, Pi, we have to show that
or that AiB t + J3A + + XiPi + Pi-Ai = 0.
But this follows, in virtue of Theorem II, from the fact that
ABC . . . XPA is a closed brokeirline.
7. Exercises.
1. Plot the points A (3,4), B (5,7) and C (7,2). Project the segments AB,
BC and CA upon the X-axis and prove that the sum of the projections is
equal to zero.
2. Proceed in a similar manner with the points P (1, 3), Q (3,2), R (5,4)
andS (7, -4).
3. Determine the length of the projections upon the X- and F-axes of the
broken line (0,0), A (4,0), P (4,4).
4. Prove that the sum of the projections upon the F-axis of the segments
AB, BC and CA of Exercise 1 is equal to zero.
5. Show that the projections upon each of the coordinate axes of the
straight line PS of Exercise 2 is equal to
the sum of the projections of the segments
of the broken line PQRS.
6. The points O (0,0) , R (1,2) and Q (-3,4)
(see Fig. 5) are the vertices of a right tri-
angle. Show that the projection of the
..... >- % hypotenuse OQ upon each of the coordinate
axes is equal to the sum of the projections
of the legs OR and RQ.
7. Plot the points A (-4,3) and B (-2,
7). Show that the projections upon each
p IG g of the axes of the line AB is equal to the
sum of the projections of AO and OB.
8. Plot the points Q (3, 7) and R(l t 3). Show that the projections
upon the coordinate axes of OQ is equal to the sum of the projections of OR
and RQ.
9. Proceed as in 8 for the points Q (4,-4) and R (-2,-3).
8. Directed angles. In the preceding paragraphs we have
enlarged the concept of line segment by attributing to it not only
magnitude, but also direction. We proceed now to extend the
concept of angle in a similar way, viz., by giving an angle sense
as well as magnitude. We distinguish between the two sides of
an angle by calling one the initial side, the other the terminal
Q
w
i
I
I
6 RADIAN MEASURE
side. The angle will then be measured by a rotation which will
bring the initial side into coincidence with the terminal side. If the
rotation is clockwise, the resulting angle is called negative j if the
rotation is counterclockwise, the angle is called positive. The
initial and terminal sides of an angle, as well as the sense of
rotation, are indicated by means of an arrow, as in Figure 6.
-270
The most common unit of measurement for angles is the degree,
which is the 360th part of the angle obtained by one complete
revolution. An angle of 180 (180 degrees) is called a straight
angle, an angle of 90 (90 degrees) is called a right angle. Angles
smaller than a right angle are called acute ; angles greater than a
right angle are called obtuse. Two angles whose sum is equal to
a straight angle are called supplementary angles; two angles
whose sum is equal to a right angle are called complementary
angles.
9. Exercises.
1. Draw angles of -45, 457, -312, 583, 1080, -630.
2. Determine the complements of the following angles: 25, 78, 23,
154, 217, -112, 325, 427, -508. Construct these angles and their com-
plements.
3. Determine the supplements of the following angles: 89, 127, -212%
195, -315, 287, 513, -459.
4. Draw the following angles: 120 + 70, 65 + 145, 180 - 25,
270 - 105, 180 + 55, 270 + 132.
10. Radian measure. The degree, in which the angles pre-
sented so far have been measured, is the unit commonly used in
practical work.* For theoretical purposes, another unit, called
the radian is to be pieferred. A radian is an angle, which sub-
* In France, the "grade," one four-hundredth part of a complete revolu-
tion, is frequently used as a unit of measurement. It has the advantage of
fitting into the system of decimal notation.
RADIAN MEASURE 7
I
tends on the circumference of any circle described with the vertex
of the angle as a center, an arc equal to the radius of that circle.
FIG. 7
Since the circumference of a circle of radius r is equal to 2 TTT,
the radius of any circle will be contained 2 TT times in its cir-
cumference. Hence the following relations hold:
360 = 2 TT radians;
1 = radians = 0.01745 . . . radians;
1 radian = = 57.29578 = 57 IT 44".81 ....
7T
By means of these relations the measurement of an angle can
be converted from radian measurement into degree measure-
ment and vice versa. The radian measurement of an angle is
frequently called the circular measure of the angle. When no
unit is indicated, it is understood that circular measure is meant.
Whenever possible, the circular measure of an angle is expressed
by means of multiples and submultiples of TT.*
The radian measurement of an angle at the center of ,a circle
bears a simple relation to the arc which this angle intercepts on
the circumference of the circle. For every radian in the angle,
we will have on the circumference an arc equal to the radius.
Hence, if the radian measurement of the angle be denoted by
and the radius of the circle by r, we shall have
length of arc = 6 r.
* A submultiple of a number is the quotient of that number by an integer;
c.g., 0/5 is a submultiple of a; ?r/3 is a submultiple of w.
8 EXERCISES
11. Exercises.
1. Determine the circular measure of the following angles: 45, 30,
-225, 330, 540. -150, 60, -270, -810, 210, 480, -650.
2. Determine the number of degrees in each of the following angles:
3 7T 4 7T 2lT 57T 7T
~2~' ~T~' If ' T' ~2~' 1>5 ' 6 '
3. Determine the circular measures of the complements and of the sup-
plements of the following angles: 27r/3, -3*-/4, 75, 77T/6, -112, -5ir/3,
135, 7T/8, 2 7T/9, -325, 7 *-/6, -3 x/2.
4. Construct the folio whig angles: ir/2, 57T/6, 7*-/3, -llir/4, ?r/6,
3 7T/4, -2 7T/3, 2 7T, -11 7T/6, ~3 7T, 2, -5.2.
6. A wheel makes 10 revolutions per second. Determine the circular
measure of the angle described by one of the spokes in 15 seconds.
6. How large an arc will a central angle of 2.5 radians subtend on a
circle whose radius is 4 feet?
7. How large an angle, at the center of a circle whose diameter is 10
feet, will subtend an arc whose length is equal to 4 feet?
8. An angle of 2 radians placed at the center of a circle intercepts an arc
of 4 feet. What is the radius of the circle?
9. If a wheel makes 50 revolutions per second, how large an angle does
its radius describe in 1 minute?
10. A carriage wheel covers a distance of 1 mile in 1000 revolutions. How
large is its radius?
CHAPTER II
THE TRIGONOMETRIC RATIOS. SIMPLE IDENTITIES
12. Standard position. For the purpose of comparing differ-
ent angles we place them in such a way that their vertices coin-
cide with the origin of a system of coordinate axes, and that their
initial sides fall along the positive half of the X-axis of this sys-
tem. Angles so placed are said to be in standard position. Ac-
cording as the terminal side of the angle in standard position
falls in I, II, III or IV, the angle is said to lie in the first, second,
third or fourth quadrant respectively.
13. Exercises.
1. Draw the following angles in standard position: 75, 7r/3, 120, Tr/2,
-165, 5 7T/6, -213, -9 ar/5, 325, 195, -3 TT, 155.
2. Draw, in standard position, the complements of the angles given in
Ex. 1.
3. Draw, in standard position, the supplements of the angles given in
Ex. 1.
14. Trigonometric ratios. On the terminal side of the angle
0, placed in standard position, an arbitrary point P is taken.
The coordinates of P and its distance
OP from O, called the radius vector of
P, are denoted by x, y and r respec-
tively; it is agreed that the direction
from to P shall always be the posi-
tive direction along the terminal side,
so that r is always a positive number.
While now the numbers x, y and r will
be different for different positions of pi G 8
the point P on the same terminal side,
it follows from the properties of similar triangles that any one
of the six ratios which subsist among these three numbers will
be the same for one position of the point P as for any other posi-
tion and will depend upon the position of the terminal side only.
10 SIGNS OF THE RATIOS
These six ratice are called the trigonometric ratios of the angle 0,
and are fundamental in the whole field of mathematics. They
are known by the following names:
- is called the sine of the angle 0, written sin 0,
- is called the cosine of the angle 0, written cos 9
- is called the tangent of the angle 9 written tan 0,
- is called the cotangent of the angle 0, written cot 0,
- is called the secant of the angle 0, written sec 0,
~ is called the cosecant of the angle 0, written cosec 0.
Notice that the two ratios occurring in each of the three groups
are obtained, one from the other, by replacing x by y and y by x,
and that interchanging x and y replaces each ratio by its co-ratio.*
15. Exercises.
1. Draw each of the following angles in standard position, measure the
coordinates and the radius vector of some point on the terminal side of each
and determine the trigonometric ratios: 25, ir/6, 2 *-/3, 145, 7 ir/5, 220,
7 7T/4, 315 3 , 11 */5, 425, 21 7r/8, 700.
2. Proceed as in Exercise 1 with the following angles: 30, *-/4, 130,
-57T/6, -210, -4T/3, -320, -5ir/4, -112, -19/6, -670, -17ir/4.
16. Signs of the ratios. Because the radius vector is always
positive, and because the signs of the coordinates of a point
depend only upon the quadrant in which the point lies, the alge-
braic signs of the trigonometric ratios of an angle are determined
by the quadrant in which the angle lies. The agreements as to the
signs of the coordinates of a point (see 3) lead to the following
table of signs for the trigonometric ratios at the top of p. 11.
We observe, furthermore, that the numerical values of the ratios
of an angle in any quadrant are equal to the ratios of some angle
in the first quadrant. A careful inspection of the diagram will
enable the student to determine, for any given angle, an angle
in the first quadrant whose ratios are numerically equal to those
* For the origin of the names of the trigonometric functions see, for in-
stance, Cajori, "A History of Mathematics," p. 109.
REDUCTION TO FIRST QUADRANT
11
I
II
in
IV
sine
+
4-
cosine
-f
__
_
4-
tangent
+
_
4.
cotangent . . .
+
4-
secant
-f
4-
cosecant
4.
+
of the given angle. By the use of a table of the trigonometric
ratios of acute angles the ratios of an arbitrary angle can there-
fore be found.
17. Exercises.
Draw each of the following angles in standard position; determine the
trigonometric ratios by measurement, as in 14. Determine an angle in the
first quadrant whose trigonometric ratios are numerically equal to those of
the given angle, and verify the results of the graphical determination by
means of a table of the trigonometric ratios of acute angles:
1. 130. 2. 215. 3. ilx/6. 4. -493. 5. 1300. 6. -9 *-/5. 7. 4 7r/5.
8. -1105. 9. 600. 10. UTT/S. 11. 310. 12. 17 ir/6.
18. Reduction to first quadrant. We turn now to the problem
of determining for every arbitrary angle 6 an angle 0' in I, such
that the ratios of 0' shall be numerically equal to the ratios of 0.
Consider, as an example, the angle $ in III, in Figure 9a. If
FIG. 9a
FIG. 9b
we construct in I an angle 0' equal in magnitude to the acute
angle between the terminal side of and the X-axis, and take a
12 EXERCISES
point P f on the terminal side of 0', such that OP' = OP, we find
by considering the two triangles OP'Q' and OPQ, that
r' = r, y' = y, and #' = x. Why?
It follows from this that the ratios of are numerically equal
to those of 9', the acute angle made by the terminal side of 6
with the X-axis.
If we construct in I an angle 0" equal to the acute angle be-
tween the terminal side of 9 and the F-axis, and take on the
terminal side of 0" a point P" such that OP" = OP (see Fig. 9b),
we find upon considering the triangles OP"Q" and OPQ, that
r = T ", 3 = _ y ", and y = - x". Why?
It follows from this that the ratios of are numerically equal
to the co-ratios of the angle 0", the acute angle made by the
terminal side of 6 with the F-axis.
These facts are summarized in the following theorem:
THEOREM I. The numerical values of the trigonometric ratios of
any angle o are equal to the ratios of the positive acute angle o' which
the terminal side of e makes with the X-axis; they are also equal to
the co-ratios of the positive acute angle 0" which the terminal side of
e makes with the Y-axis.
The algebraic signs of the ratios are determined by the quad-
rant in which the angle lies (see 16).
19. Exercises.
1. Apply the theorem of 18 to angles in I and show that the trigonometric
ratios of a positive acute angle are equal to the co-ratios of its complement.
2. Apply the theorem of 18 to angles in II and show that the trigonometric
ratios of an angle that is less than 180, are numerically equal to those of its
supplement.
3. Apply the theorem of 18 to angles in IV and show that the trigonometric
ratios of a negative acute angle are numerically equal to those of the corre-
sponding positive angle.
4. Determine all the trigonometric ratios for the following angles:
(a) 237, (b) 37T/5, (c) 11 *-/8,
(d) 338, (e) 163, (/) 17 *-/9.
6. Also for the following angles:
(a) -248, (6) -512, (c) =, (# =*, (e) (/)
RATIOS OF 30, 60, 45
13
20. Ratios of acute angles. For acute angles, the definitions
of 14 may be put in a special form equivalent to that of 14, but
frequently better adapted to the use that is to be made of them.
The abscissa, ordinate and radius vector of any point P on the
terminal side of the angle 8 are, in this Y
case, the sides and hypotenuse of a
right triangle OPQ, in which the given
angle 6 actually occurs as an acute angle.
The scaffolding, constituted by the
coordinate axes, may now be removed,
and the ratios of the* angle 6 may be
defined with reference to this triangle
OPQ instead of with reference to the
coordinate axes, by putting
"side opposite the angle " in place of y,
"side adjacent to the angle" in place of x,
"hypotenuse" in place of r.
In this way we obtain the following form for the definition of
the trigonometric ratios of acute angles:
FIG. 10
sine e
tangent e =
secant =
side opposite
hypotenuse 9
side opposite
side adjacent to e '
hypotenuse
side adjacent to
cosine e ==
cotangent o =
cosecant 9 =
side adjacent to e
hypotenuse '
side adjacent to o
side opposite e
hypotenuse
side opposite o
21. Ratios of 30, 60, 45. This special form of the definitions
Y enables us to find simple numerical
values for the ratios of certain angles:
(a) To determine the ratios of an
angle of 30, or of an angle of 60, a
' >x diagram like the one in Figure 11 is
constructed, in which QP f = PQ, and
ZOQP = ZOQP' = 90. It follows
that ZPOP'= ZOPP'- ZOP'P = 60,
so that A OPP' is equilateral. Choos-
ing PQ as the unit of measurement, we find OP = 2, PQ = 1, OQ =
V3. The definitions of 20 lead then to the following results:
FIG. 11
14
EXERCISES
THEOREM II. The trigonometric ratios of angles of 30 and 60 are
given by the following formulae:
sin 30= 1/3; cos 30 = Vg/3; tan 30 = \l V5 = V5/3 cot 30 = V3;
sec 30 = 2/V3 =[2 VS/3; cosec 30 = 3;
; cos 60 = 1/3; tan 60 = V3; cot 60 = I/ \/3 Vs/3;
sec 60= 3; cosec 60= 3/V3 = 3 Vs/3.
p
FIG. 12
(6) For the ratios of 45, a diagram like the one in Fig. 12 is
used, in which OQ is taken as the unit of measurement. The
results are summarized in
THEOREM III. The trigonometric ratios of an angle of 45 are given
by the following formulae:
sin 45 = I/ V% = V3/2 ; cos 45 = I/ V = Vg/fc ; tan 45 = cot 45 = 1 ;
sec 45 = cosec 45 = V.
22. Exercises.
1. Determine the trigonometric ratios of the following angles, without the
use of the tables:
(a) 120, 135 and 150. (6) 210, 225 and 240. (c) 300, 315 and 330,
2. Determine the value of each of the following expressions, without the
use of tables:
(a) sin 30 cos 60 + cos 30 sin 60;
(6) cos 120 cos 30 - sin 120 sin 30;
60
1 - tan 120 tan GO '
(d) cos 150 sin 120 - cos 210 sin 150;
3. Similarly for:
(a) sin 315 cos'- 45 + tan 2 30 sec 135;
(6) sin 2 240 cot 225 - cos 2 330 tan 315;*
(c) cosec 2 300 sin 60 tan 150 + sec 2 210 cot 240 cos 2 30;
(d) cot' 150 cosec 240 - tan 330 sec 3 150.
* The notation sin 2 means (sin 0)*. Similar notations are used for other
powers of the trigonometric ratios.
RATIOS OF 90, 180, 270, 360 15
23. Ratios of 90, 180, 270, 360. For the ratios of an angle
whose terminal side coincides with one of the coordinate axes, a
special consideration is necessary, because application of the
definitions of 14 would require division by 0, an operation which
is impossible. To divide a number a by would require the
determination of another number 6 such that 6 multiplied by
would yield a, which is obviously impossible, unless a were 0.
When = 0, we have x = r, y 0. Hence:
sin = 0/r = 0, cos = r/r = 1, tan = 0/r = 0,
cot = r/0 = ?, sec = r/r = 1, cosec = r/0 = ?.
For an angle of 90, x = 0, and y = r, so that we now find:
sin 90 = 1, cos 90 = 0, tan 90 = r/0 = ?,
cot 90 = 0, sec 90 = r/0 = ?, cosec 90 = 1.
Similar results are obtained for angles of 180 and 270; but
it should be noticed, that for an angle of 180, the x and the r
are equal numerically but opposite in sign, in accordance with
the agreements as to the signs of these quantities, so that we
have in this case x = r. For a similar reason, we find that
for an angle of 270, y r. Consequently the ratios of these
angles have the following values:
sin 180 = 0, cos 180 = -1, tan 180 = 0,
cot 180 = - r/0 = ?, sec 180 = -1, cosec 180 = r/0 = ?.
sin 270 = -1, cos 270 = 0, tan 270 - -r/0 = ?,
cot 270 = 0, sec 270 - r/0 = ?, cosec 270 = -1.
The question marks put after cot 0, cosec ; tan 90, sec 90,
cot 180, cosec 180; and tan 270, sec 270 are intended to indi-
cate that these ratios cannot be determined because a division
by would be involved therein.
These facts are expressed in the following theorem:
THEOREM IV. The cotangent and the cosecant of angles of and
180 do not exist; the secant and the tangent of angles of 90 and
970 do not exist. The remaining ratios of these angles are given
by the following formulae:
sin = tan = 0, cos = sec = 1;
sin 90 = cosec 90 = 1, cos 90 = cot 90 = 0;
sin 180 tan 180 = 0, cos 180 = sec 180 = -1;
sin 370 = cosec 370 = -1, cos 370 = cot 370 = 0.
16
TRIGONOMETRIC FUNCTIONS
24. Trigonometric functions. When an angle is changing, as,
e.g., the angle described by the spoke of a revolving wheel or the
angle described by a swinging pendulum, there is in general a
value of each of the trigonometric ratios corresponding to every
value of the variable angle. Whenever such a correspondence
exists between two varying magnitudes, whereby to each value
of one there corresponds a value of the other, it is said that one of
the variables is a function of the other. In the present case, the
trigonometric ratios are functions of the angle; they are called
trigonometric functions.
If we consider a variable angle 6, which while steadily remain-
ing in I, increases towards 7r/2, the variation of the tangent of 9
may best be studied by keeping either x or y fixed as the angle
varies, as in Figures 13a and 13b respectively. In either case, the
FIG. 13a FIG. 13b
ratio y/x remains positive; its numerical value increases indefi
nitely. These facts are expressed by the following statement:
"tan 6 increases indefinitely as 6 increases towards 90:
lirn tan0 = +00."
In a similar manner it is seen that if an angle decreases towards
90, while steadily remaining in II, the ratio y/x is negative
throughout, while the numerical value increases indefinitely ; i.&,
"tan decreases indefinitely as decreases towards 90
lim tan 8 oo."
The fact that lim tan 6 = + oo, and that lim tan 6 = = oo
$ __ QO - + go -f
corroborates our former conclusion that for an angle of 90 the
tangent does not exist.
PERIODICITY 17
25. Exercises.
1. Interpret and prove the following statements:
(a) lim sec 6 = -f 00 - (&) lini sec = oo.
a > 90- e > 90 +
(c) lim cot = +00. (d) lim cot = oo.
-.180 H- 9 > 180 -
(e) lim cosec = +00. (/) lim cosec = oo.
180- 0-*180 +
2. Also:
(a) lim tan = -j-oo. (6) lim tan = oo.
270 - 9 > 270 +
(c) lim sec = +00. (d) lim sec 8 QO.
- 270 4- - 270 -
(e) Jim cot = -f. (/) lim cot = oo.
0~->360+ >360~
(0) lim cosec = +00. (h) lirn cosec = oo.
__> 360 + > 360
3. Determine the value of the following expressions:
(a) sin 2 7T/2 cos 2 w/Q - tan 2 3 7r/4 sec 2 11 T/3.
(6) cos 2 7T/2 sin 2 3 vr/2 + cot 2 11 7r/4 cosec 2 11 *-/6.
(c) sin 3 TT cot 3 7r/2 tan 5 ?r/6 cos 3 TT.
(d) sec 3 TT sec 2 9 7r/4 -f cosec 3 ir/2 cosec 2 17 T/6.
