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PLANE AND SPHERICAL
TRIGONOMETRY
BY
ELMER A. LYMAN
I
MICHIGAN STATB NORMAL COLLBaS
AND
EDWIN C. GODDARD
UNITBRSITT OF MICHIGAN
U.^
ALLYN AND BACON
ISoston BVit Ci)tcajQ[o
\C\C '
I • • • » »
• !.. •-..
•• •• • .
THE NEW YOr.K
PUBLIC LIBRARY
551730 A
A:;iOR, LENOX AND
TILDEN FOUNDATIONS
COPTEIGHT, 1899, 1900,
BY ELMEB A. LYMAN
AND EDWIN 0. GODDARD.
» • * * •
• • ' • •
♦ • » ♦
• 1 • •
• « »
•*
« • • • t
.•• •
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• • • -
r
• •
« • •
• *
• • • • ' *
Norfaiootr $ret(0
3, B. duhing ft Co. — Berwick & Smitb
Norwood Mail. U.8.A.
PREFACE.
Many American text-books on trigonometry treat the solution
of triangles quite fully ; English text-books elaborate analytical
trigonometry; but no book available seems to meet both needs
adequately. To do that is the first aim of the present work, in
the preparation of which nearly everything has been worked out
and tested by the authors in their classes.
The work entered upon, other features demanded attention.
For some unaccountable reason nearly all books, in proving the
formulae for functions of a ± p, treat the same line as both posi-
tive and negative, thus vitiating the proof ; and proofs given for
acute angles are (without further discussion) supposed to apply
to all angles, or it is suggested that the student can draw other
figures and show that the formulae hold in all cases. As a
matter of fact the average student cannot show anything of the
kind ; and if he could, the proof would still apply only to combi-
nations of conditions the same as those in the figures actually
drawn. These difficulties are avoided by so wording the proofs
that the language applies to figures involving any angles, and to
avoid drawing the indefinite number of figures necessary fully
to establish the formulae geometrically, the general case is proved
algebraically (see page 58).
Inverse functions are introduced early, and used constantly.
Wherever computations are introduced they are made by means
of logarithms. The average student, using logarithms for a short
gQ time and only at the end of the subject, straightway forgets what
manner of things they are. It is hoped, by dint of much prac-
tice, extended over as long a time as possible, to give the student
a command of logarithms that will stay. The fundamental for-
^ mulae of trigonometry must be memorized. There is no substi-
tute for this. For this purpose oral work is introduced, and
there are frequent lists of review problems involving all prin-
ciples and formulae previously developed. These lists serve the
• • •
m
00
00
iv PREFACE.
further purpose of throwing the student on his own resources,
and compelling him to find in the problem itself, and not in any
model solution, the key to its solution, thus developing power,
instead of ability to imitate. To the same end, in the solution
of triangles, divisions and subdivisions into cases are abandoned,
and the student is thrown on his own judgment to determine
which of the three possible sets of formulas will lead to the solu-
tions with the data given. Long experience justifies this as
clearer and simpler. . The use of checks is insisted upon in all
computations.
For the usual course in plane trigonometry Chapters I-VII,
omitting Arts. 26, 27, contain enough. Articles marked * (as
Art. * 26) may be omitted unless the teacher finds time for them
without neglecting the rest of the work. Classes that can accom-
plish more will find a most interesting field opened in the other
chapters. More problems are provided than any student is ex-
pected to solve, in order that different selections may be assigned
to different students, or to classes in different years. Do not
assign work too fast Make sure the student has memorized and
can use each preceding formula, before taking up new ones.
No complete acknowledgment of help received could here be
made. The authors are under obligation to many for general
hints, and to several- who, after going over the proof with care,
have given valuable suggestions. The standard works of Levett
and Davison, Hobson, Henrici and Treutlein, and others have
been freely consulted, and while many of the problems have been
prepared by the authors in their class-room work, they have not
hesitated to take, from such standard collections as writers gen-
erally have drawn upon, any problems that seemed better adapted
than others to the work. Quality has not been knowingly sacri-
ficed to originality. Corrections and suggestions will be gladly
received at any time.
E. A. L., Ypsilanti.
E. C. G., Ann Arbor.
October, 1900.
CONTENTS.