26. Periodicity. In the preceding paragraph we have dis-
cussed the properties of the trigonometric functions of the vari-
able angle 6 for values of which increase or decrease towards
0, 90, 180 and 270. We consider next some other properties
of the trigonometric functions, the first to be considered being
the property of periodicity. A function is said to be periodic if
there exists a constant number a, such that the function assumes
the same value for 6 + a as for 0, no matter what value 6 may have.
The number a is called the period of the function. Examples of
periodic functions are furnished by many natural phenomena,
such as the swinging pendulum, the motion of the tides, the
movement of a vibrating string, the wave motion of sound, light
and electricity, etc. The fact that the trigonometric functions
possess this property of periodicity is one of the reasons for their
great importance in the study of natural phenomena.
If two angles which differ by an integral multiple of 360 (2 TT)
are placed in standard position, they have the same terminal
side; hence their trigonometric ratios are equal; e.g.,
sin (6 + m- 360) = sin (d + m 2w ) =
18 RELATIONS BETWEEN TRIGONOMETRIC FUNCTIONS
for every value of 6, and for every positive or negative integral
value of m. We have therefore the following theorem:
THEOREM V. The trigonometric functions are periodic; they have
a period of 2 *:
In Chapter V we shall see how the trigonometric functions, in
particular the sine and the cosine, may be used for the repre-
sentation of more general periodic functions.
27. Relations between the trigonometric functions. From the
definitions of the trigonometric ratios in 14, there follow imme-
diately the following theorems:
THEOREM VI. The sine and the cosecant, the cosine and the secant,
the tangent and the cotangent of any angle are reciprocals of each
other; i.e.:
sin e cosec = 1; cos sec = 1; tan cot = 1.
THEOREM VII. The tangent of any angle is equal to the quotient
of its sine by its cosine; the cotangent of any angle is equal to the
quotient of its cosine by its sine; i.e.:
Sfr -S-
Since (see Fig. 8), the abscissa, ordinate and radius vector of
any point P are respectively the legs and the hypotenuse of a
right triangle, we have for any angle 6:
Dividing both sides of this equality in succession by r 2 , by # 2
and by ?/ 2 , and making use of the definitions given in 14, we
obtain:
THEOREM VIII. The sum of the squares of the sine and the cosine of
any angle is equal to unity; the square of the secant of any angle
diminished by the square of the tangent, and the square of the cose-
cant of any angle diminished by the square of the cotangent are
each equal to unity; i.e.:
sin 2 + cos 2 = 1; sec 2 tart 2 = 1; cosec 2 - cot 2 = 1.
Theorems VI, VII and VIII may be used to determine all the
trigonometric ratios of an angle as soon as one of them is known.
By means of these theorems a great many other relations may be
proved to hold between the trigonometric functions.
EXERCISES 19
28. Exercises.
1. Given sin = f . Determine the remaining ratios of 0.
Since sin is negative, may lie in III or in IV. In either case, we can
draw the terminal side of 0, by determining_a point for which y = 2 and
r = 3. We find then x = Vs or x = V5 according as is in III or IV.
Knowing x, y, and r, we can obtain the trigonometric ratios of 0.
Proceeding without direct application of the definitions, we can use the
formula sin 2 -f- cos 2 1 to find cos 0; after that the other ratios can all
be determined by means of Theorems VI and VII of 27.
2. Determine all the ratios of 0, when it is known that tan = 4.
2. Similarly, when cos = i and lies in IV.
4. Also when cot == 7 and lies in II.
Prove that the folio whig identities result from the formulae of 27 S
6. sin as tan cos 0.
6. cot = cos cosec 0.
7. tan 2 -f- sin 2 + cos 2 = sec 2 0.
8. cosec sin = cot cos 0.
9. sin 2 sin 4 sin 2 cos 2 0.
10. sec 2 -h cot 2 = cosec 2 + tan 2 0.
11. sin 2 - cos 2 = sin 4 - cos 4 0.
12. tan -H cot = sec cosec 0.
13. sec 2 -f cosec 2 = sec 2 cosec 2 0.
14. cosec 2 sec 2 = cot 2 sec 2 4- cosec 2 tan 2 0.
IK 1 + sin _ cos __ 4/1 + sin .
cos ~~ 1 sin ~~ V l sin
16. 2 (cos 8 -f sin 6 0) - 3 (cos 4 + sin 4 0) -1.
,,- tan a -f tan
17* 7 : r-r = tan a tan 0.
cot a H- cot
cos _ sin 1 cos
+ cos ~" 1 + cos ~" sin
CHAPTER III
LOGARITHMS
29. Theory of Exponents. In order to facilitate the calcula-
tions involved in the application of the trigonometric ratios to
the solution of problems, we take up the study of logarithms. We
recall first the following parts of the theory of exponents.
(a) A continued product, all of whose factors are equal to the
same number, a, is called a power of that number. The number
a is called the base of the power; the number of factors in the
product is called the exponent of the power. Powers are classified
according to their exponents; they are represented in abbreviated
form by the base and the exponent.
Thus a a a a a, the 5th power of a, is represented by a 5 ;
a being the base of the power and 5 the exponent.
(6) Besides powers whose exponents are positive integral num-
bers, defined in (a), we consider powers whose exponents are zero,
negative or fractional. Such powers are defined as follows:
The zero-th power of any number is equal to 1; e.g., a = 1.
A power of a whose exponent is negative is the reciprocal of the
power of a, whose exponent is the corresponding positive number:
- = JL.
a ~ a*'
A power of a positive number, a, whose exponent is a unit frac-
tion is the real positive root of a whose index is equal to the de-
nominator of the fraction: a 1/fl = \Ta.
Thus, e.g., 2~ 5 = 1/32; 36* = 6; 7 = 1.
(c) From these definitions are derived the following funda-
mental theorems, usually referred to as the "Laws of Exponents":
I. The product of two powers of a number a is equal to that
power of a, whose exponent is equal to the sum of the exponents
of the factors: a? X a* = a v+a .
II. The quotient of two powers of a number a is equal to that
power of a, whose exponent is equal to the exponent of the dividend
diminished by the exponent of the divisor: a* -f- a q =
20
DEFINITION OF THE LOGARITHM 21
III. A power of a power of a number a is equal to that power
of a whose exponent is equal to the product of the exponents of
the given powers: (a?) q = a? a .
Thus, e.g.: \/2 X J = 2* X 2~ 2 = |H =
25 ^ -dta = 5 2 -f- 5^= 5 5 = 3,125.
30. The use of exponents in calculation. The fundamental
idea underlying the use of logarithms in calculations is that
the laws of exponents may be utilized for the purpose of mul-
tiplication and division of numbers, for raising numbers to a
power, or for extracting roots. If, e.g., we had a list of the suc-
cessive positive integral powers of 2 from 2 1 to 2 150 , the product
and quotient of two of them, their powers and roots, could,
within certain limits, be found by addition, subtraction, multi-
plication and division. For example, we would have:
049 1
08 v 9 17 9+17 _025. _ 049-63 _ O-14 _ L .
Z X Z ^ 4 , 2^ 4 4 ,
(2 17 ) 6 = 2 l7x6 2 102 * A/2" 44 = (2 144 ) ~fa 2 12
If, therefore, we could write every number as a power of some
fixed number, all multiplications and divisions could be reduced
essentially to additions and subtractions, while powers and roots
of numbers could be found by simple multiplications and divisions.
31. Definition of the logarithm. Accordingly, we lay down
the following definition :
DEFINITION I. The logarithm of any number p with respect to
the base a is the exponent of that power of a which is equal to p.
This logarithm is designated by the symbol log a p. We shall con-
sider only logarithms of positive numbers with respect to bases
which are positive.*
As a consequence of this definition, we can say:
Iog2 16 = 4, because 2 4 = 16; Iog 3 = 2, because 3~ 2 = T J ;
logs V5 = ^, because 5^ = V5; Iog 7 1=0, because 7 C = 1.
* It does not fall within the scope of this book to consider whether for any
number p and any base a, there always exists an unique logarithm of p with
respect to the base a. We take for granted that if a and p are positive, there
always exists uniquely a real number loga p. The further study of this ques-
tion belongs to the Theory of Functions.
22 FUNDAMENTAL THEOREMS ON LOGARITHMS
32. Exercises.
1. Determine Iog 3 27, logs sV, logg v'S, Iog 3 1, Iog 3 ^27, logs 3.
2. Determine logio 10, Iog 10 1000, logio .01, logio .0001, log io 1.
3. Determine Iog 2 8, log* v^, Iog 2 A, Iog2 l/V^, Iog2 4/\/8.
4. Determine logs 4, Iog 4 8, Iog 9 27, Iog2 7 J, logic i-
5. It is known that logio 17 = 1.23045, logio 29.5 = 1.46982, logio 83 = 1.91908.
Determine 10 1 - 46 * 2 , lO 1 - 23 ^, 10 1 - 91908 , 10 3 - 23046 , 10- 46982
6. Determine 10 Iogl 7 , 5 log6 7 , 2 Iog ' 7 , a log 7 .
33. The fundamental theorems on logarithms. From the
" Laws of Exponents," quoted in 29 (c), we derive now, by the aid
of the definition of 31, the following fundamental theorems on
logarithms, in which we prove what was merely suggested in the
last paragraph of 30.
THEOREM I. The logarithm of the product of two numbers with
respect to the base a is equal to the sum of the logarithms of the
factors:
lOga (pq) = lOgap + loga q*
Proof. The theorem evidently says nothing else than that the
exponent of that power of a which equals pq is equal to the sum
of the exponents of the powers of a which are equal to p and q*
For, let
loga p = x, and log a q = y;
i.e., let a x = p, and a v = q.
Then a x+v = pq; i.e., log a pq = x + y = log a p + log a q.
THEOREM II. The logarithm of the quotient of two numbers with
respect to the base a is equal to the logarithm of the dividend dimin-
ished by the logarithm of the divisor:
lOga - = lOgapJ- lOga q.
THEOREM III. The logarithm of a power of a number with respect
to the base a is equal to the exponent of the power multiplied by the
logarithm of the number:
lOg a p n = n lOga p.
The proofs of Theorems II and III are left to the student.
We proceed to illustrate the use of these theorems:
(a) To determine logio 14 when it is known that logio 2 == .30103
and logio 7 = .84510.
EXERCISES 23
From Theorem I, it follows that logio 14 = logio 2 + logio 7
= 1.14613.
14 X 25
(6) To determine logio ^7 > when 18* 2 ^ - 30103 > lgio 3
= .47712, logio 5 = .69897 and logio 7 = .84510.
By the use of Theorems I and II, we find:
logio 27 ^ logl 14 + logl 2 ' 5 "~ l glo;27 -
But, from Theorem III it follows that logio 25 = logio 5 2 = 2
logio 5 = 1.39794, and that logio 27 - log :0 3 3 = 3 logio 3 = 1.43136.
' logio M ^ 25 = logic 14 + 2 logio 5-3 log w 3 = 1.11271.
(c) To determine log V/- 1 -^ , when logio 2 = .30103, logio 3
V ol
= .47712 and log lo 5 = .69897.
By the use of Theorems I, II and III, we find:
logio V -Q rj = 3 [2 logio 5 + 5 logio 2 4 logio 3]
o JL
= I [1.39794 + 1.50515 - 1.90848] = I .99461
= .33154.
34. Exercises.
Given, that logio 2 - .30103, logio 3 * .47712, and logio 7 = .84510, determine
, i n , t 5 / 27 X 49
1. logio 5. 9. logio y gQ
2. logio 21.
3. logio 35. 10. f - 81X64
4. logio 70.
* i 49
5. logio ^r
6. logio \/32.
125 X 32
7. logio
8. logi
27
/28 X 25
81
24
COMMON LOGARITHMS
35. Common logarithms. It will now be recognized that the
plan suggested in 30 can actually be carried out, provided we can
" write every number as a power of some fixed number/' i.e.,
provided we can find the logarithm of every number with respect
to some fixed base. The Differential Calculus furnishes methods
for solving this problem and for constructing a " table of
logarithms." For the purpose of computation, logarithms with
respect to the base 10, called common logarithms are most useful.
Without considering here the problem of constructing a table of
common logarithms, we shall see how a comparatively small table
of logarithms can be made to serve our purpose.*
From now on, the symbol "log" will be used to designate logio,
i.e., common logarithm.
We know then that log 1000 - 3, log 100 = 2, log 10 = 1,
log 1=0, log .!=-!, log .01 = - 2, log .001 = - 3, etc.; i.e.,
if we arrange numbers in the geometrical progression of scale A,
their common logarithms will form the arithmetical progression of
scale B.
Scale A
N
.001
.01
.1
1
10
100
1000
Scale B
log N
-3
-3
-1
1
2
a
We assume now, without proof, that as the number N increases,
its logarithm will also increase. It will then follow that if a num-
ber N lies between two successive terms of scale A, log N will lie
between the two corresponding integers of scale B. Hence if we
have determined the place of a number N in scale A, we shall know
the integral part of log N, and conversely, if we know the integral
part of log N, we shall know the position of N in scale A.
The integral part of the logarithm of a number is called the
characteristic; the fractional part, taken positively, is called the
mantissa. Here it is to be understood that if the logarithm of a
number is negative, it will be written as the sum of a negative
integer (the characteristic) and a positive fraction (the mantissa).
For example, log = = log 10~~^ = | will be written in the
* Any set of five-place tables of logarithms of numbers and of trigonometric
functions may be used in connection with this text.
COMMON LOGARITHMS 25
form -1 + .50000, or 9.50000 - 10; the characteristic is -1 (or
9-10), the mantissa is .50000.*
We conclude that the position of a number in scale A and the
position of its logarithm in scale B are related to each other in
the way expressed by the following theorem:
THEOREM IV. The characteristic of the common logarithm of a
number TV is the lesser of the two integers in scale B, corresponding
to the two numbers in scale A between which N lies; conversely, a
number N will lie between those two numbers in scale A, which cor-
respond to the characteristic of log N and the next greater integer
in scale B.
This theorem may be reduced at once to the following working
rule:
The characteristic of log N is equal to the integer in scale B, corre-
sponding to that number in scale A which is pointed off like A r ; con-
versely, N is to be pointed off like that number in scale A, which
corresponds to the characteristic of log N in scale B.
It remains to determine the mantissa of a logarithm; it is found
from a table. It is important to observe however that the
mantissa of the logarithm is the same for all numbers which differ
only in the position of the decimal point. Let us compare, for
instance, log 475.3 and log .04753. Since 475.3 = 10 4 X .04753,
it follows that log 475.3 = 4 + log .04753. But the addition of
4 to log .04753 will simply increase its characteristic by 4, with-
out affecting the mantissa. In general, if A and B differ only in
the position of the decimal point, the greater one, say A, can be
obtained by multiplying the smaller one, B, by a power of 10, i.e.
A 10* X B, where k is a positive integer. Therefore log
A = k + log B, so that log A is obtained from log B by adding
to it the integer k. This however will merely increase the charac-
teristic by fc and will not alter the mantissa. This fact finds expres-
sion in the following theorem:
THEOREM V. The mantissa of the common logarithm of a num-
ber is determined by its sequence of digits only; conversely, the
sequence of digits of a number is determined by the mantissa of its
logarithm.
* When a five-place table is used, all mantissas are written out to five
places of decimals.
26 USE OF A TABLE OF LOGARITHMS
36. Use of a table of logarithms. Theorems IV and V enable
us to find from a five-place table of logarithms the logarithm to
five places of any number of five significant figures; * and to
determine five significant figures of a number whose mantissa is
given to five places. It only remains to become familiar with the
arrangement of the tables. This is best explained by means of
examples.
Example 1. To find log 47.316.
Since 47.316 lies between lO^and 100, (or, since 47.316 is pointed
off like 10) the characteristic of log 47.316 is 1. To determine the
mantissa we look up in the tables the sequence of digits 47316.
The first three digits are found in the left-hand column of the
table of logarithms of numbers, usually headed N, the fourth digit
is found in the line at the top or bottom of the page. In this way
we find that
for 47320 the mantissa is 67504; while
for 47310 the mantissa is 67495.
For an increase of 10 in the number, there is therefore an in-
crease of 9 in the mantissa. Now we approximate to the value
of the mantissa for 47316 by assuming that as the number in-
creases by ten equal steps from 47310 to 47320, the mantissa
will also increase by ten equal steps from 67495 to 67504. We
must then add to 67495 six tenths of 9. The table of proportional
parts (usually headed P.P.) shows that T <V of 9 is equal to 5.4,
which we round off to 5. We conclude that the mantissa for
47316 is 67500 and that log 47.316 = 1.67500.
Example 2. To find log .089327.
Since .089327 lies between .01 and .1, (i.e. since .089327 is pointed
off like .01) the characteristic of the logarithm is 2 or 8 10.
The mantissa for 89330 is 95100; the mantissa for 89320 is
95095. A difference of 10 in the number makes a difference of
5 in the mantissa.
From the table of proportional parts, we find that T V of 5 is 3.5,
which we round off to 4, so that the mantissa for 89327 is 95099
and therefore log .089327 = 8.95099 -10.
* The figures of a number significant for its logarithm are the digits left
after the ciphers at the beginning and end of the number have been removed.
USE OF A TABLE OF LOGARITHMS 27
Example 3. Given log N = 2.23130. To determine N.
We now reverse the process followed in the preceding examples
and begin by searching among the mantissas given in the table
for 23130. We find that
23147 is the mantissa for 17040, and that
23121 is the mantissa for 17030, so that a difference of
26 in the mantissa makes a difference of 10 in the number.
The given mantissa 23130 exceeds 23121 by 9; from the table
of proportional parts, we find that fo of 26 equals 7.8, while
T \ of 26 equals 10.4. Of these, the former is closer to 9 than
the latter; hence 23130 is the mantissa for 17033, which must there-
fore be the sequence of digits for the required number N. Since
the characteristic is 2, N must be pointed off like 100 so that the
decimal point must be placed between the and the 3. We con-
clude that the required number N is 170.33.
Example 4. Given log N = 9.07025 - 10. To determine N.
We find that 07041 is the mantissa for 11760, while 07004 is
the mantissa for 11750, so that a difference of 37 in the mantissa
makes a difference of 10 in the number. Hence a difference of 21
in the mantissa corresponds to a difference of 6 in the number;
the sequence of digits of N must therefore be 11756. Since the
characteristic is 1, the number must be pointed off like .1, so
that we conclude that N = .11756.
The table of the logarithms of the trigonometric functions gives
directly the logarithms of the trigonometric ratios, both charac-
teristic and mantissa, except that from the logarithms of sines
and cosines, and from the logarithms of tangents of angles less
than 45 and of cotangents of angles greater than 45, the term
10 is usually omitted. The arrangement of the table will be
readily understood from the following examples.
Example 5. To determine log tan 37 23' 45".
We find log tan 37 23' = 9.88315 - 10 and log tan 37 24'
= 9.88341 10, so that a/ difference of 1 minute in the angle
causes a difference of 26 in the logarithm of its tangent. Accord-
ingly a difference of 45" in the angle will cause a difference of
20 in the logarithm of the tangent. Hence log tan 37 23' 45" =
9.88335 - 10.
28 CALCULATION BY MEANS OF LOGARITHMS
Example 6. To determine log cos 54 29' 13".
We find log cos 54 29' = 9 .76413 - 10, while log cos 54 30'
= 9.76395 10, so that as the angle increases by 1 minute, the
logarithm of its cosine decreases by 18. Consequently for an
increase of 13" in the angle there will be a decrease of H of 18,
i.e., of 4 in the logarithm of the cosine; hence log cos 54 29' 13"
= 9.76409 - 10.
Example 7. Given log sin 6 = 9.47468 10. To determine
We find that log sin 17 21' = 9.47452 - 10, while log sin 17 22'
= 9.47492 10, so that a difference of 40 in the logarithm of the
sine corresponds to a difference of 1 minute in the angle. Conse-
quently, a difference of 16 in the logarithm of the sine corresponds
to a difference of if of a minute, i.e., of 24" in the angle. We
conclude that 6 = 17 21' 24".
37. Exercises.
Determine:
1. log 56.387. 6. log 978.94. 9. log cos 47 58' 15".
2. log .084923. 6. log .00073299. 10. log cot 15 47' 50".
3. log 1.0576. 7. log sin 27 15' 20". 11. log sin 78 29' 40".
4. log .20458. 8. log tan 68 37' 35". 12. Jog cos 36 35' 45".
Determine the number N, when:
13- logN = 1.65783. 16. log AT = .27586. 17. log AT= 7.80880 -10.
14. logJV = 9.04987-10. 16. log N = .09675. 18. log N =3. 97538.
Determine the angle 6, when:
19. log tan 6 = .27725. 22. log cot = 8.83225 - 10.
20. log cos 6 = 9.88247 - 10. 23. log cos = 9.68552 -10.
21. log sin 9 - 9.48030 - 10. 24. log tan = 9.96795 -10.
38. Calculation by means of logarithms. We are now prepared
to apply the results of the preceding articles. In all calcula-
tions, it is important to arrange the work in such a way as to
secure the greatest possible accuracy with least effort. This is
accomplished best by making a plan of the entire calculation
before looking up the logarithms, as illustrated in the following
examples:
CALCULATION BY MEANS OF LOGARITHMS 29
Example L To determine N = i/ 47 * 321 *
v .y/Oi
015732
9763
First we apply Theorems I, II and III; in this way we find
that log N = [log 47.321 + log .015732 - log .9763],
Accordingly we make the following plan for the calculation:
log 47.321 1
log .015732 = 8 -10
log .9763 = 9 -10
S
3-
log N = .
N = .
After having completed the plan, we turn to the tables to deter-
mine the mantissas and to complete the calculation; this gives
the following results:
log 47.321 = 1.67505
log .015732 = 8.19678 - 10
19.87183 - 20
log .9763 = 9.98958 - 10
logN = 29.88225 -30
3 _
log N = 9.96075 - 10
N = .91358.
In order to make the subtraction possible without introducing
a negative mantissa, we wrote the characteristic of the minuend
in the form 19 20 instead of 9 10; to make the division possible
without introducing a negative fraction, we wrote the mantissa
of the dividend in the form 29 - 30, instead of 9 - 10.
m . . , r T356.12 X (.56836) 2 T
Example 2. To determine N = '
L. OA.
30 EXERCISES
We find that log N = 5 [log 356.12 + 2 log .56836 -(log 51.834
+ 1 log .0843)], so that the calculation is carried out in the fol-
lowing form:
log 356.12 = 2.55159
log .56836 = 9.75462 - 10; 2 log .56836 = 19.50924-20
2.06083
log 51.834 = 1.71461
logj.0843 = 28.92583 -30; f log .0843 = 9.64194 - 10
A
1.35655 1.35655
8
.70428
log N = 3.52140
N = 3322.0
7 o rr, , . . Ar .075869 X sin 47 15' 36"
Example 3. To determine N = ^ ^,
log N = log .075869 + log sin 47 15' 36" - log tan 68 23'.
log .075869 = 8.88006 - 10
log sin 47 15' 36" = 9.86596 - 10
39.
Cal<
1.
4.
R <
log tan 68 23'
log AT
N
Exercises.
julate:
8.74602 - 10
= .40201
= 8.34401 - 10
= .022081
e 4/14.03 X 1.028
V 105
* V (.005734)*
4.3857 X .1296
V 507
4.357 X (.08356) 8
7 * .00097435
4/6.7248 X .098376
(.4316) 1 X 52.07
V 57.849 X .0001574
51.86 X sin 35 18'
(.083) 2
' sin 49 23'
10. tan 28 37' 15" X cos I
f/54.7 X .0287
15.8 cosec 75 20' 20".
EXERCISES 31
11. 9.63 X sin 42 17' X sin 18 lfi 4/57.8 X 67.845 X 23.593
29' 30". V 98.627
.1875 X tan 65 34' 10" j /.0024856 X .57321 X .009847
sin 17 54' 20" 17 ' V sin 9 15' 15"
10 4 74.268 X sin 81 20' , ft j /.08734 X .09586 X .06792
13< V tan 14 38' 18 ' V ^5647
F78.643 X cos 21 17H* lfl .5487 X sin 38 27' 15" ,
L9.5064 X cot 68 23'J " .98346
IK 1.0567 X cos 28 43' 50" ft 3.75 X sin 53 27' 30"
2.3981 ' sin 145 35' 40"
CHAPTER IV
THE SOLUTION OF RIGHT TRIANGLES. APPLICATIONS
40. The right triangle. If enough elements of a right triangle
are given to determine it completely, every other magnitude con-
nected with the triangle can be determined by means of the trigo-
nometric ratios of the angles. A right triangle is completely
determined by any one of the following sets of data:
(a) two legs, (6) one leg and the hypotenuse, (c) one acute
angle and a leg, (d) one acute angle and the hypotenuse.
In the case of the right triangle the remaining elements can be
determined in each case by the use of the special form of the
definitions for acute angles, given in 20. It is customary to
denote the lengths of the sides of a triangle by small Roman
letters, the vertices of the opposite angles by the correspond-
ing Roman capitals, the vertex of the right angle and the hy-
potenuse in a right triangle usually being denoted by C and c,
respectively.
In cases (a) and (b) the knowledge of the sides enables us to
find one of the ratios of either of the acute angles, and hence
these angles themselves. In cases (c) and (d) we proceed in the
opposite manner: from the given side and appropriate trigono-
metric ratios of the given angle the other sides are found. The
second acute angle is the complement of the given angle.
From the definitions given in 20 we obtain immediately the
following theorem:
THEOREM I. In a right triangle, the following relations hold be-
tween the sides and angles:
the side opposite an acute angle hypotenuse X the sine of the
angle;
the side adjacent to an acute angle = hypotenuse X the cosine of
the angle;
the side opposite an acute angle = the adjacent side X the tangent
of the angle.
32
ACCURACY OF THE CALCULATION 33
41. Example (see Fig 14).
Given. ZC = 90, LA * 47 13', b = 14.35.
Required. B, a, c.
Solution. We find Zfi = 90 - 47 13' = 42 47'.
Moreover, we have by Theorem I,
B
a = 6 tan A ;
also, 6 = c cos A,
whence c *= 6/cos A.
The unknown elements of the triangle have
now been expressed in terms of the known
elements. It remains to calculate a and c. A
This may be done by the use of logarithms FIG. 14
(see 38) as follows:
log 6 = log 14.35 = 1.15685; log 6 = log 14.35= 1.15685
log tan A = log tan 47 13'= .03364;
log cos A = log cos 47 13' = 9.83202 - 10
A S
loga= 1.19049 logc= 1.32483
a ==15.506 c = 21.127
Using tables giving the values of the ratios themselves (the
so-called natural values) instead of those of their logarithms, we
obtain :
a = 14.35 X tan 47 13' - 14.35 X 1.0805 = 15.505.
b = 14.35 -v- cos 47 13' = 14.35 + .6792 = 21.128.
42. Accuracy of the calculation. Checking the results. The
values of the logarithms and of the trigonometric ratios found in
a table are correct only to within the limit of accuracy of the table,
i.e., to within .000005 if 5-place tables are used, or .00005 in the
case of 4-place tables.
Therefore, since the sum and difference of two approximate
numbers are more readily determined and are frequently more
nearly accurate than their product or quotient, it follows that in
most cases the calculation by means of logarithms is to be pre-
ferred, particularly if the data of the problem are themselves
approximations.*
* A comparison of the advantages of calculations with and without the use
of logarithms requires a much more detailed discussion than can here be de-
voted to it. Most of the calculations in this book have been made by means
of logarithms, on account of the greater convenience of this method.
34
ISOSCELES TRIANGLES
To secure a higher degree of certainty as to the correctness of
the final numerical results of a calculation, these results should
be checked. A rough check can be obtained by drawing the
figure to scale by use of ruler and compass and then measuring
the required elements. A more accurate, numerical check in-
volves the testing of the results, together with the data, in other
relations between the sides and angles of the triangle than those
used in the solution of the problem.
For the right triangle, the Pythagorean relation c 2 = a 2 + & 2
serves the purpose especially well, since it involves all three sides
of the triangle. If written in the form
a* = c 2 - 6 2 = (c - 6) (c + 6) or 6 2 = c 2 - a 2 = (c - a) (c + a),
it is well adapted to logarithmic calculation. For the example of
the preceding section, it furnishes the following check:
c = 21.127
a = 15.506
c + a = 36.633
c-a= 5.621
log (c + a) = 1.56388
log (c - a) = .74981
A
log (c + a) (c - a) - 2.31369
log b = 1.15685
This value should agree to within four units of the last decimal
place with the value of log b derived from the value of b as given.
43. Isosceles triangles. An isosceles triangle is divided into
two congruent right triangles by a perpendicular from the vertex
to the base. The methods explained in 40, 41 and 42 suffice
therefore for the treatment of such triangles.
fr
FIG. 15
Example (see Fig. 15).
Given. Z.P = Z.Q = 37 28', a
Required. V, h, b.
.23152.
EXERCISES 35
Solution. We find Z F/2 = 90 - 37 28' 52 32'; .-. Z V
= 105 4'.
Furthermore
sin P = A/a, i.e., h = a sin P,
and cos P = f 6/a, i.e., J 6 = a cos P,
whence b = 2 a cos P.
log a = 9.36459 - 10 log a = 9.36459 - 10
log sin P = 9.78412 - 10 log cos P - 9.89966 - 10
log h = 9.14871 - 10 log 2 = .30103
h = .14087 log b - 9.56528 - 10
6= .36752
Check. W = a 2 - (6/2) 2 - (a + 6/2) (a - 6/2)
a = .23152
6/2 = .18376
a + 6/2 = .41528; log (a + 6/2) - 9.61834 - 10
a - 6/2 = .04776; log (a - 6/2) = 8.67906 - 10
A
log [a 2 - (6/2) 2 ] = 8.29740 - 10
2
log h = 9. 14870 -10
44. Exercises.
Calculate the unknown elements of the triangles ABC in which Z C = 90
and in which the following elements are given; check the results:
1. a = 373, 6 = 526. 7. A 84 35', c 378.
2. a = .1432, 6 = .0756. 8. A = 44 35', c = 378.
3. a - 2.146, c = 4.292. 9. A = 45 3', a - .08512.
4. b - 13.071, c - 19.603. 10. a - .06891, c - .09004.
5. A = 68 25', b = 8732. 11. b 13.683, a - 3.9857.
6. B = 27 13', 6 = .06315. 12. B - 5 2', c = 1.0059.
Determine the remaining elements of the following isosceles triangles (the
notation being in accordance with Fig. 15) and check the results:
13. b = 26.804, P = 57 13'. 17. b ** 24.192, h - 16.387.
14. b = 35.96, V = 128 46'. 18. h - .05831, P 10 19',
16. a 12.05, /i - 8.041. 19. 6 - 9.0834, a = 9.9457.
16. h - 1.0203, 7 - 44 52'. 20. a - 6.8032, P - 25 27'.
21. a = 16584, V - 90.
36
PROJECTION
45. Projection. The projection of a directed segment AB of
a directed line I upon another directed line w, which makes with
I an angle can now be readily determined. Let the length of
AB (see Fig. 16) be equal to a; then, since the direction of AB
FIG. 16
is opposite to the positive direction upon I, AB = a. The
projection of AB upon m is AiBi. Through A draw a line m'
parallel to ra, cutting BB l in C. Then AC = AiBi (Why?),
and Z. CAD = 6 (Why?). Moreover, cos0 = -cosfl' (see 18).
Since now AiBi is in the positive direction upon m, we have
AiBi = AC = a cos 0' = a cos 6 = AB cos 8.
This result finds expression in the following theorem:
THEOREM II. The projection of a directed segment AB of a directed
line / upon a directed line m, which makes with / an angle 0, is equal,
In magnitude and direction, to AB cos 0.
FIG. 17
This theorem is illustrated by the following examples:
(a) In Figure 17, I makes with m an angle of 240; since AB
is a negative segment of the directed line Z, we have AB = 4.
Hence, A& -4 cos 240 = -4 -| = 2.
APPLICATION OF THEOREM II
37
(6) In Figure 18, 1 makes with ra an angle of 135 and AB = 3.
-V2
Hence AiBi = 3 cos 135 = 3 ^ = -2.1.
135 .
FIQ. 18
46. Application of Theorem II. We return now to Theorem II
of 6 and calculate the projections of the segments by means of
Theorem II of the present chapter. Consider, for example, the
equilateral triangle ABC of Figure 19, of which the side AC
FIG. 10
makes an angle of 15 with the positive X-axis.
The side AB then makes with OX an angle of 75,
the side BC makes with OX an angle of 45,
the side CA makes with OX an angle of 195.
If the length of the side of the triangle be denoted by a and the
projections of the vertices upon the X-axis by A\, B\, and Ci, then
we find:
AA + ftCi + CiAi = a (cos 75 + cos (-45) + cos 195)
a (cos 75 + cos 45 - cos 15)
- a (.2588 + .7071 - .9659) = 0.
38
APPLICATIONS
47. Exercises.
1. Determine the projection upon the X-axis of a segment, 5 feet long, of
a line which makes with the Jf-axis an angle of 30.
2. Determine the projection of the same line upon the F-axis.
3. Determine the projections upon the X- and F-axes of a segment, 7 feet
long, of a line parallel to the F-axis.
4. Show that the sum of the projections upon the X-axis of the sides of
the square of Figure 20 is equal to zero. Show that the same result holds
for the projections upon the F-axis.
6. (See Fig. 21.) Show that the projection of AC upon the X-axis (F-axis)
is equal to the sum of the projections of AB and BC upon the X-axis (F-axis).
FIG. 20
FIG. 21
48. Applications. Numerous problems in different fields of
science can be solved by means of the methods developed in the
preceding paragraphs. To solve such problems there is required,
however, besides a knowledge of trigonometry, an understanding
of the technical terms used in those different fields. We explain
below a few technical terms which will be used in the exercises of
the following paragraph and in Chapter VII. For further appli-
cations the student is referred to books on surveying, navigation,
astronomy, artillery, etc.
The angle made by the line along which an object is sighted,
with a horizontal line through the point of observation and in
the same vertical plane as the line of sight, is called the angle of
elevation or the angle of depression of the object, according as the
object is higher or lower than the point of observation.
The surveyor frequently designates the direction of a line by
means of its bearing, i.e., the acute angle which ~ the line makes
with a N. S. line through the point of observation, indicating at
the same time whether the line runs to the east or the west of the
EXERCISES
39
N. S. line. The bearings of the lines OA } OB and OC in Figure
22 are denoted as N 23 E (read 23 degrees East of North),
FIG. 22
FIG. 23
N 47 W (read 47 degrees West of North) and S 18 W (read
18 degrees West of South) respectively.
The points of the compass, as used by the navigator, are indi-
cated on Figure 23.
49. Exercises.
1. The angle of elevation of the top of a mountain from a point A, situ-
ated in a plane 1500 feet below the top, is 19 27'. Determine the distance
from A to the top in an air line; also the horizontal distance from A to the
foot of the mountain.
2. From a lightship L, at a distance of 500 feet from a point A on shore,
the angle of elevation of a water tower vertically above A is 28 33'. Deter-
mine the height of the tower above the level of the ship.
3. The angle of depression of a point P as seen from an airship 1800 feet
above the ground is 63 27'. What is the straight line distance of the air-
ship from P?
4. An observation tower T is 40 feet high and stands 20 feet from the
edge of a vertical cliff, whose top is 400 feet above sea level. A ship S is in
the vertical plane through T and on a line at right angles to the shore line;
its angle of depression from T is 9 23' 16". How far is S from the shore ?
6. From a point A on one side of a stream and 5 feet above the ground,
the angle of elevation of the top of a tree straight across the stream is 20 13'.
The height of the tree above the level of A is estimated to be 15 feet. How
wide is the stream at that point?
40 EXERCISES
6. A vessel is observed directly south from a lighthouse L and S 43 27'
W from a lighthouse M known to be 50 miles due East from L. What are
the distances of the vessel from each of the lighthouses?
7. From points P and Q, 150 feet apart and both lying in the same ver-
tical plane through a spire S, the angles of elevation of the spire are observed
to be 12 13' and 28 36' respectively. How high is the spire if P and Q are
on the same side of the spire; if P and Q are on opposite sides of the spire?
8. Two points, P and Q, known to be a mile apart on a level road which
lies in a vertical plane through the top of a mountain A are observed from A.
The angles of depression of P and Q are 23 17' and 32 27' respectively.
How high is the mountain top above the road?
9. The bearing of the line CA is N 12 E; the bearing of the line CB is
S 78 E, while A bears N 15 W from B. The distance CB is known to be
257 feet. Determine the distances from A to B and to C.
10. From a point B on one side of a stream and 5 feet above the ground
the angle of elevation of a point P directly across on the opposite shore is
found to be 34 13'. A level line BC, 100 feet long is laid off at right angles
to the stream, and from C the angle of elevation of P is found to be 20 43'.
Determine the height of P and the width of the stream.
11. A railroad track consists of a horizontal piece followed by a down-
ward grade 4 miles long, making an angle of 12 with the horizontal. It is
proposed to replace the 12 grade by a 4 grade. How much of the hori-
zontal track must be removed to accomplish this change?
12. How much track mileage would be saved by the change proposed hi
Problem 11?
CHAPTER V
THE GRAPHS OF THE TRIGONOMETRIC FUNCTIONS
60. Graphs of sin 6 and cos 9. Many properties of the trigo-
nometric functions appear in a new light if we consider their
graphical representation. In this chapter we shall be concerned
principally with the graphical representations (usually called the
graphs) of the functions sin and cos 6 and of the sine and cosine
functions of expressions of the form aO + 6, i.e., of sin (aB + 6)
and cos (aB + 6). For their construction we make use of the
graphical representation of pairs of numbers used in Chapter I.
Angles 6 measured by an arbitrary unit will be represented by
distances along the X-axis (abscissae); with each abscissa will
be associated as ordinate a line representing the value of that
function of the angle 6 whose graph we wish to obtain. The
determination of these ordinates can be carried out graphically
by the following simple device.
After having established the unit of measurement on the X-
and F-axis, we describe a circle whose radius is equal to this
,r
Sin
FIG. 24
unit and whose center is on the X-axis (see Fig. 24). At the
center of this circle we construct an angle whose initial side lies
in the positive direction along the X-axis. From the point P,
where the terminal side of 6 meets the circle, we drop a perpen-
41
42
EXAMPLES OF GRAPHS
dicular PQ upon the X-axis. Since for P the radius vector is
equal to unity, the sine of angle 6 is measured in magnitude and
direction by the ordinate of P, i.e., by the line QP. Through P we
draw a line PP', parallel to the X-axis. We then lay off on the
X-axis a distance OA, representing the angle in magnitude and
in sense, through A we draw a line A A' parallel to the F-axis,
meeting the line PP' in jB; this point B is then a point on the
graph of sin 0.
If the angle 6 be laid off with its initial side in the positive
direction along the F-axis and on the vertical diameter of the
FIG. 25
circle (see Fig. 25), we obtain an ordinate QP which measures in
magnitude and direction the cosine of 6.
51. Examples of graphs.
(a) Construct the graph of sin 6 for 6 varying from to 360 (see
Fig. 26).
y
Yir 2
/x\\
7TA p 2
FIG. 26
At the center of the auxiliary circle we lay off angles of 0, 15,
30, . . . 330, 345, 360, with initial sides in the positive direc-
tion along the X-axis and with terminal sides cutting the circle
EXAMPLES OF GRAPHS
43
in points P , Pi, P*> etc. On the X-axis we lay off distances OA Q
= 0, OAi = 7T/12, 0-42 = 7T/6, etc., representing the radian meas-
urements of these angles. Through P , Pi, Pz, etc. we draw lines
PoXo, Pi-Xi, PA, etc. parallel to the X-axis; through A , -4.1, ^2,
etc. we draw lines A F , A\Y\, A^Y^ etc. parallel to the F-axis.
The intersections JS , BI, -62, etc. of the lines A F and PoXo,
J-iFi and PiXi, AzYz and P 2 X 2 , etc. respectively, are points whose
^-coordinates measure angles and whose ^/-coordinates measure
the sines of these angles. A smooth curve drawn through the
points BQ, BI, B%, etc. is the graph of the function sin 6 for 6 vary-
ing from to 360.
(6) Construct the graph of cos 6 for varying from to 360 (see
Fig. 27).
Q Ol
,p
&
x\ \
yt->^
"/]
^
*
/\ \\
/ / ^J.
i
S^
f
/\^\\\
/ //^\
\
s
l~~"^$4
1^^^'^
S^
f
V
\^^<^
fes^^_
o,
AI-
42
Sy
f
-u
V^ xX ' //i
Y^\j\^y
^
s
\///
\\\/
X
s
N/. /-
\ i
FIG. 27
We construct at the center of the auxiliary circle angles of 0,
15, 30, . . . 345, 360 with initial sides in the positive direc-
tion along the Y-axis and with terminal sides cutting the circle
in points Qo, Qi 9 Qt, etc. On the X-axis we determine the points
A Ai, A 2 , etc. as in example (a). Lines QoX , QiXi, $2X2, etc.
parallel to the X-axis will meet lines A F , AiFi, -4.2F 2 , etc.
parallel to the F-axis in points B > BI, B^ etc, whose x-coordi-
nates measure angles and whose ^/-coordinates measure the cosines
of these angles. A smooth curve drawn through these points
Bo, BI, B 2 , etc. is the graph of the function cos for varying
from to 360.