Chapter I. Angles — Measurement of Angles.
PAGS
Angles; magnitude of angles . . • . ... . . . 1
Rectangalar axes ; direction 2
Measurement; sexagesimal and circular systems of measurement;
the radian 3
Examples 6
Chapter II. The Trigonometric Functions.
Function defined .8
The trigonometric functions . • . . ... • . 9
Fundamental relations • . 11
Examples 14
Functions of 0°, 30% 45°, 60% 90** 15
Examples • • • • 18
Variations in the trigonometric functions . . • • • 19
Graphic representation of functions ....••• 22
Examples 27
Chapter HE. Functions of any Angle — Inverse
Functions.
Relations of functions of - ft 90** i ft ISO*' ± ft 270'* ±6 to the
functions ot$ 29
Inverse functions 35
Examples ,36
Review . . . . • 38
Chapter IV. Computation Tables.
Natural functions 40
Logarithms . • • • 40
Laws of logarithms • . 42
Use of tables 45
Cologarithms . • • • • • • • • • .49
Examples • • • 50
V
vi CONTENTS.
Chapter V, Applications.
PAOK
Measurements of heights and distances •••••• 51
Common problems in measurement • • 52
Examples 54
Chapter VI. General FoRMULiE — Trigmjnometric
Equations and Identities.
Sine, cosine, tangent oia± P 56
Examples 59
Sin $ ± sin ff>, $ ± cos <l> . 61
Examples 62
Functions of the double angle 63
Functions of the half angle 64
Examples 64
Trigonometric equations and identities ..•••• 66
Method of attack 66
Examples 67
Simultaneous trigonometric equations 69
Examples 70
Chapter VII. Triangles.
Laws of sines, tangents, and cosines i 72
Area of the triangle 76
Solution of triangles 76
Ambiguous case 78
Model solutions ; 80
Examples 83
Applications 84
Review 86
Chapter "Viil. Miscellaneous.
Incircle, circumcircle, escribed circle 92
Orthocentre, centroid, medians 94
Examples 96
Chapter IX. Series.
Exponential series 97
Logarithmic series 99
Computation of logarithms 100
De Moivre's theorem 103
Computation of natural functions 104
Hyperbolic functions 109
Examples 110
CONTENTS. vii
Chapter X. Spherical Trigonometry.
PAGE
Spherical triangles 112
General formulae 114
Right spherical triangles 123
Area of spherical triangles 125
Examples 128
Chapter XL Solution of Spherical Triangles.
General principles 129
Formulae for solution 130
Model solutions 131
Amhiguous cases 132
Right triangles 134
Species 135
Examples 137
Applications to Geodesy and Astronomy 138
PLANE TRIGONOMETRY.
>>»Co<i-
CHAPTER I.
ANQLISS— MEASUREMENT OF ANQLES.
1. Angles. It is difficult, if not impossible, to define an
angle. This difficulty may be avoided by telling how it
is formed. If a line 'revolve ahovi one of its points^ an angle
is generated^ the magnitude of the angle depending on the
amount of the rotation.
Thus, if one side of the angle d, as OjR, be originally in
the position OX^ and be revolved about the point to the
position in the figure, the
angle XOB is generated.
OXia called the initial line,
and any position of OB the.
terminal line of the angle ,v^^ r x
formed. The ansfle is , ,--''
considered positive if gener- -'' ft i
ated hy a counter-clockwise
rotation of OR^ and hence negative if generated hy a clockwise
rotation. The magnitude of depends on the amount of
rotation of Oi2, and since the amount of such rotation may
be unlimited, there is no limit to the possible magnitude of
angles, for, evidently, the revolving line may reach the posi-
tion OR by rotation through an acute angle ^, and, likewise,
by rotation through once, twice, •••, w times 860®, plus the
acute angle 0, So that XOR may mean the acute angle
0,0 + 360^ + 720^ ..., ^ + n • 360^
1
2 PLANE TRIGONOMETRY.
In reading an angle, read first the initial line, then the
terminal line. Thus in the figure the acute angle XOR^ or
2T, is a positive angle, and ROX^ or nr, an equal negative
angle.
Ex. 1. Show that if the initial lines for ), ), ^, — }, right angles are
the same, the terminal lines may coincide.