(c) Construct the graph of sin (2 6 30) for 6 varying from
to 360. We construct as in example (a) the lines PoXo, PiXi,
?2X2, etc. and A F -AiFi, A 2 F 2 , etc. By means of these lines
we determine a point B Q , whose ^-coordinate measures an angle
of and whose ^-coordinate measures the sine of 2 30,
44
EXERCISES
i.e., the sine of 30; a point ft, whose z-coordinate measures
an angle of 15, and whose y-coordinate measures the sine of
2 15 30, i.e., the sine of 0; a point ft, whose x-coordinate
measures an angle of 30 and whose ^-coordinate measures the
sine of 2 30 30, i.e., the sine of 30, etc. A smooth curve
through the points B Qy BI, ft, etc. is the required graph.
(d) Construct the graph of cos (6/3 + 45) for 6 varying from
to 360 (see Fig. 28).
Y
FIG. 28
We construct, as in example (6) the lines QoJSo, QiXi,
etc. and A Q Y Q , AiYi, A^Y^, etc. By means of these lines we
determine a point B , whose re-coordinate measures an angle of
and whose y-coordinate measures the cosine of 0/3 + 45,
i.e., cos 45; a point BI whose ^-coordinate measures an angle
of 15 and whose ^-coordinate measures the cosine of 15/3
+ 45, i.e., cos 50; a point B 2 , whose z-coordinate measures an
angle of 30, and whose y-coordinate measures the cosine of
30/3 + 45, i.e., cos 55, etc. A smooth curve drawn through
the points JS , ft, ft, etc. is the required graph.
52. Exercises.
Construct the graphs of the following functions, for 9 varying from to
360:
1. cos 20.
2. cos 0/2.
3. sin 20.
4. sin 0/2.
6. 2cos0.
6. 2sin0.
$sin0.
i cos 0.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
sin 3 0.
3 sin 0.
3 cos 0/3.
cos 3 0.
i cos 3 0.
3 sin 0/3.
| sin 3 0.
4 cos 3 0.
17.
18.
19.
20.
21.
22.
23.
24.
cos (0).
sin (0 + 90).
cos (0 - 180).
cos (0 + 180).
cos (B - 90).
sin (0 - 90).
cos (0 -f 270).
sin (0 - 270).
OPERATIONS ON GRAPHS 45
53. Operations on graphs. The graph of the function sin 3
associates with any particular value of 6 the same ordinate that
the graph of sin associates with 3 0, a point three times as far
from the origin as 0. Hence to every point A on the graph of
the function sin there corresponds a point B on the graph of the
function sin 3 0, three times as near to the F-axis as the point A.
It follows from this that the graph of the function sin 3 may
be obtained by contracting the graph of the function sin0 in
the direction of the X-axis in the ratio 3:1. If we apply these
same considerations to the general case, we obtain the following
theorem:
THEOREM I. The graphs of the functions sinad, cosctf may be
obtained by contracting the graphs of the functions sin e and cos o
respectively in the direction of the X-axis in the ratio a : 1, if a is a
positive rational number.
Here it is to be understood that if a < 1, the contraction
becomes an enlargement in the ratio 1 : I/a.
The graph of sin (0) associates with each value of 6 the
same ordinate that the graph of sin associates with 0. Hence
to every point A on the graph of sin there corresponds a point B
on the graph of sin (0) obtained by reflecting A in the F-axis
as a mirror.* This leads to the following theorem:
THEOREM II. The graphs of the functions sin (-a), cos (-0) may
be obtained by reflecting the g aphs of the functions sin e and cos e
respectively in the Y-axis as a mirror.
The graph of the function sin (20 + 180) associates with any
particular value of the same ordinate that the graph of sin 2
associates with + 90, a point on the X-axis 90 farther to the
right than 0. Hence, to every point A on the graph of sin 2 there
will correspond a point B on the graph of sin (20 + 180), a dis-
tance of 90 to the left of A. It follows from this that the graph
of sin (2 + 180) can be obtained by translating the graph of
sin 2 in the direction of the negative X-axis a distance of 90.
In the same way, we see that the graph of sin (2 180) may
be obtained by translating the graph of sin 2 in the direction of
the positive X-axis a distance of 90.
* B is said to be the reflection of A in the F-axis as a mirror if the Y-axis,
is the perpendicular bisector of the line AB.
46
OPERATIONS ON GRAPHS
If we carry through the same considerations for the general
case, we obtain the following theorem:
THEOREM III. The graphs of the functions sin (ad -f 6), cos (aB -f- 6)
may be obtained by translating the graphs of sin ao and cos ad respec-
tively a distance of | b/a |,* in the direction of the negative X-axis, if b/a
is positive; in the direction of the positive X-axis, if b/a is negative.
Theorems I, II and III may be used to obtain the graphs of the
functions sin (aO + b) and cos (a0 + 6), where a and b are arbi-
trary rational numbers by translating, reflecting and contracting
(enlarging) the graphs of the functions sin 8 and cos 6 respectively.
This will now be illustrated by examples.
(1) To obtain the graph of the function sin (3 + 30), we first
contract the graph of the function sin 6 horizontally in the ratio
3:1, obtaining in this way the graph of the function sin 3
(see Fig. 29), in virtue of Theorem I.
Y
FIG. 29
We now translate the graph of sin 3 a distance of 10 to the
left and obtain in this way, in virtue of Theorem III, the graph
of sin (3 [6 + 10]) = sin (30 + 30).
Y
[XT
FIG. 30
(2) To construct the graph of the function sin (90 0), we re-
flect the graph of sin in the F-axis as a mirror, which gives us
the graph of sin (0), in virtue of Theorem II (see Fig. 30).
* The symbol | b/a \ designates the numerical value of b/a. Thus, if a = 1,
b = 90, | 6/0 | = | -90 | = 90; similarly, | -4 | = 4, | -f | = f.
APPLICATIONS OF GRAPHS 47
Translating this latter graph a distance of 90 to the right, we
shall obtain, in virtue of Theorem III, the graph of sin ( [0 90])
= sin (90 - ff).
64. Exercises.
Construct the graphs of the following functions:
1. cos (90 + 0). 7. cos (0 + 180). 13. cos (3 0-60).
2. sin (90 - 0). 8. cos (180 - 0). 14. sin (6 - 180).
3. cos (90 - 0). 9. sin (270 + 0). 15. sin (3 -K45).
4. sin (0 - 90). 10. sin (270 - 0). 16. cos (90 - 2 0).
6. sin (90 + 0). 11. sin (2 0+60). 17. sin (60 +30).
6. cos (0-90). 12. cos (2 0-90). 18. cos (180 - 40).
56. Applications of graphs* The graphs of the trigonometric
functions sin (ad + 6) and cos (aO + 6), which we have learned
to construct in the foregoing paragraphs, will now be used to
illustrate and verify some of the important properties of the func-
tions sin 6 and cos 6.
(1) The functions sin ax and cos ax are periodic functions
whose period is equal to 2 TT/ | a { , where a represents any rational
number. (See footnote on p. 46 and 26.)
(2) The graphs of the functions sin (6) and sin 6 are symmet-
ric with respect to the -Y-axis, i.e., for any value of 0, the corre-
sponding values of these functions are equal but opposite in sign.
We conclude therefore:
(1) sin (-0) = -sln0, for every value of 6.
The graphs of the functions cos ( 6) and cos coincide, so
that we have:
(2) cos (-0) = cos 0, for every value of 0.
Formula (1) shows an analogy between the sine function and an
odd power of a variable; for we know that, e.g., ( 0) 15 = 15 .
On the other hand formula (2) shows an analogy between the
cosine function and an even power, for, e.g., ( 0) 16 = 16 . On
account of this analogy, formulae (1) and (2) are frequently
expressed in the form " the sine is an odd function," and " the
cosine is an even function."
(3) The graph of the function sin (90 0) coincides with the
graph of cos 0; from this we conclude:
(3) sin (90 - 0) = cos 0, for every value of 0. (See 19, Ex. 1.)
48
GRAPHS OF TAN 6 AND COT 6
Formula (3) as well as most of those which appear in the exer-
cises below can also be obtained by the use of Theorem I in Chap-
ter II. The advantage gained by deriving them by the present
method lies in the fact that it emphasizes more sharply the fact
that these relations hold for every value of the variable, i.e., that
they are properties of the trigonometric functions.
66. Exercises.
Prove graphically:
1. sin (180 6) = sine.
2. sin (0 - 180) - -sin*.
3. sin (B + 180) - -sin*.
4. cos (90 + *) = sin 0.
5. cos (0 90) - sin 0.
6. cos (180 +0) = -cos*.
7. sin (90 + *) = cos B.
8. sin (e 90) = cos*.
9. cos (90 -0) = sin*.
10. cos (180 - 0) = -cos*.
11. cos (0 - 180) - -cos $.
12. sin (0 + 270) -cos 0.
13. cos (0 -f270) =sinfl.
14. sin (270-*) - -cos*.
15. cos (270 - 0) - -sin *.
16. cos (0 - 270) - -sin 0.
17. sin (0 - 270) - cos 0.
18. sin (0 - 45) + sin (0 -f 45) - >/2 sin 9.
19. sin (0 + 45) - sin (0 - 45) - V5 cos *.
20. sin -f cos =* V2 sin (0 -}- 45).
67. Graphs of tan0 and cot0. We proceed now to a brief
consideration of the graphs of he functions tan and cot 6. For
this purpose we begin by developing a graphical method for
determining an ordinate which measures the tangent of a given
angle 0, analogous to the method developed for the sine in 60.
V
We return to the auxiliary circle used in 50 and 61 and draw a
line TT' tangent to this circle at the right hand extremity of
its horizontal diameter (see Fig. 31"). At the center of the circle
GRAPHS OF TAN AND COT $
49
we draw an angle 6 whose initial side lies in the positive direction
along the X-axis. From the point S, where the terminal side
of 6 or, for angles in II and III, the terminal side of produced
through the origin, cuts the tangent line TT, we draw a perpen-
dicular SR upon the X-axis. Since for S the abscissa is equal
to unity the tangent of 6 is measured by the ordinate of S, i.e., by
the line RS. Through S we draw a line SS' parallel to the X-axis.
Furthermore, we lay off on the X-axis, as in 50, a distance OA,
representing the angle 6} through A we draw a line AA' parallel
to the F-axis, meeting the line SS' in B] this point B is then a
point on the graph of tan 6.
Now, following the general method explained in 61, we readily
construct the graph of the function tan (aO + 6).
To obtain an ordinate measuring the cotangent of 0, we draw a
line UU' tangent to the circle at the left hand extremity of its
horizontal diameter. We lay off the angle at the center of the
circle, with its initial side in the positive direction along the
F-axis, and determine the point S where the terminal side of 0, or
the terminal side produced through the origin, meets this tangent
line; from S we drop a perpendicular SR upon the X-axis. It is
clear (see Fig. 32) that if we consider this position of as the
U
u
\ ,. i\
FIG. 32
standard position, the ordinate for the point S is equal to unity,
so that the cotangent of B is measured by the abscissa, i.e., by
the line RS, no matter how large the angle may be. Through
S we draw a line SS' parallel to the X-axis; we lay off, as before,
a distance OA on the X-axis, measuring the angle 0, draw the
line A A 1 parallel to the F-axis and determine the point B, in
50 GRAPHS OF TAN AND COT (aB+b)
which the lines SS f and A A' meet. This point B is a point on
the graph of cot B. The graph of the general function cot (aB + 6)
is now constructed by the methods used in 61.
58. Exercises.
Construct the graphs of the following functions, for B varying from to
360:
1. tan 26. 7. tan 0/3. 13. 2tan0.
2. cot 0/2. 8. cot 30. 14. 3 tan 0/3.
3. tan (0 -{- 90). 9. cot (0 -f 270). 16. tan (2 -f 90).
4. cot (0-90). 10. tan (0-270). 16. cot (2 - 180).
6. tan (0 + 180). 11. tan (180 -0). 17. cot (180 + 2 0).
6. cot (0 + 180). 12. cot (180 - 0). 18. tan (180 - 2 0).
69. Graphs of tan (aB + b) and cot (aB + 6). Mere repetition
of the arguments of 63 will enable us at once to establish the
following theorems, analogous to Theorems I, II, III of 53.
THEOREM la. The graphs of the functions tana0, cota0 may be
obtained by contracting the graphs of the functions tan and cot
respectively in the direction of the X-axis in the ratio a : 1, if a is a
positive rational number; if a < 1, the contraction becomes an en-
largement in the ratio 1 : I/a.
THEOREM Ha. The graphs of the functions tan (0), cot (0) may
be obtained by reflecting the graphs of the functions tan and cot
respectively in the Y-axis as a mirror.
THEOREM Ilia. The graphs of the functions tan (o0 + 6), cot
(ad + b) may be obtained by translating the graphs of the functions
tan a0 and cot o0 respectively a distance of | b/a \ , in the direction of
the negative X-axis, if b/a is positive; in the direction of the positive
X-axis, if b/a is negative.
These theorems enable us to obtain the graphs of functions of
the form tan (aB + 6) and cot (aB + fr), where a and b are arbitrary
rational numbers, by translating, reflecting and contracting (en-
larging) the graphs of tan B and cot B respectively.
(a) To construct the graph of tan (2 B + 45), we first contract
the graph of tan0 horizontally in the ratio 2 : 1, so as to obtain
APPLICATIONS
51
the graph of tan 2 0. Then we translate the latter graph a dis-
tance of 22| in the direction of the negative X-axis. (See Fig, 33.)
FIG. 33
(6) To construct the graph of cot (180 - 2 0), we reflect the
graph of cot in the F-axis as a mirror, and contract the latter in
the ratio 2 : 1 in the direction of the X-axis, obtaining in this
way the graph of cot (20). If we now translate this graph a
distance of 90 in the direction of the positive X-axis, we obtain
the graph of the function cot (-2 [0 - 90]) = cot (180 - 20),
which proves to be identical with the graph of cot (20). (See
Fig. 34.)
Y
FIG. 34
60. Applications. The graphs of the functions tan (ad + b)
and cot (aO + 6) may now be used to derive some of the prop-
erties of the functions tan and cot 0, by the method explained
in 55. We obtain in this way the following results:
(1) the functions tan ad and cot a0 are periodic functions whose
period is TT/| a |.
52 GRAPHS OF SEC 6 AND COSEC 6
It is to be observed that the period of the tangent and cotangent
functions is half as large as that of the corresponding sine and
cosine functions.
(2) the tangent of w/2 and 3 if/2, and the cotangent of and TT
do not exist (see Theorem IV, Chapter II). Moreover
Lim tan = + oo . Lim tan = oo .
Lim cot 6 = oo . Lim cot = +00 . (See 24, 26.)
0-180- 6 >1SQ+
(3) the tangent and cotangent are odd functions, i.e.,
tan ( 0) = -tan 0, cot ( 0) cot e.
(4) tan (180 - 0) = - tan 0, etc.
tan (90 0) = cot 0, etc. (See exercises below.)
61. Exercises.
Construct the graphs of the following functions:
1.' tan (90 -f 0). 6. tan 2 0. 9. tan (3 6 - 90).
2. cot(-0). 6. tan (180 +0). 10. tan (0/2 -f- 90).
3. tan (90 - 6). 7. cot (0 - 180). 11. cot (2 - 90).
4. cot (90 +0). 8. cot 0/2. 12. tan (90 -20).
Prove graphically:
13. tan (0 -f 180) = tan 0. 19. tan (0 - 180) = tan 0.
14. cot (0 -f 180) = cot 0. 20. cot (270 -f 0) - -tan 0.
15. tan (90 + 0) -cot 0. 21. tan (270 - 0) - cot 0.
16. cot (90 - 0) = tan 0. 22. cot (0 - 270) = -tan 0.
17. tan (0 -f 270) = -cot 0. 23. tan (45 + 0) = cot (45 - 0).
18. cot (90 + 0) = -tan 0. 24. tan (45 - 0) cot (45 + 0).
62. Graphs of sec and cosec 0. For the sake of complete-
ness, we add a method for the construction of the graphs of the
functions sec 6 and cosec 6. For the former, we use the diagram
of Figure 31. Since for the point S, the abscissa is equal to unity,
the secant of is measured by the radius vector MS. If we
describe an arc with M as center and MS as radius we may deter-
mine on the vertical diameter of the circle a point Si, whose
GRAPHS OF SEC0 AND COSEC
53
perpendicular distance from the X-axis is equal in magnitude
and direction to sec 0.* Through Si we draw a line SiSi parallel
to the X-axis meeting the line AA f in a point B, which is a point
on the graph of sec 0. The graph is completed by means of the
method of 51 (see Fig. 35).
FIG. 36
For the graph of the cosecant function we use a diagram similar
to the one used for the cotangent (see Fig. 32). The ordinate of S
* Here it is to be understood that 1 is to be determined on the positive
half of the diameter, if S lies on the terminal side of 0; on the negative half of
the diameter if S lies on the terminal side of 6 produced through the origin.
54 EXERCISES
being equal to unity, the cosecant of 6 is measured by the radius
vector MS. Describing an arc with center at M and radius
equal to MS, we find on the vertical diameter of the circle a point
Si whose distance from the X-axis is equal in magnitude and
direction to the cosecant of 6 (see footnote on p. 53). Drawing
a line SiSi parallel to the X-axis, we determine upon the line
A A', drawn parallel to the F-axis, a point B which is a point of
the graph of cosec 0. Repeating this construction for various
values of 6, as explained in 51, we obtain the graph of the function
cosec (see Fig. 36).
For the functions sec 6 and cosec 6 we can now develop theorems
analogous to theorems I, II, III of 53, by means of which the
graphs of the general functions sec (aB + b) and cosec (ad + b)
can be obtained by translating, reflecting and contracting (enlarg-
ing) the graphs of sec 6 and cosec 0. This in turn enables us to
prove graphically various properties of these functions.
63. Exercises.
Construct the graphs of the following functions:
1. sec 2 e. 6. cosec 0/2. 9. cosec 3 0.
2. sec (90 + 0). 6. cosec (180 + 0). 10. sec (0 + 270).
3. cosec (-0). 7. cosec (0 - 90). 11. cosec (0/2 + 90).
4. sec(-0). 8. sec (3 - 270). 12. sec (180 - 0).
Prove graphically:
13. The secant function is an even function.
14. The cosecant function is an odd function.
15. sec (90 - 0) = cosec 0. 23. sec (180 - 0) = - sec 0.
16. cosec (90 - 0) = sec 0. 24. sec (0 - 180) = - sec 0.
17. sec (180 + 0) = - sec 0. 25. sec (270 + 0) = cosec 0.
18. cosec (180 4. 0) - cosec 0. 26. cosec (270 -f 0) = - sec 0.
19. sec (90 + 0) ~ - cosec 0. 27. cosec (180 - 0) = cosec 0.
20. sec (0 - 90) = cosec 0. 28. cosec (0 - 180) = - cosec 0.
21. cosec (90 + 0) sec 0, 29. sec (0 - 270) = - cosec 0.
22. cosec (0 - 90) * - sec 0. 30. cosec (0 - 270) sec 0.
CHAPTER VI
THE ADDITION FORMULAE
64. A special case. In the preceding chapter we have proved
formulae like sin (0 + 90) = cos 6; cos (90 - 0) = sin 0, etc.
We next inquire how the trigonometric functions of the sum and
difference of any two angles may be expressed in terms of the
functions of these angles; i.e., we ask in the first place for formulae
for sin (a + 0), sin (a - 0), cos (a + 0) and cos (a - 0) in terms
of sin a, cos a, sin and cos 0.
We begin by deriving in a slightly different manner some of
the formulae mentioned above. From the graphs of the functions
sin (a + 90) and cos a; cos (a + 90) and sin a, we have already
concluded, that
(1) sin ( + 90) = cos ; and that (2) cos ( + 00) = -sin a.
If in these formulae we replace a by a 90, we obtain
(3) sin a = cos (a - 90), and (4) cos a = -sin (a- 90),
which may also be derived directly from the graphs.
65. Addition formulae for the sine and the cosine. We pro-
ceed now to the general case. We place the angle a in standard
position (see Fig. 37) and let the initial side of coincide with the
terminal side, OP, of a. Then the angle a + will be in standard
position and its terminal side will coincide with the terminal side
of 0.
Let the circle of unit radius about cut the terminal side of
in A; let the perpendicular from A upon OP (or upon PO pro-
duced through 0) meet the line OP in B, and let the ordinate and
abscissa of A be A' A and OA f respectively. Then we have, since
OA = 1,
(1) cos (a + 0) = OA' sin (a + 0) = A'A,
(2) cos = OB sin = BA.
OB is a segment of the directed line OP and BA is a segment of
the directed line BQ which makes with OP an angle of +90.
55
56 ADDITION FORMULAE FOR THE SINE AND THE COSINE
From the Corollary to Theorem II in Chapter I, it follows that
the projection of OA upon each of the coordinate axes is equal to
the sum of the projections of OB and BA. In virtue of Theorem
II of Chapter II the projections of these different lines may be
FIG. 37
obtained by multiplying them by the cosines of the angles between
the directed lines on which they lie and the axes upon which they
are projected. Now, the projections of OA upon the X- and
F-axes are respectively equal to OA' and A' A, i.e., to cos (a + ft)
and sin (a + 0), in virtue of (1). Moreover:
OP makes with the -XT-axis an angle equal to a.
OP makes with the Y-axis an angle equal to a 90.
BQ makes with the X-axis an angle equal to a + 90.
BQ makes with the F-axis an angle equal to a.
Hence, we have, making use also of (1) ad (2) :
cos (a + 0) = cos ft cos a + sin cos (a + 90),
sin (a + 0) = cos |3 cos (a 90) + sin /3 . cos a.
If we now make use of formulae (2) and (3) of 64, we obtain
the following important result:
(3) cos (a H- 0) = cos a cos ft sin a sin 9
sin (a + ft) = sin a cos ft + cos a sin 0;
EXERCISES 57
i.e., the cosine of the sum of two angles is equal to the product of the
cosines of these angles diminished by the product of their sines;
the sine of the sum of two angles is equal to the product of the sine
of one by the cosine of the other plus the product of the sine of the
other by the cosine of the first.
Formulae (3) are known as the addition formulae for the sine
and cosine functions. The proof given here is entirely inde-
pendent of the quadrant in which the angles lie. The student
should however, carry the proof through for various positions of
the terminal sides of the angles.
These addition formulae may now be used in the first place to
derive some of the other results of Chapter V. We have, for
instance:
sin (a + 180) - sin a cos 180 + cos a sin 180 - -sin a,
cos (a + 180) = cos a cos 180 - sin a sin 180 = -cos a.
66. Subtraction formulae. In the second place we use the
addition formulae to express the sine and cosine of a ft in terms
of the sine and cosine of a. and ft. We know from 55, (2) that the
sine is an odd function and that the cosine is an even function, i.e.,
that
sin ( ft) = sin /3, cos ( ft) = cos ft.
We replace ft by ft in the addition formulae, and make use of
the above formulae. In this way we find:
COS (a 0) = cos [a + ( 0)1 = cos a cos (48) sin a sin ( /3)
= cos a cos 4- sin a sin /3,
sin (a |3) = sin [a -f ( /3)] = sin a cos ( 0) + cos a sin ( 0)
= sin a cos ft - cos a sin 0.
67. Exercises.
1. Express sin 75 and cos 75 in terms of the ratios of 45 and 30, and
use the results for calculating the ratios of 75.
2. Calculate the ratios for 105.
3. Calculate the ratios for 15. Compare the results with those obtained
from the tables.
4. Determine the angles in the third quadrant whose ratios may be
calculated without the use of tables, by means of the addition and subtraction
formulae.
58 ADDITION AND SUBTRACTION FORMULAE
Verify the following formulae by means of the addition and subtraction
formulae: %
6. cos (180 - 0) = -cos 6. 7. cos (270 db 0) =t sin 0.
6. sin (180 - 0) = sin 0. 8. sin (270 - 0) = - cos 0.
9. sin (0 + 45) + sin (0 - 45) = V2 sin 0.
10. cos (B 4- 45) + cos (0 - 45) - V2 cos 0.
11. sin (0 + 30) + cos (6 + 60) = cos 0.
12. sin (0 - 60) = -cos (0 + 30).
13. cos (45 + 0) sin (45 -0) = } V (cos - sin 0).
14. sin (45 + 0) = cos (45 - 0) = jV2 (cos -f sin 0).
15. cos (a -f 0) COS (a 0) = COS 2 a -f cos 2 1.
16. sin (a + 0) sin (a - 0) = sin 2 a sin 2 = cos 2 cos 2 a.
17. cos (nir + 0) = ( !) cos0. 19. cos (nir - 0) = ( 1) cos0.
18. sin (WTT + 0) = (-1)* sin 0. 20. sin (WTT - 0) = (-1JW+ 1 sin 0.
21. cos [(2 n + 1) T/2 - 0] = (-!) sin 0.
22. sin [(2 n + 1) r/2 - 0] = (-!) cos 0.
23. cos [(2 n + 1) 7T/2 + 0] = ( ~ D* +1 sin 0.
24. sin [(2 n + 1) 7r/2 + 0] = (-!) cos 0.
68. Addition and subtraction formulae for the tangent and
cotangent. Since tan = sin 0/cos and cot = cos 0/sin
(Why?), the addition and subtraction formulae for the sine and
cosine enable us to obtain corresponding formulae for the tangent
and cotangent, viz.,
, , ON sin (a + P) sin a cos /3 + cos a sin
tan (a + - - - - -
v
- , , ^r - 5 - : - : - ^>
cos (a + ft) cos a cos p sin a sin /?
and
COS (Q! + ff) _ cos ce COS g sin a sin j8
( . _
^ a
sin (a + j8) sin a cos )8 + cos a sin |
The results may be expressed in terms of tangents alone by
dividing the numerators and the denominators of each of these
fractions by cos a. cos 0, or in terms of cotangents alone by
dividing them by sin a sin ft. In this manner we find:
sin a cos , cos a sin
, / \ o\ cos a cos cos a cos __ tan -f tan
an (a + 0; - cos a cos ^ ^ sin a sin "~ 1 - tan a tan 0*
cos a cos cos a cos
DOUBLE AND HALF-ANGLE FORMULAE 59
and
cos a cos ff sin a sin ff
__
t ( 4- ti\ s * n a s * n ff s * n a sm ft _ Cot a COt /9
sin a cos ff , cos ex sin ff cot a + cot
sin <a sin /3 sin a sin /3
Starting with the formulae
and proceeding in exactly the same way as above, we find:
*/ rt tan a - tan fl cot a cot g + 1
and **(-*)
69. Exercises.
1. Determine can 75 and cot 75 by means of the addition formulae for
the tangent and cotangent.
2. Determine Ian 15 J and cot 15 by means of the subtraction formulae.
3. Determine tan 105 and cot 105. (It would not be advisable to
write 90 4- 15 in place of 105 in this problem. Why not?)
1 -f tan 6 cot -f 1
4. Prove that tan (45 + 0) = cot (45 - 6)
5. Also that tan (45 - 0) = cot (45 + 0) =
1 tan cot
1 tan cot 1
tan cot -f 1
6. Express tan (a -f- ft) in terms of the cotangents of a and 0.
7. Express cot ( + 0) in terms of tan a and tan /3.
8. Prove that tan (30 + 0) = cot (60 - 0) = * * 3tang .
v 3 tan
9. Prove that tan (60 + 0) = cot (30 - 0) = 1 + 3co **.
cot - v 3
70. Double angle formulae and half-angle formulae. From
the addition formulae we derive, by putting a = # = 0, the fol-
lowing formulae, by means of which our knowledge of the ratios
of any angle enables us to find the ratios of an angle twice as
large; in other words, formulae which express the ratios of any
angle in terms of the ratios of an angle half as large; we find:
60 DOUBLE AND HALF-ANGLE FORMULAE
COS 2 6 = cos (0 -f 0) = cos cos sin sin = COS 2 6 sin 2 0.
sin 2 6 = sin ((9+0) = sin cos + cos sin = 2 Sin COS 0,
tan 4- tan 2 tan
tan 3 =
1 - tan tan 1 - tan 2
If we put = </2, and therefore 2 = 0, we obtain an equiva-
lent form of these formulae, bringing out more vividly the fact
that they express the ratios of an arbitrary angle in terms of the
ratios of one half of that angle, viz.,
sin 6 = 2 sin 6/2 cos 0/2, cos 6 = cos 2 6/2 - sin 2 0/2, tan =
where we have again written 6 in place of </>.
The second of these formulae leads to another important set of
results, in the following manner:
To the two members of the identity
cos = cos 2 0/2 - sin 2 0/2,
we add the corresponding members of the identity
1 = C os 2 0/2 + sin 2 0/2,
yielding the result 1 + cos 6 = 2 cos 2 0/2. (1)
If we subtract the first of these identities from the second we
obtain:
1 - cos = 2 sin 2 0/2. (2)
Upon solving the resulting identities (1) and (2) for cos 0/2
and sin 0/2 respectively, we obtain:
9 t/
! 2 = V'
I 4- cos
formulae which express the ratios of one half of an angle in terms
of the ratios of that angle. The plus or minus sign is to be used
according to the quadrant in which 0/2 falls.
Dividing the second of the latter formulae by the first, we find:
stn0 _ 1 cos
+ cos ~ 1 + cos ~ sin '
the last two forms being derived from the first by multiplying
the numerator and the denominator of the fraction under the
radical sign by 1 + cos and 1 cos respectively. In the last
two expressions the double sign is not necessary, because 1 + cos
FACTORIZATION FORMULAE 61
and 1 cos are always positive, while tan 0/2 and sin always
have the same sign.
71. Exercises.
1. Determine the functions of 30 by means of those of 60.
2. Determine the ratios of 30 by means of those of 15, found in 67, 3.
3. From the ratios of 15, derive those of 7 30'.
4. Obtain a formula which expresses cot 6/2 rationally in terms of sin 9
and' cos 0.
6. Prove that cosec 2 = J sec cosec 0.
* xi x i I 2 sec
6. Prove that sec x = y j-
+ sec
7. Derive the double angle formula for the tangent from the double
angle formulae for sine and cosine.
8. Derive the half angle formula for the tangent from the double angle
formula for the tangent.
72. Factorization formulae. We return once more to the addi-
tion formulae for the sine and cosine in order to derive from them
a set of formulae which will enable us to convert the sum (or the
difference) of the sines or cosines of two angles into a product.
Such a conversion is of great importance, as is evident from
Theorem I of Chapter III, whenever we wish to carry out loga-
rithmic calculations with expressions which involve sums or differ-
ences of sines or cosines, and also for many theoretical purposes.
We know:
sin (a + ft) = sin a cos ft + cos a sin ft,
and sin (a ft) = sin a cos ft cos a sin ft.
Adding the corresponding members of these two identities, we
find:
(1) sin (a + ft) + sin (a ft) = 2 sin a cos ft.
Subtracting them, we find:
(2) sin (a + ft) sin (a: ft) = 2 cos a sin ft.
These formulae can be put in a slightly different form, more
useful for the purpose for which we are deriving them, by putting
62 EXERCISES
a + ft = 6 and a ft = <, whence we obtain by addition and
subtraction
a - 9 .* and fl- g -*.
"""
Substitution of these values f or a + ft, a , a and /3 in formulae
(1) and (2) gives us the following results:
sin -f sin $ 2 sin -.
Q _ A
and sin sin = 2 cos -^-2 sin
<
We now proceed in exactly the same manner with the formulae
cos (a + 13) = cos a cos ft sin a sin &
and cos (a ft) = cos a cos /? + sin a sin 0,
and find the formulae:
__
cos 4- cos = 2 cos ' cos
i ~m < m <
and cos - cos - 2 sin ' ' sin *
A A
73. Exercises.
Convert the following sums and differences into products:
1. sin 55 + sin 65. 4. cos 87 + cos 42.
2. cos 75 - cos 15. 6. cos 312 - cos 252.
3. sin 27 - sin 18. 6. sin 213 + sin 237.
Calculate by means of logarithms:
7. (sin 49+ sin 35) (cos 37 -cos 51). 9. (cos 137 + cos 84J'.
sin 57 + sin 24 , 1(K V cos 306 - cos 246.
sin 57 - sin 24
Prove the following identities:
sin + sin A . 6 -\- <j> ,6
11. - r ~ = tan ........ ' cot
- r ~ ........
sni B sin <f> 2 2
> _
cos ^ -f cos 2
13 sin- + sin 4. ^>
cos 0-f cos 2
4 ,. C080
MISCELLANEOUS EXERCISES 63
74. Summary of formulae proved in Chapter VI.
1. cos (a + ft) = cos a cos ft sin a sin 0, see 66.
2. sin (a + 0) = sin a cos + sin /3 cos a, see 66.
3. cos (a 0) = cos a cos + sin a sin 0, see 66.
4. sin (a /3) = sin a cos /3 sin cos a, see 66.
R J- / 1 ON <* + CO
5. tan (a + j8) = ; - 1 - 1 5: > see 68.
v y 1 tan a tan '
f* j. / r>\ tan a tan /3 -
6. tan (a /3) = r - 7 ^~, see 68.
v 7 1 + tan a tan
7. cos 2 (9 = cos 2 sin 2 0, see 70.
8. sin 2 = 2 sin cos 0, see 70.
9. tan20== 2t * n * see 70.
1 tan 2 _
10. cos 0/2 = d=V(l + cos"0)72, see 70.
11. sin 0/2 = V(l cos 0)/2, see 70.
10 1 cos sin _ A
12. tan^r = - . . = - : - r, see 70.
2 sin 1 + cos 0'
t\ i j ^ ^^^ i
13. sin + sin < = 2 sin "T y cos y y see 72.
^ ^
^ I i n _ i
14. sin sin = 2 cos - sin - , see 72.
Li
f\ I i f\ __ i
15. cos + cos </> = 2 cos ^-^ cos r-^, see 72.
A A
/) I JL /J _ JL
16. cos0 cos^ = 2 sin r-^sin r-^, see 72.
75. Miscellaneous Exercises on Chapter VI. (The more diffi-
cult examples are marked with a *.)
1. Determine the ratios of 165 (a) by means of the addition formulae;
(6) by means of the half-angle formulae.
2. Prove that tan 7r/8 = V2 1.
3. Prove the identity: sin (60 -f 0) cos (30 -f 6) = sin 0.
A - . . (sin 67 - sin 34) sin 23
4. Calculate: 7^3-^
cos 17
Prove the following identities:
6. cos 3 = 4 cos 3 3 cos
6. sin 3 6 3 sin 4 sin 3 9.
Q . sin 2 a sin 2
*7. tan (+)- -, ^ -
sin a cos a sin /3 cos /3
8. sin 2 3 sin 2 2 = sin 5 sin 0.
64 MISCELLANEOUS EXERCISES
3 tan tan 3
9. tan 3 = '
1 3 tan 2 6
11. sin (a 4- /3 4- 7) = sin a cos ft cos 7 + sin ft cos <x cos 7 + sin 7 cos a
cos j3 sin a sin ft sin 7.
12. cos (a 4- ft 4- 7) = cos a cos cos 7 sin a sin cos 7 sin sin 7
cos a sin a sin 7 cos /3.
*13. tan a tan + tan ft tan 7 -j- tan 7 tan a = 1, provided a + ft + 7 = 90.
14 sin 2 a cos __ , a
1 4* cos 2 a 1 + cos a 2
16. sin 4- sin (0 -f 2 /3) + sin (04-4 *-/3) = 0.
17. cos + cos (0 + 2 7T/3) + cqs (0+ 4 *-/3) = 0.
18. cos (a + 0) + cos (a 0) -f cos (-or + 0) + cos (-a 0) = 4 cosa COS0.
l-tan 2 /2
19 * 1 4- tan 2 a/2
*20. tan 2 a tau - ,
cos a 4- cos 3 a
*21. sin 2 a 4- sin 2 4- sin 2 7 = 4 sin a sin sin 7, if a 4- + 7 180.
22. sin 3 = 4 sin sin Or/3 4- 0) sin ( ff /3 - 0).
23. Calculate the functions of 3 -rr/S.
*24. tan ( v /4 4- 0/2) + cot (ir/4 -j- 0/2) = 2 sec 0.
*26. cos 2 a 4- cos 2 4- cos 2 7 = 1 4 cos a cos ft cos 7, if a 4- +
7 = 180.
CHAPTER VII
THE SOLUTION OF TRIANGLES
76. The Law of Sines ; the area of a triangle. We consider
in this chapter the problem of "solving an arbitrary triangle
ABC" i.e., the determination of the unknown elements of a tri-
angle ABC of which sufficient elements are given. The problem
is solved by the use of various relations subsisting between the
sides and angles of a triangle.
Denoting the length of the perpendicular dropped from the
vertex A upon the opposite side BC by h a , and using analogous
notations for the other perpendiculars, we have, in virtue of
Theorem I of Chapter IV (see Figs. 38 and 38a) :
a
FIG. 38
(1) h a = csinB = frsinC, hb = asinC = csinA,
h c = b sin A = a sin B*
From these formulae we derive:
(a) THEOREM I. The sines of the angles of a triangle are propor-
tional to the sides opposite the angles.* (LAW OF SINES.)
* It is a familiar theorem of plane geometry that of two unequal sides of
a triangle the greater side lies opposite the greater angle. The law of sines
may be looked upon as completing this theorem by stating how the unequal
sides are related to the angles opposite them.
65
66 TWO ANGLES AND ONE SIDE
To prove this law, we divide the two expressions for h a in equa-
tions (1) by 6c, those for A& by ac. In this way we obtain:
sin B/b = sin C/c = sin A/a,
which was to be proved.
(b) THEOREM II. The area of a triangle is equal to one half of the
product of any two sides multiplied by the sine of the angle Included
by them*
Proof. Denoting the area of triangle ABC by A, we have:
A = a/2 h a = 6/2 h b = c/2 h c .
If in these expressions we substitute for the altitudes ha, hi> and
h c the values which they have in equations (1), we find
A = | ab sin C = \ be sin A = J ca sin B t
which was to be proved.
77. Two angles and one side. A triangle is determined when
two angles and a side are given. The law of sines suffices to deter-
mine the remaining elements in this case. For, suppose that A,
Bj and a are given; we have then C = 180 (A + B).
Moreover sin B/b = sin A/a and sin C/c = sin A/a; hence
, a sin B , a sin C
5 = . an( j c __ __ ,
sin A sin A
which completes the solution.
If the given elements are measured by short numbers the cal-
culation may be carried out directly by means of a table of natural
values; the expressions for b and c, however, are well adapted to
calculation by means of logarithms.
Example.
Given, a = 43.257, A = 57 23', C = 49 47'.
Required. B, b and c.
Solution. B = 180 - (A + B) = 180 - 107 10' = 72 50'.
6 a . a sin 5 43.257 sin 72 50'
i.e., o =
sinB sin A' ' sin A sin 57 23'
EXERCISES
67
- 7? = - T, i.e., c -
sm C sin A
log 43.257= 1.63606
log sin 72 50'= 9.98021-10
a sin C = 43.257 sin 49 47'
sin A sin 57 23'
log 43.257= 1.63606
log sin 49 47' = 9.88287 - 10
11.61627-10
log sin 57 23'= 9.92546-10
log 6= 1.69081
6=49.069
11.51893-10
log sin 57 23'= 9.92546-10
log c = 1.59347
c =39.216
8
78. Exercises.
L A = 39 27', B
2. C = 30
108 51', b = .43215. Determine C, a, c.
: 45, c = 123. Determine B, a, 6.
3. = 27 45' 15", C = 89 19' 20", c - 14.302. Determine A, a, 6.
4. A = 37 12' 30", C = 58 26' 40", a - 103.47. Determine B, 6, c.
6. We wish to determine the distance from a point A to a point B, situ-
ated in a marsh, visible but not accessible from A (see Fig. 39). For thia
"" 750 feet u
FIG. 39
purpose, we measure the distance from A to a point C, from which both A
and B are visible and we measure the angles BAG arid ACS. Calculate AB t
if AC = 750 feet, Z.BAC = 32 27' and Z.ACB = 47 12'.
6. Devise a method for finding the distance from a point A on one bank
of a river to a point B on the opposite bank.
T
FIG. 40
7. To determine the height of a tower TS (see Fig. 40) we measure the
angle of elevation of its top T from two points A and B lying on the same
aide of T, in the same vertical plane with T and a known distance apart.
68 TWO SIDES AND AN ANGLE OPPOSITE ONE OF THEM
Calculate the height of the tower, if we find that AB = 257 feet, Z. TAB
19 43' and Z TBS = 47 29'. (Compare the present method of solving
this problem with the method used for solving problem 10 in 49.)
8. Solve problem 10 in 49 by the method used in the preceding problem.
79. Two sides and an angle opposite one of them. If two
sides of a triangle and an angle opposite one of them are given,
the triangle is not completely determined. Let there be given
the sides a and b and the acute angle A (see Fig. 41). To con-
struct the triangle ABC we lay off a line equal to b on one of the
legs of the angle A. The vertex C is then located. We then
strike an arc with C as a center and a as a radius. The intersec-
tion of this arc m with the second leg AX of angle A determines
the third vertex of the triangle, B.
It is now clear that if a is shorter than the perpendicular dis-
tance p from C to AX, as in Figure 42a, then the arc m will not
FIG. 42a
PIG. 42b
intersect AX at all, so that no triangle can be constructed. If a
is equal to the perpendicular p, as in Figure 42b, we obtain a single
right-angled triangle. If a is greater than p, but less than b (see
Fig. 42c), then the arc m will cut AX in two points, B and B' 9
TWO SIDES AND AN ANGLE OPPOSITE ONE OF THEM 69
and we will obtain two triangles, ACB and ACB', both of which
will satisfy the requirements of the problem. Since CB = CJS',
Z CB'B Z. CBB'j and therefore the angles B and E' occurring
in triangles ABC and AB'C respectively are supplementary angles.
If a > b (see Fig. 42d) the arc m will meet AX again at two points
l
FIG. 42d
FIG. 42e
B and B', but one of these points will fall on XA produced, so
that the triangle AB'C, to which it gives rise, does not contain
the given angle A. Hence in this case there is only one triangle
which satisfies the requirements of the problem, viz., triangle ABC.
If the given angle A is obtuse the construction of triangle ABC
proceeds in the same way as before (see Fig. 42e). It is evident
that in this case there is no solution possible, unless a > b. We
observe moreover that in all the cases which we have considered
the perpendicular p is equal to 6 sin A. Hence we can put the
results of this discussion in the following form:
THEOREM III. If two lines and an angle, such as a, 6, A are given,
then there may be no triangle, one triangle or two triangles of which
the given lines are sides and of which the given angle is an angle
opposite one of these sides, viz.,
if a < b sin A, there is no triangle;
if a = b sin A and A is acute, there is one triangle;
if a > 6 sin A 9 a < b and A is acute, there are two triangles;
if a = 6 and A is acute, there is one triangle;
if a > 6, there is one triangle.
The different cases mentioned in this theorem are illustrated
in the examples which follow. The solution proceeds in each case
according to the following plan:
70 TWO SIDES AND AN ANGLE OPPOSITE ONE OF THEM
r) mn >4
From the law of sines, we find: sin B = , which enables
ct
us to determine B. Next we can calculate C, since C = 180
(A + B). Finally we use the law of sines to find c, viz., c = : -j- -
If a < b sin A, we find that sin B > 1; hence no angle B can
be found, and therefore no triangle exists. If a = b sin A, then
sin B = 1, B = 90 and we have a single right-angled triangle.
If a > 6 sin A, then sin B < 1 and we can determine angle B
from the tables. But, besides the acute angle B found from the
tables, we must also consider the supplement of J3, whose sine is
equal to the sine of B] we have therefore B r = 180 B. More-
over if a < by then we must have B > A (see footnote on p. 65);
hence both angles, B and B f , can be used, if A is acute, while
neither can be used if A is right or obtuse. If, on the other hand,
a > b, then we must have B < A, so that only the acute angle B f
found from the tables, can be used.
Example 1.
Given, a = 4.73, b = 18.65, A = 43 27'.
Required, c, B, C.
Solution.
sin B sin A xu f . D 6 sin A 18.65 sin 43 27'
^ = therefore sin B = = . ,_ . >
6 a ' a 4.73
log 18.65 = 1.27068.
log sin 43 27' = 9.83741 - 10,
A
log b sin A = 11.10809 - 10,
log a = log 4.73 == .67486.
We notice that log b sin A > log a; therefore a < 6 sin A.
Hence no triangle can be found with the given elements.
Example 2.
Given, a = 14.73, b = 18.65, A = 43 27'.
Required, c, B, C.
TWO SIDES AND AN ANGLE OPPOSITE ONE OF THEM 71
Solution. We have now
. D 6 sin .4. 18.65 sin 43 27'
sin > = = ., . ,_ >
a 14.73
log 18.65 = 1.27068,
log sin 43 27' = 9.83741 - 10,
A
log 6 sin A = 11.10809 - 10,
log 14.73 = 1.16820,
8
log sin B = 9.93989 - 10.
Here a > 6 sin A, a < b and A is acute. Hence there are two
triangles in this case. Wejind:
First solution Second solution
B = 60 32' 43". B' = 119 27' 17".
We know A = 43 27'. We know A = 43 27'.
Hence C - 180 -(4 + 5) Hence C" = 180 - (A + B)
= 760'17". = 17 5' 43".
Furthermore : Furthermore :
sin C sin A , , , , sin C' sin A . , . ,
= j and therefore ; = > and therefore
c a c a
a sin C = 14.73 X sin 76 17" , a sin C' 14.73 X sin 17 5' 43"
C sin A sin 43 27' " ' sin A sin 43 27'
log 14.73 = 1.16820 log 14.73 = 1.16820
log sin 76 17" = 9.98691 - 10 log sin 17 5' 43" = 9.46829 - 10
A A
1 1 . 1551 1 - 10 10.63649 - 10
log sin 43 27' = 9.83741-10 log sin 43 27' = 9.83741-10
s s
logc= 1.31770 logc'= .79908
c = 20.783 c'= 6.2963
72 TWO SIDES AND AN ANGLE OPPOSITE ONE OF THEM
Figure 43 is a drawing to scale of the triangles ABC and AB'C.
A B 7 B
FIG. 43
Example 3.
Given, a = 24.73, b = 18.65, A = 143 27'.
Required, c, B, C.
7 . . _> & sin A 18.65 sin 143 27'
solution. smB = = - 94770
log 18.65 = 1.27068
log sin 143 27' = log sin 36 33' - 9.77490 - 10
log b sin A = 11.04558 - 10
log 24.73 = 1.39322
8
log sin B = 9.65236 - 10
B = 26 41' 14"
Since a > b and A is obtuse, there is only one solution and B
must be acute. Hence the value found from the table is the
only one that can be used in this case.
C = 180 - (143 27' + 26 41' 14") = 9 51' 46"
a sin C 24.73 sin 9 51 ' 46"
c =
sin A sin 143 27'
log 24.73 = 1.39322
log sin 9 51' 46" = 9.23373 - 10
A
log a sin C = 10.62695 - 10
log sin 143 27' = 9.77490 - 10
8
log c = .85205
c= 7.1130
THE LAW OF COSINES
73
80. Exercises.
1. a - 42.3, 6 - 57.03, A - 35 35'. Determine c, B, C.
2. c = 507.8, 6 = 751.3, B - 23 47'. Determine a, A, C.
3. b = 5.5, c = 4.3, C = 75 29'. Determine a, A, B.
4. a = 3.207, c 7.831, C = 137 18'. Determine 6, A, B.
6. a = 37.052, b = 49.312, A = 19 25'. Determine c, B, C.
6. c .047031, 6 = .047031, C - 28 31'. Determine a, A, B.
7. An island, known to be 75 miles wide, subtends an angle of 40 17'
from a point P, 40 miles distant from one extremity of the island. How far
is P from the other extremity of the island?
8. A flagpole, 10 feet high, subtends an angle of 2 37' from a point A.
If A is 200 feet from the foot of the pole, how far is it from the top?
9. Two lighthouses, M and L are 40 miles apart. At 8.30 A.M. a ship S
leaves M and travels at the rate of 12 miles per hour. At 11 A.M. the distance
between M and L subtends an angle of 33 from the ship. How far is S
from L at that moment?
81. The Law of Cosines. Two sides and the included angle.
Three sides. If three sides of a triangle are given, and also if
two sides and the included angle are given, the triangle is deter-
mined. The law of sines does not suffice, however, to calculate
the unknown elements in these cases. We therefore proceed to
derive a new relation between the sides and angles of the triangle.
FIG. 44
With the notations indicated in Figures 44 and 44a, we have
BD = ai = c cos By EG = a, h a = c sin B.
Using Theorem I of Chapter I, we find, for both Figure 44 and
Figure 44a, CD + DB + BC = 0. Therefore:
02 = CD = - DB - BC = BD - BC = c cos B - a. (1)
Moreover 6 2 = W + a 2 2 = c 2 sin 2 B + a 2 2 . (2)
74 THE LAW OF COSINES
Substituting (1) in (2), we find:
b 2 = c 2 sin 2 B + (c cos B a) 2
= c 2 sin 2 B + c 2 cos 2 B + a 2 - 2 ac cos B;
i.e., b 2 = c 2 + a 2 2 ac cos B,
We have therefore proved the following theorem:
THEOREM IV. The square of one side of a triangle is equal to the
sum of the squares of the other two sides diminished by twice the
continued product of these two sides and the cosine of the angle
included by them. (LAW OF COSINES.)
By dropping perpendiculars from each of the other two vertices
of triangle ABC and proceeding in an exactly similar manner, we
obtain two formulae analogous to the one derived above, viz. ;
c2 = a 2 + b 2 - 2 ab cos C and a 2 = 6 2 + c 2 - 2 be cos A.
In the form in which they are here given, these formulae enable
us to calculate the third side of a triangle of which two sides and
the included angle are known. If we solve them for the cosines
of the angles, we obtain:
A 6 2 +c 2 -a 2 ca-> -c , ON
cos A = 777- - , cos B = - - , cos C= ^ r > (3)
2 be 2 ca 2 ab
in which form they are well adapted to determining the angles of a
triangle whose sides are given.
We observe that none of the formulae derived in the present
section are suited to logarithmic calculation. For this reason
they are useful only for computations which involve short numbers.
Example 1.
Given, a = 14, b = 27, C - 35.
Required, c, A, B.
Solution. By means of Theorem IV we find:
c2 = a * + &2 _ 2 ab cos C - 14 2 + 27 2 - 2 14 27 cos 35;
c 2 196 + 729 - 756 X .81915 = 925 - 619.27740 305.72260.
c = V 305.72260 = 17.484.
THE LAW OF COSINES 75
To complete the calculation, we use the law of sines, and find:
A a sin C , . 6 sin C
sin A = - 1 and sin B = -
c c
log 14= 1.14613 log 27= 1.43136
log sin 35 = 9.75859 - 10 log sin 35 = 9.75859 - 10
- A - A
10.90472 - 10 11.18995 - 10
log 17.484 = 1.24264 log 17.484 = 1.24264
_ a _ a
log sin A = 9.66208 - 10 log sin B = 9.94731 - 10
A = 27 20' 28" or 152 39' 32" and B = 62 20' 34" or 117 39' 26".
Since a < c, we must have A < C y ; hence A = 27 20' 28".
Since b > a and 6 > c, we must have B > A and B > C; both
values of B satisfy this condition; it is clear, however, that with
the smaller value of B, the condition that A + B + C = 180
would not be satisfied; therefore B = 117 39' 26".
Example 2.
Given, a = 10, b 15, c = 19.
Required. A, B, C.
Solution. Using formula (3), we find:
+ ^ ~ # _ 100 + 225 - 361 _ -36 _
roq _
COS C ~
2 oh ~ 300 ~ 300
Since cos C is negative, /. C is obtuse and cos (180 C) =
-cos C = .12000.
We find from the tables, that 180 - C = 83 6' 29"; and hence
C = 96 53' 31".
Furthermore,
c 2 + a 2 - 6 2 361 + 100 - 225 236
COS B = 2m - = - 380 - = 380
therefore B = 51 36' 27".
And finally,
52 + c 2 - a 2 225 + 361 - 100 486
COS A = - 2Tc - = - 570 - = 570
whence we obtain A = 31 30' 4".
Check. A+B + C = 180 0' 2".
76 EXERCISES
82. Exercises.
l.o = 120, b = 150, C = 60. Determine c, A, B.
2. p = 1.3, q = 1.4, r = 1.5. Determine P, Q, 12.
3. a = 17, 6 = 15, c = 29. Determine A, B, C.
4. r = .45, s = .78, T 7 = 45. Determine t, R, S.
6. To determine the width of a lake, the distances of its extreme points
A and B from a point P and the angle subtended by AB at P are measured.
It is found that AP = 750 feet, BP = 600 feet, and ZP = 32.
6. It is desired to make a triangle out of sticks that are 5, 8, and 9 inches
long. What angle should the first two of these sticks make, in order that
the third one may be just long enough to join their free ends?
7. Prove that a 2 = & 2 + c 2 - 2 be cos A.
8. Prove that a 2 -f 6 2 -f c 2 = 2 ab cos C + 2 6c cos A + 2 ca cos B.
9. Prove that cos C = -
83. Summary and critique. In the preceding sections we have
learned to calculate the unknown parts of triangles of which are
given (a) one side and two angles (77); (6) two sides and an
angle opposite one of the sides (79) ; (c) two sides and the included
angle (81); or (d) three sides (81). We know, moreover, from the
study of plane geometry, that if a triangle is to be determined
by sides and angles, then the given elements must form one of
the four sets (a), (6), (c) or (d) enumerated above. Hence the
general problem of " solving a triangle" has been solved in sec-
tions 77-81, in so far as it relates to triangles determined by
means of sides and angles. There are however two criticisms to
be made of the theory developed so far, viz. :
(1) The methods developed in 81 for cases (c) and (d) are not
suited to the use of logarithms and are not very useful, therefore,
in problems involving long numbers.
(2) There are no convenient methods for accurately checking
the calculations in cases (a), (6) and (c).
In order to meet these criticisms some further relations between
the sides and angles of a triangle will now be derived.
84. The law of tangents. From Theorem I, we conclude that
sin A __ a f
SnuB ~6' W
THE LAW OF TANGENTS 77
Adding 1 to both sides of this equation, we obtain:
sin A + sin B a + b
sin B b '
subtracting 1 from both sides of equation (1) gives us:
sin A sin B __ a b
sin B b
(2)
(3)
If we divide the sides of (2) by the corresponding sides of (3), we
find:
sin A + sin B a -f b ...
* ____ . (4)
sin A sin B a b
We now make use of formulae (13) and (14) of 74 to factor
respectively the numerator and denominator on the left hand
side of (4) ; we also divide the numerator and denominator by 2
and obtain:
a + b = sin $ (A + B) cos \(A-B)
a - 6 ~~ sin $ (A - B) cos | (A + E)
= tan 1 (A + B) = cot|C
tan %(A-E) tan f (A - -B)"
To justify the last step we observe that A + B + C = 180.
Hence
\ (A + B) = 90 - * C. (5)
/. tan }(A + B) = tan (90 - i C) = cot i C (see 60, (4)).
The result is most conveniently written in the following form,
tanJ(^-B) =2^| cot JC,
which formula is expressed in the following theorem:
THEOREM V. The tangent of one half the difference of two angles
of a triangle Is equal to the quotient of the difference of the sides
opposite these angles by their sum, multiplied by the cotangent of
one half the angle included by them. (LAW OF TANGENTS.)
78 MOLLWEIDE'S EQUATIONS
85. Mollweide's Equations. If we multiply both sides of equa-
tions (2) and (3) of 84 by the corresponding sides of the equation
sin B
n , , t ,
= _ we fi nc i that
'
m
w
_ -_ _
sm C c '
sin A + sin B a + 6
sin C c
and
sin A sin B __ a b
sin C Y c
(2)
We factor the numerators on the left hand sides of these equa-
tions by means of Formulae (13) and (14) of 74. The denomina-
tors we change by writing sin C = 2 sin C cos C, which is a
consequence of formula (8) of 74. In this way equations (1) and
(2) will assume the following form:
sin \ (A + B) cos $ (A - B) = a + b^ ,
sin % C cos ^ C c
and
sin \ (A. B) cos \ (A + B) _ a b ^ ,..
sin | C cos \ C c
Moreover, from equation (5) of 84, it follows that
sin \ (A + B) = cos -| C
and cos % (A + B) = sin J C, (see 19 and 55).
Consequently, equations (3) and (4) may be simplified to the
final form:
a + b cos \ (A - B) , a - b _ sin | (A - B)
. - ._. and 5 -^
c sin f C c cos f C
These equations are known as Molliveide's equations.
86. Exercises.
1. Prove: tan } (C-A) = -^ cot i B.
C "T" Gt
-. sinHS-C) 6 - c
2. Prove: , . - =
cos \ A a
3. Derive a proof of the law of tangents from Mollweide's equations.
TWO SIDES AND THE INCLUDED ANGLE
79
87. Two sides and the included angle. The law of tangents
enables us to meet the criticism brought forward in 83 with refer-
ence to case (c). Moreover, Mollweide's equations are well
adapted to serve for checking the calculations in cases (a), (6)
and (c), because they involve all the sides and all the angles of
the triangle.
Example.
Given, a = .4503, b = .7831, C = 43 48'.
Required, c, A, B.
Solution. We use the law of tangents to determine angles A
and B:
tan %(B - A) = cot J C = cot 21 54'.
log .3328 = 9.52218 -10
log 1.2334 = .09110
9.43108-10
log cot 21 54' = .39578
8
but,
log tan \ (B - A) = 9.82686-10
|(B-A) = 33 52' 11",
= 90- C = 90-21 54' = 68 6'
A and S.
Therefore,
Z B =
and
Z A =
Furthermore:
(B + A) + J (B - A) = 101 58' 11"
(B + A)-l(B-A)= 34 13' 49".
a sin C .4503 X sin 43 48'
C """ sin A "" sin 34 13' 49"
log .4503 = 9.65350-10
log sin 43 48' = 9.84020-10
19.49370-20
log sin 34 13' 49" = 9.75014-10
log c = 9.74356 -10
c= .55406
S
80 THE HALF-ANGLE FORMULAE
cos I A a
i (B - O = 29 5' 6", \A = 17 6' 55", 6 - c= .22904
log sin i (B-C) = 9.68673 - 10 log (6 - c) = 9.35992 - 10
log cos \A = 9.98032 - 10 log a = 9.65350 - 10
S
9.70641 - 10 9.70642 - 10
88. Exercises.
1. c 27.04, b = 84.31, A = 112 44'. Determine o, B, C.
2. a = 3152, c = 4281, B - 88 27'. Determine 6, A, C.
3. p - .0432, q = .0586, # = 47 36'. Determine r, P, Q.
4. s = .8132, z = .5817, F = 120. Determine y, X, Z.
6. A hill slopes at an angle of 17. A point P is 39.3 feet up the hillside;
a point Q is in the plane, 173.5 feet from the foot of the hill in a line at right
angles to the edge of the hillside. How far is P from Q?
6. The distance from the earth to the sun is approximately 92.9 million
miles, the distance from the earth to the moon approximately 239,000 miles.
What is the distance from the sun to the moon at a moment when the line
sun-moon subtends at the earth an angle of 24 36'?
89. The half-angle formulae for the angles of a triangle. We
start with the first formula for tan \ A given in 70 and multiply
numerator and denominator of the fraction under the radical
sign by 2 be. In this way we find :
, . A l\ cos A A /2 be 2 be cos A ,^
tan \ A = I/ - ; T- = \/ ~~i ^r-r r (I)
2 V 1 + cos A V 2 be + 2 be cos A
If we multiply both sides of formula (3) of 81 by 2 6c, we obtain:
2 be cos A = b 2 + c 2 - a 2 . (2)
We substitute (2) in (1); this gives us:
/2 be + a 2 - 6 2 - c 2 = . /a 2 - (b 2 -"2 be + c 2 )
tan 2 A "" V 2 be + 6 2 + c 2 - a 2 V (6 2 + 2 6c + c 2 ) - a 2
v/
(a + 6 c) (a b + c)
(6 + c + a) (b + e - a) '
THREE SIDES 81
Now, we introduce the abbreviation 2 s for the perimeter of
the triangle; i.e., a + 6 + c = 2s. Subtracting in turn 2 a, 2 6
and 2 c from both sides of this equality, we obtain:
-a + 6 + c = 2 (s - a),
a b + c = 2 (s 6),
and a + b c = 2 (5 c).
These expressions are now substituted in equation (1), which
then becomes:
tan * A - 4 A* -*)(*- <0 = * i(* - o) (a - b) (a - c) _
2 V $ ( a) s a
where -'- 0(f)
In a similar way, we obtain the analogous formulae:
tan i B = - and tan C = -
s 6 s c
These formulae will be referred to as the half-angle formulae
for the triangle.
90. Three sides. The results of the preceding section make it
possible to remove the one remaining point contained in the
criticism of 83, viz., to devise a treatment for case (d) adapted to
calculation by logarithms.
Example.
Given, a = 14.931, 6 = 16.902, c = 24.315.
Required. A, B, C.
Solution. a = 14.931
6 = 16.902
c - 24.315
- A
2 s = 56.148 and therefore s = 28.074;
s
s
s
-b =
C =
13.143
11.172
3.759
and
and
and
log
log
log
(s
(s
(s
-) =
-b) =
-c) =
1.
1.
,11870
,04813
,57507
A
<?
s =
28.074
logs =
2
1.
.74190
,44830
1.29360
2 _
* It is well to use this check upon the calculation of the quantities s,
s a, s 6, and s c before proceeding with the rest of the work.
82 INSCRIBED AND CIRCUMSCRIBED CIRCLES
Since p = V (s-a)(s-b)(s-c) ^
s
Subtracting log (s a), log (s 6) and log (s c) in turn from
log p, we find by means of the half -angle formulae:
log tan A/2 = 9.52810 - 10, log tan B/2 = 9.59867 - 10,
log tan C/2 = .07173;
/. A/2 = 18 38' 33", B/2 = 21 38' 52", C/2 = 49 42' 37",
and A = 37 17' 6", B = 43 17' 44", C = 99 25' 14".
Check. A + B + C = 180 0' 4".
91. Exercises.
l.o=* .96834, b = .94572, c = .95902. Determine A, B, C.
2. a = 453.67, h = 112.34, c = 369.85. Determine A, B, C.
3. x = 1.004, ?j = 1.705, 2 = 1.526. Determine .Y, F, Z.
4. a = 4500, 6 = 5400, c = 6300. Determine A, B, C.
6. A triangular piece of land is to be staked off, so that its sides measure,
respectively, 73.84 rods, 68.701 rods, and 32.503 rods. How may this be
done?
6. Two vessels leave the same harbor at the same moment, both going
at the rate of 12 miles per hour. After 2i hours, the vessels are 45 miles
apart. If one of the vessels was sailing in a due easterly course, in what
direction was the other vessel going?
92. Inscribed and circumscribed circles. Area. The results
of the preceding articles enable us to derive other important
relations, viz. :
(1) A formula for the radius of the inscribed circle (see Fig. 45).
The points of tangency of the inscribed circle of radius r divide
the total perimeter of the triangle into six parts, which are equal
in pairs. Hence the sum of the segments of any set of three, of
INSCRIBED AND CIRCUMSCRIBED CIRCLES 83
which no two are equal, must be equal to s, one half of the peri-
meter; e.g.,
AF + BD + DC = s or AF + a = s;
hence AF = EA = s a.
In similar manner we show that FB = BD = s b and DC
= CE = s - c.
Moreover A AOF is a right-angled triangle (Why?) and LOAF
= A/2. (Why?)
rru < 4. A OF r
Therefore tan -^ =
2 AF s - a
On the other hand, we have from the half-angle formulae for
the angles of a triangle: tan -5- = ~ . From this we conclude
s a
that r = p, i.e.:
THEOREM VI. The radius of the inscribed circle of a triangle ABC
Is given by the formula:
__ 4 / (s a) (s b) (s c)
where s designates one half of the perimeter of the triangle*
(2) A second formula for the area of a triangle.
We see from Fig. 45, that
A ABC = A AOB + A BOG + A CO A
= r c/2 + r a/2 + r 6/2 = r a - C = rs.
Using the formula for the radius of the inscribed circle found
above we get:
THEOREM VII. The area of a triangle ABC is given by the formula:
A = V s (s - o) (s - 6) (s c).
This formula, known as Hero's formula, expresses the area of
the triangle in terms of the sides only; it is well known from the
study of plane geometry. Together with the formula derived
84
SUMMARY OF RESULTS OF CHAPTER VII
in 76, it suffices to determine the area of the triangle in all cases
which we have considered.
(3) A formula for the radius R of the circumscribed circle.
We draw (see Fig. 46) the diameter AOD and connect C with D.
FIG. 46
Then Z ACD is a right angle (Why?) and Z ADC = Z ABC
= 5 (Why?). Hence sin B = sin ADC = -^ ^ = ^rrs From this
6
it follows hat 72 =
sin
By means of the law of sines, we
derive from this wo other expressions for R, yo that we can now
state the following theorem:
THEOREM VIII. The radius of the circumscribed circle of a triangle
ABC is given by the formulae:
R = a _. * __.
ZslnA SsinB 2sinC
93. Summary of the results of Chapter VII.
a b c
1. Law of sines:
sin A sin B sin C
2. Law of cosines: a 2 = b 2 + c 2 2 6c cos A,
6 2 = c 2 + a 2 - 2 c-a cos 5.
c2 a 2 + 52 _ 2 ab cos C.
see 76.
see 81.
SUMMARY OF RESULTS OF CHAPTER VII 85
3. Law of tangents: tan \ (A B) = , cot \ C, see 84.
a -f-
tan
tan
C
4. Half -angle formulae for the triangle:
tan J A = ^ . see 89.
s -a
tan | B = -2 ,
s o
s
see
, A /(s a) (s b) (s c)
where /> = y - - - -
5. Mollweide's equations:
a + b cos \(A - B) a-b sin -J- (A - B)
- = - . ^^^__ , _____ = ----- -
c sin f C c cos | C
b + c = cos ^ (B - C) b -c = sin % (B - C)
a ~ sin | A ' a cos | A
c + a _ cos | (C A) c _ a _ sin | (C A)
b ~ sin | ^ ' 6 "~ cos | B
6. The area of triangle:
a. A = | a6 sin C = \ be sin A = | ca sin B, see 76.
6. A = V s (s a) (s 6) (s - c), see 92.
7. The radius of the inscribed circle:
. /(s a) (s 6) (s c) --
r = if ^ ^ '- 1 see 92.
T S
8. The radius of the circumscribed circle:
P __ a __ b _ c
Lv ~ r A r r> r\ !?v ' 866 /Mt
2 sm A 2 sm B 2 sm C
86 MISCELLANEOUS EXERCISES AND APPLICATIONS
9. Methods for solving triangles:
Solved by use of:
Case
Checked by use of:
Without logs
With logs
Two angles and one
Formulae 1, 6a
Formulae 1, 6a
Formulae 5
side.
Two sides and an
Formulae 1, 6a
Formulae 1, 6a
Formulae 5
angle opposite one
of them.
Two sides and the
Formulae 2, 6a
Formulae 3, 6a
Formulae 5
included angle.
Three sides.
Formulae 2, 6b
Formulae 4, 6b
Z.A + ZB + /.C
= 180
94. Exercises.
1. Prove the law of sines by means of tne method indicated in 92 (3).
2. Show, in Figure 45, that BD + CE -f EA = s.
3. Prove, in Figure 45, that CE = DC = s - c.
4. Use 92, (3) to show that the area of a triangle is equal to j-p .
Determine the area and the radius of the circumscribed circle for each of
the following triangles:
6. a = .473, b = .586, C = 23 47' 12".
6. x = 3.045, Y = 47 28', Z = 65 34'.
7. a = 41.35, b 36.78, A = 35 27'.
8. c = 632, a = 571, B = 30.
Determine the area and the radius of the inscribed circle for each of the
following triangles:
9. a = 40.37, b = 31.56, c = 27.08.
10. a = .971, b = .506, c = .683.
11. a = 437, c = 856, C = 38 41'.
12. a = .0456, 6 = .0731, C = 74 26'.
95. Miscellaneous exercises on Chapter VII and applications.
1. The area of a triangular piece of land is 43 acres. One side measures
440 yards, and the angle at one extremity of this side is 43. What must the
remaining sides and angles be? (1 acre = 4840 sq. yds.)
MISCELLANEOUS EXERCISES AND APPLICATIONS 87
2. A lighthouse is observed N 15 W from a vessel which is sailing 15
miles an hour in a due northerly course. Half an hour later the bearing of
the same lighthouse is N 37 W. How far is the lighthouse from the second
position of the vessel and how long will it be before the lighthouse is sighted
due West?
3. From the top of a mountain the angles of depression of two consecu-
tive milestones in the same vertical plane with the top of the mountain are
10 and 15. How high is the mountain ?
4. A forester observes that the angle of elevation of an observation tower
from a point P is 5; after walking towards the tower along a horizontal road
for a distance of 500 feet, he observes that the angle of elevation has changed
to 35. How much farther will he have to go to reach the foot of the tower?
Calculate the unknown parts and the area of each of the triangles indicated
in Exs. 5-7:
6. a = 47.032, 6 = 35.614, A = 27 45' 16".
6. a = 3, b = 5, c = 7.
7. a = 23, c = 17, C = 42 23'.
Determine the area and the radii of the inscribed and circumscribed
circles in each of the triangles indicated in Exs. 8-10:
8. a = 256, C = 17 13', B = 45 16'.
b = 2.25, c = 1.75, A = 54.
a = 15, b = 17, c = 25.
11. The ratio of the lengths of two sides of a triangle is 5 : 8, the included
angle is 35. Determine the other angles of the triangle.
12. A wireless tower is built on the edge of a cliff. From a boat at sea,
the angle of elevation of the top of the tower is 30. After rowing towards
the shore for a distance of 400 feet it is found that the angles of elevation
of the bottom and top of the tower are 45 and 57 respectively. How high
is the cliff and how high is the tower ?
9.
10.
13. To determine the distance between two points A and B, situated on
the surface of a lake, two points C and D are selected hi such a manner that
both A and B are visible from C and D (see Fig. 47). It is found that CD
88 MISCELLANEOUS EXERCISES AND APPLICATIONS
= 200 feet and that LACE 33, /BCD 54, </AZ)B - 40 and
//>r^ **o ou *u * ,I/Y 200 sin 65 , ,, . Drt 200 sin 75
LCD A = 65 . Show that AC = : ^^ , and that BC
sin 28 ' ""* """" ~ sin 21 '
Then calculate the distance AB.
14. A vessel is sailing due east at the rate of 20 miles per hour. At 10
A.M. a lighthouse L is bearing N 10 W, while a second lighthouse M is bear-
ing N 40 E; at 2 P.M. the bearings of L and M are N 50 W and N 5 E
respectively. How far are L and M apart?
16. Determine also from the observations recorded in Problem 14 the
direction of the line from L to M.
16. To measure the height of a mountain AF, above a horizontal plane P,
we measure the distance between two stations B and C (see Fig. 48), so selected
FIG. 48
that at least one of them, say B, lies in the plane P, that each is visible from
the other and that A is visible from both. The angles ACS, ABC and the
angle of elevation a of A as seen from B are measured. It is found that
BC = 500 feet, / ACB = 85 25', Z ABC = 84 33' and a = 40 17'. Deter-
mine the height of the mountain above the plane P.
17. The angle of elevation of a church steeple T from a point R is 17 25'.
At a point , 250 feet from R, the line TR subtends an angle of 73 47', while
from R the line TS subtends an angle of 65 8'. Determine the height of T
above the horizontal plane through R.
18. A flag staff on top of a monument subtends an angle of 3 at a point
P, 400 feet above the ground and at a horizontal distance of 300 feet from
the foot of the monument. From the same point P the monument itself sub-
tends an angle of 43. Determine the height of the flag staff and the height
of the monument.
19. One side of a triangle is 75 feet and the angle opposite this side is 34.
The sum of the other sides is 125 feet. Determine all the sides and angles of
this triangle.
20. A building 20 feet high is surmounted by a steeple 30 feet high. How
far from the foot of the building must an observer stand in order that the
building and steeple may subtend the same angle at his eye, which is 5 feet
above the ground?
21. A pole 10 feet tall is divided into two parts, a lower part of 6 feet and
an upper part of 4 feet. Frt>m a point P, 7 feet above the bottom of the pole,
MISCELLANEOUS EXERCISES AND APPLICATIONS 89
these two parts subtend the same angle, the pole being held vertically. How
far is the pole from P?
22. The angle of elevation of a church steeple from a point A, due south
of it, is 27; and from a point B, due west of the steeple, and in the same hori-
zontal plane as A, the angle of elevation is 35. Moreover the distance AB
is 150 yards. Determine the height of the steeple above the plane AB.
23. From a point A the angle of elevation of the top T of a flagpole which
stands on top of a building is 37 47'. From a point B, 100 feet nearer to the
building and lying in the vertical plane through T and A, the angle of eleva-
tion of T is 47 32'. How high is T above the ground?
24. The angle of elevation of the bottom of the flagpole described in
Problem 23, as seen from B, is 45 25'. Determine the height of the pole.
26. To determine the height XY of a wireless-tower X above a horizontal
plane P two points A and B are selected in this plane P (see Fig. 49). The
FIG. 49
angles XAY, XBY, YAB and ABY and the distance AB are measured. It
is found that Z XAY = 67, Z XBY = 58 3'.1, Z YAB = 27, Z YBA = 18
and AB = 81.6 feet. Determine the height XY and check the calculations.
CHAPTER VIII
INVERSE TRIGONOMETRIC FUNCTIONS. TRIGONO-
METRIC EQUATIONS
96. Inverse functions. The concept function which we have
used in preceding chapters will now be studied in somewhat
greater detail. The concept may be defined for our purpose in
the following way:
DEFINITION I. If a relation between two variables, x and y, is given
in such a way that to every value of either there correspond one or
more values of the other, then x is a function of r, and y is a function
of x.
If these two functional dependences of y upon x and of x upon
y are written down explicitly, we obtain two functions, one in
terms of x, yielding t/, the other in te Das of y, yielding x. In
the first of these functions x is the independent variable and y the
dependent variable; in the second y is the independent variable
and x the dependent variable. Two such functions which result
from the same relation between the two variables are called
inverse with respect to each othe . This may be expressed as
follows:
DEFINITION II. If a relation between two variables, x and y, be
solved in turn for A in terms of y and for y in terms of v, we obtain
two functions which form a pair of inverse functions.
The important connections be ween hem are recognized more
readily if one let er be used for the independent variable in both
functions and another letter for the dependent variable in both
functions.
Example 1. The relation between the variables x and y de-
termined by the equation
3z + 2t/-7 = (1)
gives rise to the functions
(2) and * = (3).
90
INVERSE FUNCTIONS
91
The graphical representation of the relation between x and y
can readily be obtained from equations (2) or (3); in either case
we will obtain the straight line of Figure 50.
2 $1 8
FIG. 50
Denoting the independent variable by t and using u tp denote
the dependent variable in both cases, equations (2) and (3) will
become
7-3* 7-2*
u = an( j u _. .
FIG. 51
Here we have a pair of inverse functions, whose graphs are repre-
sented in Figure 51.
92
GRAPHS OF INVERSE FUNCTIONS
Example 2. The relation y* x = gives rise to the pair of
inverse functions
u = t 2 and u = V?,
whose graphs are given in Figure 52.
FIG. 52
In the first function of this pair, one positive value of u corre-
sponds to every value of t] but in the other function two values of u
correspond to every positive value of t, and no value of u to negative
values of t. The function u = t 2 is a single-valued function of f,
defined for all values of t; the inverse function u = V7 is a two-
valued function of t defined for positive values of t only. It may
happen that the inverse function of a single-valued function is
3-valued, or, in general, multiple-valued.
97. Graphs of inverse functions. Let the functions u = f (t)
and u = g (f) be a pair of inverse functions whose graphs are the
curves drawn in Figure 53. If the point P (a, 6) belongs to the
graph of u = f (0, then the point Q (6, a) must be on the graph of
u ~ 9 (0; because the two functions may be thought of as having
been obtained from one and the same equation between x and y,
x and y being replaced by t and u in one case, and by u and t in
the other. Now it is readily seen that two such points P (a, 6)
and Q (6, a) are symmetrically situated with respect to the 45
line MN (see Fig. 54), i.e., that MN is the perpendicular bisector
of the line PQ. For As ROP and QOS are congruent. (Why?)
GRAPHS OF INVERSE FUNCTIONS
93
Hence PO = QO and Z ROP = Z SOQ. Consequently Z POT
= Z QOT (Why?) and A POT 7 and A QOT are congruent.
(Why?) Therefore PI 7 = QT and Z PTO = Z QTO = 90, which
FIG. 53
FIG. 64
was to be proved. Consequently from points A, B on the graph
of any function, points A', B', . . . on the graph of the inverse
function may be obtained by determining A', B', . . . as points
which are symmetrically situated with A, J3, . . . with respect
to the 45 line MN, i.e., by reflecting the points A, B, . * . in
94
THE INVERSE SINE FUNCTION
M N as a mirror*. These results may be summarized in the fol-
lowing theorem:
THEOREM I. If u =/ (t) and u = g (t) are a pair of inverse func-
tions the graph of either function may be obtained by reflecting the
graph of the other function in the 45 line as a mirror.
98. Exercises*
Determine the inverse function associated with each of the following func-
tions:
1. u = 4 t - 3. 2. u = t*/3. 3. u = logio t. 4. u t 9 .
Construct the graphs of the following pairs of inverse functions:
6. u = Vj 4, u = 2 -h 4. 6. w = 2/3, w = 3 .
7 - 4* -7
Q^ V^ SSS
99. The inverse sine function. The equation y = sin # estab-
lishes a relation between the variables x and ?/, represented
graphically by the sine curve (see dashed curve in Fig. 55). For
u
- JL.
2
-1
-IT
a
a i
\.U -
JSL.
a
FIG. 55
every value of x there is one corresponding value of y. If the
equation be solved for x in terms of y, and x and y be replaced by
u and t respectively, we obtain the inverse sine function; it is
expressed in the form u = arc sin t } which is read "u is the angle
* See footnote on page 45.
THE OTHER INVERSE TRIGONOMETRIC FUNCTIONS 95
whose sine is t." The graph of the inverse sine function of t, i.e.,
of arc sin t, may be obtained, in virtue of Theorem I, by reflecting
the sine curve in the 45 line as a mirror. In this way we obtain
the full-drawn curve of Figure 55.
The inverse sine function of t cannot be expressed in a simple
manner in terms of algebraic or trigonomelric functions of t. It
must be regarded as an entirely new function, defined as the in-
verse of the sine function. From this definition its properties
will be derived by the aid of its graph.
We observe that the function u = arc sin t gives an indefinitely
large number of values of u for every value of t y which lies between
1 and 1, and that it gives no value of u at all for values of t
which lie outside the range ( 1, 1). The inverse sine function
is an infinitely multiple- valued function, defined only for values of
t between 1 and 1.
So, e.g., arc sin \ = 30, 150, 510, -210, . . . just as sin 30
= sin 150 = sin 510 = sin (-210 = = J.
DEFINITION III. The symbol Arc sin t, called the principal value
of arc sin t, Is used to designate the numerically smallest angle
whose sine is equal to t.
For example Arc sin \ = 30 = Tr/6.
It is clear that Arc sin t is a single-valued function of t, defined
for values of t between 1 and +1. Its graph is given by the
heavily drawn portion of the graph of arc sin t in Figure 55.
When t is between and 1, Arc sin t will be between and -;
when t is between 1 and 0, Arc sin t will be between = and 0.
2i
100. The other inverse trigonometric functions. In a similar
manner the equations u = arc cos , u = arc tan t, u = arc cot ,
u arc sec t and u = arc cosec t represent the inverse functions
associated with the remaining trigonometric functions cos , tan t,
cot t, sec t, and cosec t respectively. These equations are read
"u is the angle whose cosine is equal to J," etc. The graphs of
the inverse trigonometric functions are obtained from the graphs
of the associated trigonometric functions by the use of Theorem I,
i.e., by reflecting the latter in the 45 line as a mirror. In Fig-
ures 56 and 57, the graphs of the functions arc cos t and arc
tan t are indicated by the full drawn curves. We observe that
both of these functions are infinitely multiple-valued for those
96 THE OTHER INVERSE TRIGONOMETRIC FUNCTIONS
U
-1
\
FIG. 56
-*~r
m CO!
arc tanf
FIQ. 57
EXERCISES 97
values of the independent variable for which they exist at all; i.e.,
for every value of t between 1 and 1 there is an infinitely large
number of values of arc cos t] while for every value of t there is
an infinitely large number of values of arc tan t.
If cot u = t, then tan u = l/; hence arc cot t = arc tan l/t,
i.e., an angle whose cotangent is equal to t is equal to an angle
whose tangent is equal to l/t. For an analogous reason we have
arc sec t = arc cos l/t and arc cosec t = arc sin l/t. Since the
functions arc cot t, arc sec , and arc cosec t are so readily expres-
sible by means of the functions arc tan t, arc cos t> and arc sin t
respectively, we shall limit our study to the latter three functions.
As in the case of arc sin t, we define single-valued functions
corresponding to the multiple-valued functions arc cos t and arc
tan t.
DEFINITION IV. The symbol Arc cos t, called the principal value
of arc cos t, designates the least positive angle whose cosine Is equal
to t.
DEFINITION V. The symbol Arc tan t, called the principal value
of arc tan t, designates the numerically smallest angle whose tangent
Is equal to t.
The graphs of the single-valued functions Arc cos t and Arc
tan t are given by the heavily drawn portions of Figures 56 and
57 respectively.
When t lies between and 1, Arc cos t is between and 7r/2;
when t lies between 1 and 0, Arc cos t is between w/2 and TT;
when t is positive, Arc tan t is between and 7r/2;
when t is negative, Arc tan t is between ?r/2 and 0.
101. Exercises.
1. Determine arc cos i; Arc tan 1; Arc cot V3; Arc cos 1; arc sin 0; Arc
cos 1/V2.
2. Evaluate sin (arc sin $); cos (arc cos f); tan (arc tan V?).
3. Determine: cos (arc tan l/v^S); tan (arc sin Vj); sec (arc cos |);
sin (arc cos .4321); cos (Arc sin 0); cot (Arc tan V3).
4. Evaluate :_ cos (Arc cos J -f- Arc sin }); tan (Arc sin 1/V2 Arc tan 1);
sin (Arc tan V3 + Arc cos 1/V5).
6. Construct the graphs of the functions u = arc sin 2 t; u = arc sin 2/3;
u arc cos t/3.
98 MULTIPLE- AND SINGLE-VALUED INVERSE FUNCTIONS
102. Relations between multiple-valued and single-valued in-
verse functions. There is a simple algebraic relation between
the multiple-valued function arc sin t and the single- valued func-
tion Arc sin t. If u = Arc sin t, we know that
t sin u = sin(n 180 + w), when n is an even integer, and that
t = sin u sin(n 180 u), when n is an odd integer (see Chap-
ters V and VI).
Hence arc sin t = n 180 + Arc sin t, when n is even,
and arc sin t = n 180 Arc sin t, when n is odd.
These two equalities, which may be derived directly from the
graph of arc sin t (see Fig. 55), are combined in the single formula:
arc sin t n 180 + (-l) n Arc sin t.
In entirely analogous manner we obtain the further formulae:
arc cos t = n 360 Arc cos t,
arc tan t = n 180 + Arc tan t.
103. Trigonometric equations. Let us consider the following
problem: To construct a rectangle whose perimeter is 10" and
whose diagonal is 4".
FIG. 58
If ABCD be the required rectangle and 6 the angle between the
diagonal AC and the side AB (see Fig. 58), we have:
AB = AC cos = 4 cos 6
and
#<7 = AC sin = 4 sin 0.
The conditions of the problem are therefore expressed by the
equation
8 cos + 8 sin = 10.
This equation involves trigonometric functions of the unknown
0; it is called a trigonometric equation. Equations of this sort
SOLUTION OF TRIGONOMETRIC EQUATIONS 99
occur in various branches of mathematics and in its applications.
We consider their solution.
104. Solution of trigonometric equations. " To solve a trigo-
nometric equation' 7 means to determine all possible angles which
satisfy the equation. It is important to notice the difference
between a trigonometric identity, which is true for all angles for
which it has a meaning, and a trigonometric equation, which is
satisfied by certain angles only. The determination of these
special angles constitutes the " solution of the trigonometric
equation/'
The work of solving a t.igonometric equation may be divided
into three parts, viz. :
(1) To express all the trigonometric ratios occurring in the
equation in terms of a single ratio of a single angle,
(2) To solve the resulting equation obtained in (1), as an
algebraic equation for hat ratio, and
(3) To determine all the angles corresponding to the values
for that ratio found in (2).
Returning to the example of 103, we proceed as follows:
(1) We express cos 6 in terms of sin 6 by means of the formula:
cos 6 = Vl sin 2 6. Substituting this in the equation derived
in 103, we obtain the equation
sin + Vl - w^~d = ,
or
Vl - sin 2 e = | - sin 0.
(2) We square both sides of this equation, collect terms and
clear of fractions. In this way we obtain the following quadratic
equation in terms of sin 6:
32 sin 2 - 40 sin + 9 = 0.
This equation we solve by means of the quadratic formula,
which gives us:
- 40 =*= ^1600 - 1152 _ 40 A/448 _ 40 d= 21.1660
- _ _
~ 64 64 64
/. sin 6 = 61.1660/64 or 18.8340/64.
(3) From this we conclude :
= arc sin 61.1660/64 or arc sin 18.8340/64.
100 FURTHER EXAMPLES
By means of the tables, we find
Are sin 61.1660/64 = 72 53' and Are sin 18.8340/64 = 17 7'.
.'. 6 = n 180 + (-!) 72 53' or n 180 + (-1)* 17 7'.
The nature of the problem is such that only acute angles
are applicable, so that we need only consider the principal values
of 6. Using these, we find:
AB = 4 cos = 1.177 or 3.823,
and BD = 4 sin = 3.823 or 1.177;
we verify that 2 AB + 2 BD = 10.
105. Further examples.
1. To solve the equation
tan 6 tan 2 6 + cot 6 + 2 = 0,
we use the formulae:
(1) tan 2* = - and cot 9 = (see 74, 9).
Substituting these expressions in the original equation, it
becomes :
1 - tan 2 d ' tan '
(2) This is an algebraic equation in terms of tan 6} we clear
the equation of fractions and collect terms. This leads us to the
following quadratic equation:
tan 2 - 2 tan 6 - 1 = 0.
If we solve this equation for tan by means of the quadratic
formula, we find:
. n ; r
= 1 d= V2 = 2.4142 or -.4142.
(3) It remains to determine arc tan 2.4142 and arc tan .4142.
From the tables, we find: Arc tan .4142 = 22 30', and therefore
Arc tan (-.4142) = -22 30', whence follows arc tan (-.4142)
= n 180 -22 30'.
FURTHER EXAMPLES 101
Furthermore Arc tan 2.4142 = 67 30', and therefore arc tan
2.4142 = n 180 + 67 30'.
2. To solve the equation
3 sin 2 8 4 sin 4 = 0.
In this problem the object of step (1) has already been accom-
plished, since the equation in its original form is an algebraic
equation in terms of sin 0.
Solving it by means of the quadratic formula, we find:
sin0 = 2 or -f = -.6667.
There are no angles whose sine equals 2 ; we have only to deter-
mine therefore arc sin ( .6667). From the tables, we find that
Arc sin .6667 = 41 49'; hence Arc sin (-.6667) = -41 49',
from which we obtain the final result:
= arc sin (-.6667) = n 180 + (-1) : (-41 49').
For: n = 1, we find 6 = 180 + 41 49' = 221 49';
n = 2, = 360 - 41 49' = 318 11' ;
n = -1, e = -180 + 41 49' = -138 11';
w = -2, 0= -360' - 41 49' = -401 49';
n = 0, = -41 49'; etc.
The method outlined above for the solution of trigonometric
equations is not always the shortest or the most convenient one.
Frequently other special methods enable us to obtain a solution
more quickly or in better form. The student will do well to bear
this in mind, particularly in cases in which the algebraic equa-
tion, to which the method described above leads, is such that he
cannot solve it. One such special method is illustrated by the
following example.
3. To solve the equation
cos 2x sin 2 x = 1.
Since tan 45 = 1, we may write this equation in the form
tan 45 cos 2 x sin 2 x 1,
from which we derive, by multiplying both sides of the equation
by the corresponding sides of the equality cos 45 = 1/V2,
sin 45 cos 2 x - cos 45 sin 2 x = I/ V2,
pr sin (45 -2 a?) = 1/V.
102 FURTHER EXAMPLES
From this we conclude that we must have
45 - 2 x = arc sin l/A/2 = n 180 + (-1)" 45,
i.e., 2 x = 45 - n 180 - (-1)* 45,
>IKO / i \n . J.KO
or finally : a; = ( ij 4a - n 90.
2i
For n = 0, 1, 1, 2, 2, we obtain the special values
x = 0, x = -45, x = 135, x = -180, x = 180 respec-
tively.
The same method may be applied in solving any equation of
the form
a cos m 6 + 6 sin m = c,
provided
We divide both sides of the equation by 6 and determine the
angle a by the formula a = Arc tan a/6. Then tan a = a/6, and
cos a s- .- ', the square root being taken with the sign of 6;
v a' 2 4-& 2
then the equation will takethe following form:
tan a cos m + sin m 6 = c/b.
We now multiply both sides of the equation by cos a and make
use of the addition formula for the sine; in this manner we find
. / , M c COSQ!
sin (a + m 6) = r -
Since now \c = Va 2 + 6 2 , the right hand side of this equation is
always numerically less than 1. It may therefore be solved for
a + m ti, so that we obtain the final result:
i r c i
= - a + arc sin /=== h
m L \/aM r 6 2 J
where a = Arc tan a/6, and where the square root is to be taken
with the sign of 6.
EXERCISES 103
106. Exercises.
1. Solve the equation: tan x -f- 3 cot x = 4.
2. Solve the equation: sin + cos = V2/2.
3. Solve the equation: tan 6 3 cos 8 -j- sec = 0.
4. Solve the equation: cos 2 x = 3 sin # + 2.
5. Solve the equation: sin $ = tan 0/2.
6. Solve the equation: cos 2 A 2 sin 2 A.
7. Solve the equation: 4 sec 2 = 3 (tan + 2).
8. Determine all the angles whose tangent is the reciprocal of their sine.
9. Determine all the angles whose secant is the reciprocal of their tangent.
10. The graphs of the functions sin and tan are drawn with respect to
one set of axes and units. Determine all those values of for which the
ordinate on the tangent curve is twice as long as the corresponding ordinate
on the sine curve.
11. Construct an isosceles triangle whose perimeter is 20 inches and
whose altitude is 6 inches. (HINT. Take the base angle of the triangle as
the unknown.)
12. Construct a right triangle of which the perimeter is 36 inches and one
leg is 12 inches.
13. Determine a point P on the circumference of a circle of radius 5 inches,
the sum of whose distances from the extremities of a fixed diameter AB is
equal to 14 inches.
14. Construct a triangle ABC such that the perimeter is equal to 30
inches, the side c is equal to 10 inches and the angle B is equal to 60.
15. Resolve a force of 100 pounds into two mutually perpendicular com-
ponents, whose sum is equal to 120 pounds.
16. Construct a triangle ABC of which the perimeter is 60 feet the side
a \s equal to 20 feet and the angle B is equal to 45.
ANSWERS
Art. 4, Page 3
2. (3,3 VI). 3. (2, -2). 6. (0,0). 7. (3,5); (5,0); (1,1). 8. 5, V5T,
V2. 9. db4; =fc5; impossible;
Art. 9, Page 6
2. 65, 12, 113, -64, -127, 202, -235, -337, 508. 3. 91, 53,
392, -15, 495, -107, -333, 639.
Art. 11, Page 8
1. /4, 7T/6, -57T/4, llir/6, 37r, -5/6, A -37T/2, -9*/2, 7ir/6,
8 T/3, -65 r/18. 2. 270, -240, 72, 225, -90, 270/7r, 414 %. 5.
300 tr. 6. 10 feet. 7. 144 %r. 8. 2 feet. 9. 6000 TT. 10. .84 feet.
Art. 22, Page 14
2a. 1/2. 26. -V/2. 2c. 0. 2d. (-3 + V3)/4. 3o. -7V2/12. 36.
3/2. 3c. (~2 + V)/3, 3d. 46/9.
Art. 25, Page 17
3a. -13/4. 36. 4. 3c. -l/Vg. 3d. -6.
Art 28, Page 19
2. sin^ = =fc 4/VI7, cos0 -dsl/Vi?, cot^ - 1/4, seed Vl7, cosecfl
-dbVir/4.
Art. 32, Page 22
1. 3, -3, 1/3, 0, 3/5, 1. 4. 2/3, 3/2, 3/2, -2/3, -1/4.
Art. 34, Page 23
8. .46831. 9. .28452. 10. 2.21638. 11. .60061. 12. -.93404.
Art. 39, Page 30
1. 3.1335. 2. 1.9851. 3. .0052026. 4. 5712. 6. .63014. 6. 4.2266.
7. 583.35. 8. 8.524. 9. 39.478. 10. .31365. 11. 2.0550. 12. 1.3425.
13. 4.0198. 14. 7.3605. 15. .38638. 16. 30.628. 17. .000087240. 18.
.060285. 19. .34697. 20. 5.3321.
105
106 ANSWERS
Art. 44, Page 35
1. A = 35 20' 28", B = 54 39' 32", c = 644.83. 2. A - 62 10' 8",
B = 27 49' 52", c = .16193. 3. 4= 30, B = 60, 6 = 3.717. 4. A =
48 10' 51", B = 41 49' 9", a = 14.609. 5. B = 21 35', a = 22,073, c =
23,737. 6. A = 62 47', a - .12279, c = .13808. 7. = 5 25', a =
376.32, 6 = 35.682. 8. B 45 25', a = 265.33, 6 = 269.22. 9. B = 44
57', b .08497, c = .12027. 10. A = 49 56' 5", B = 40 3' 55", b =
.057956. 11. A = 16 14' 26", B = 73 45' 34", c 14.252. 12. A =
84 58', a = 1.0020, b = .088254. 13. V = 65 34', a = 24.751, ft = 20.809.
14. P = 25 37', a = 19.940, ft = 8.621. 15. P = 41 51' 32", V = 96
16' 56", 6 17.9496. 16. 6 = .84246, a = 1.1038, P = 67 34'. 17. P =
53 34', V = 72 52', a = 20.368. 18. b = .64064, a .32559, F == 159
22'. 19. P = 62 49' 45", V - 54 20' 30", ft = 8.8480. 20. 6 = 12.2860,
h = 2.9235, V = 129 6'. 21. 6 - 23,454, ft = 11,727, P = 45,
Art. 47, Page 38
1. 4.3301 feet. 2. 2.5 feet. 3. feet; 7 feet.
Art. 49, Page 39
1. 4,504.8 feet; 4,247.7 feet. 2. 272.04 feet. 3. 2,012.2 feet. 4.
2,641.4 feet. 5. 27.155 feet. 6. 52.781 miles; 72.705 miles. 7. 53.868 feet;
23.246 feet. 8. 1.3313 miles. 9. 566.09 feet; 504.38 feet. 10. 85.21 feet;
125.30 feet. 11. 7.980 miles. 12. .059 miles.
Art. 67, Page 67
1. (Vg + V2) /4, (Vg - V2)/4.
Art. 69, Page 59
1. 2 + V3, 2-V.
Art. 71, Page 61
3. V8;-2Va-2V2/4, Vs + 2 V6 -f 2 V2/4, (V2 - 1) (V3 - V2).
Art. 73, Page 62
1. 2 sin 60 cos 5. 2. -2 sin 45 sin 30. 3. 2 sin 4 30' cos 22 30'.
4. 2 cos 64 30' cos 22 30 X . 5. 2 cos 12 sin 30. 6. -2 sin 45 cos 12.
7. .22490. 8. 2.8834. 9. -2.5073. 10. .99726.
Art. 75, Page 63
la. ( Vg - V2)/4, ( V6 + V2)/4, V3 - 2. 16. V2 - V/2, V^-f V/2,
V - 2. 4. .14762.
ANSWERS 107
Art. 78, Page 67
l.o- .29015, c - .23995, C 31 42'. 2. a 173.95, 6 = 237.62, B =
105. 3. o = 12.736, 6 = 6.6607, A - 62 55' 25". 4. 6 = 170.27,
c = 145.80, B = 84 20' 50". 6. 559.40 feet. 7. 137.18 feet.
Art. 80, Page 73
1. c = 72.612, B = 51 40' 30", C = 92 44' 30"; c' = 20.149, B' = 128
19' 30", C' = 16 5' 30". 2. a = 1187.5, .1 - 140 23' 56", C = 15 49'
4". 3. No solution. 4. b = 5.1662, 4 = 16 7' 27", 5 = 26 34' 33",
6. c = 79.735, B = 26 15' 36", C = 134 19' 24"; c' = 13.280, B' = 153
44' 24", C' = 6 50' 36". 6. a = .082650, A = 122 58', B = 28 31'-
7. 100.91 miles. 8. 203.87 feet or 195.72 feet. 9. 61.671 miles.
Art 82, Page 76
1. c = 137.48, A = 49 6' 20", B = 70 53' 35". 2. P = 53 7' 50",
Q = 59 29' 25", R = 67 22' 50". 3. A - 26 44', B = 23 23' 12", C =
129 52' 48". 4. t = .56081, R = 34 34' 6", S = 100 26'. 5. 399 feet.
6. 84 15' 37".
Art. 88, Page 80
l.o = 97.988, B = 52 31' 18", C = 14 44' 42". 2. b = 5247.1, A =
36 54' 20", C = 54 38' 40". 3. r = .04343, P = 47 16' 5", Q 85 7'
55". 4. ?y = 1.2136, X - 35 28' 24", Z = 24 31' 36". 6. 211.40 feet.
6. 92.685 million miles.
Art. 91, Page 82
1. A = 61 6' 29", B = 58 46' 4", C = 60 7' 24". 2. A = 132 45' 4",
= 10 28' 36", C = 36 46' 26". 3. X = 35 39' 50", F = 81 56' 20",
Z = 62 23' 50". 4. A = 44 24' 50", B = 57 7' 20", C - 78 27' 50".
6. Angles should be 85 50' 24", 68 7' 4", 26 2' 28". 6. N 42 50' 40" E
or S 42 50' 40" E.
Art. 94, Page 86
5. A =*055898, R = .3033. 6. A = 3.3797, R = 1.6544. 7. A 697.4,
R = 35.648. 8. A = 90,218, R = 316.89. 9. A = 426.6, r = 8.6172.
10. A - .16379, r = .15166. 11. A = 157,375, r - 128.71. 12. A
.0016056, r = .016573.
108 ANSWERS
Art 95, Pages 86-89
1. 1,387.1 "yards, 1,106.7 yards, 121 16' 7", 15 43' 63". 2. 5.1818
miles; 16.6 min. 3. 2,722.7 feet. 4. 71.393 feet. 6. c = 75.527, B -
20 38' 55", C = 131 35' 49", A = 626.3, 6. A = 21 47' 12", B = 38
12' 46", C * 120, A = 6.4951. 7. b = 23.962, A = 65 47' 0", B 71
50' 0", A = 185.75; 6' = 10.016, A' 114 13' 0", ' = 23 24' 0", A' =
77.642. 8. A 7,768.8, r = 28.433, R - 144.33. 9. A = 1.5928, r =
.54269, R = 1.1556. 10. A = 124.44, r = 4.3665, R = 12.807. 11. 36
17' 58", 108 42' 2". 12. 239.93 feet, 129.53 feet. 13. 300.92 feet. 14.
106.84 miles. 16. N 75 E. 16. 1,849.6 feet. 17. 109.34 feet. 18. 16.07
feet, 346.40 feet. 19. 96.742 feet, 28.258 feet, 133 50' 14", 12 9' 46".
20. 15 V7 feet. 21. V23 feet. 22. 61.699 feet. 23. 266.88 feet. 24.
19.04 feet. 26. 84.01 feet.
Art. 98, Page 94
1. u = (t + 3)/4. 2. u = VS. 3. u = I. 4. u v^
Art. 101, Page 97
1. db60, 420, etc.; 45, 225, etc.; 30; 0; 0; 180; etc.; 45. 2. 1/3;
2/5; V7. 3. =fcV"/2; 1; 4; .9018; 1; 1/V3. 4. 0; 0; (V3 + D/2V2.
Art. 106, Page 102
1. 45 + n 180, 71 33' 54" + n 180. 2. nw + (-1) V/6 -7r/4.
3. n- 180 -f (~l) n 4148'39". 4. nw + (-!)" + l ir/6, nir -f (-l) 7 ** 1 ^.
5. 2 HTT, (2 n + 1) r/2. 6. riTri ( - 1) V/6. 7. n 180* + 49
36' 34 ",n- 180 -23 2' 39". 8. n- 360 51 49' 30". 9. n 180 + (-l) n
38 10' 30". 10. mr, 2nw *-/3, 2 mr =t 2 T /3. 11. Base angle = 61
55' 40"; base = 6.4. 12. Sides are 15 and 9; angles are 36 52' 12" and
53 7' 48". 13. Chords are 8 and 6. 14. A = 60, a = b = 10. 16. 97.42
and 22.58 pounds. 16. b = 16.7964, c = 23.204, C = 77 39'.
INDEX
(The Numbers refer to Articles)
Abscissa, 3
Acute angle, 8
Angle, 8, 48
of depression, 48
of elevation, 48
Area of a triangle, 76, 92
Axis, 2
of abscissae, 3
of coordinates, 3
of ordinates, 3
Base of a power, 29
of a system of logarithms, 31
Bearing, 48
Characteristic of logarithms, 35
Checking calculations, 42, 83
Circle, 92
Circumscribed circle, 92
Circular measure, 10
Clockwise rotation, 8
Common logarithms, 35
Compass, 48
Complimentary angles, 8
Co-ratio, 14
Cosecant, 14, 62
Cosine, 14, 51, 81
Cotangent, 14, 57
Counterclockwise rotation, 8
Degree, 8
Dependent variable, 96
Depression, angle of, 48
Directed angle, 8
Elevation, angle of, 48
Equations, trigonometric, 103, 104
Exponent, 29
Fractional exponent, 29
Function, even, 55
inverse, 96, 97
multiple-valued, 96
odd, 55
single-valued, 96
trigonometric, 24
Grade, 10
Graphing, 50
Graphs, 51, 57, 62, 97, 100
Independent variable, 96
Initial side, 8
Inscribed circle, 92
Inverse cosecant, 100
cosine, 100
cotangent, 100
function, 96
secant, 100
sine, 99
tangent, 100
Law of cosines, 81
Laws of exponents, 29
logarithms, 33
Law of sines, 76
Law of tangents, 84
Logarithm, 30
Mantissa of a logarithm, 35
Mollweide's equation, 85
Negative exponents, 29
Obtuse angle, 8
Ordinate, 3
Origin, 2
of coordinates, 3
109
110 INDEX
Period, 26 Submultiple,, 10
Periodicity, 26 Supplementary angles, 8
Power. 29 ._. . . _,_ _ .
Principal value, 100 angent 14 ' 57 ' 84
Projection, 6, 45 Terminal side, 8
Trigonometric equations, 103, 104
Quadrant, 3 Trigonometric functions, 24
ratios, 14
Radian, 10 Unit distance, 2
Radius vector, 14
Right angle, 8 Variable, 96
Rotation, clockwise, 8 x-coordinate, 3
counterclockwise, 8 X-axis 3
Secant, 14, 62 y-coordinate, 3
Sine, 14, 51, 76 Y-axis, 3
Standard position, 12
Straight angle, 8 Zero-th power,
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