2. Name four other angles having the same initial and terminal lines
as I of a right angle ; as } of a right angle ; as ) of a right angle.
2. Rectangular axes. Any plane surface may be divided
by two perpendicular straight lines XX' and TY' into four
_ portions, or qttadrants.
XX' is known as the axiods^
TV as the y-axis^ and the two
together are called axes of refer-
X ' L X ence. Their intersection is the
origin^ and the four portions of
the plane surface, XOY^ TOX\
y" X'0T\ Y'OX, are called respec-
Yyq,, 2. tively the firsts second^ thirds and
fourth quadrants. The position of
any point in the plane is determined when we know its dis-
tances and directions from the axes.
3. Any direction may be considered positive. Then the
opposite direction must be negative. Thus, if AB represents
any positive line, BA is an equal nega-
tive line. Mathematicians usually
consider lines measured in the same direction as OX or OT
(Fig. 2) as positive. Then lines measured in the same direc-
tion as OX' or OT' must be negative.
The distance of any point from the y-axis is called the
abscissa^ its distance from the 2;-axis the ordinate^ of that
point ; the two together are the cod'rdinates of the point,
usually denoted by the letters x and y respectively, and
written (re, y).
p"
o
N
\"t
Y'
Fio. 3.
ANGLES — MEASUREMENT. 3
When taken with their proper signs, the codrdinates define completely
the position of the point. Thus, if the point P is + a units from YT ,
and + h units from XX' ^ any convenient
unit of length being chosen, the position of
P is known. For we have only to measure
a distance ON equal to a units along OX,
and then from N measure n distance h ,
units parallel to OF, and we arrive at the
position of the point P, (a, 6). In like
manner we may locate P', ( — a, ft), in the
second quadrant, P", (—a, — 6), in the
third quadrant, and P'", (a, — 6), in
the fourth quadrant.
Ex. Locate (2, -2); (0,0); (-8, -7); (0, 5); (-2, 0); (2, 2);
(m, n).
4. If OX is the initial line, 6 is said to be an angle of the
firnt^ second^ thirds or fourth quadrant^ according as its ter-
minal line is in the first, second, third, or fourth quadrant.
It is clear that as OR rotates its quality is in no way affected,
and hence it is in all positions considered positive^ and its ex-
tension through 0, 0R\ negative.
The student should notice that the initial line may take any position
and revolve in either direction. While it is customary to consider the
counter-clockwise rotation as forming a positive angle, yet the condi-
\R X X* li *^®"^ ^^ ^ figure may be such
•^ \ • • / ^hat a positive angle may be
generated by a clockwise rota-
tion. Thus the angle XOR in
each figure may be traced as
a positive angle by revolving
the initial line OX to the posi-
tion OR. No confusion can result if the fact is clear that when an
angle is read XOR, OX is considered a positive line revolving to the
position OR. OX' and OR' then are negative lines in whatever direc-
tions drawn. These conceptions are mere matters of agreement, and the
agreement may be determined in a particular case by the conditions of
the problem quite as well as by such general agreements of mathema-
ticians as those referred to in Arts. 3 and 4 above.
Fio. 4.
5. Measurement. All measurements are made in terms
of some fixed standard adopted as a unit. This unit must
4 PLANE TRIGONOMETRY.
be of the same kind as the quantity measured. Thus, length
is measured in terms of a unit length, surface in terms of a
unit surface, weight in terms of a unit weight, value in terms
of a unit value, an angle in terms of a unit angle.
The measure of a given quantity is the number of times it
contains the unit selected.
Thus the area of a given surface in square feet is the
number of times it contains the unit surface 1 sq. ft. ; the
length of a road in miles, the number of times it contains
the unit length 1 mi. ; the weight of a cargo of iron ore in
tons, the number of times it contains the unit weight 1 ton ;
the value of an estate, the number of times it contains the
unit value $1.
The same quantity may have different measures, according
to the unit chosen. So the measure of 80 acres, when the
unit surface is 1 acre, is 80, when the unit surface is 1 sq. rd.,
is 12,800, when the unit surface is 1 sq. yd., is 887,200.
What is its measure in square feet ?
6. The essentials of a good unit of measure are :
1. That it be invariable^ i.e. under all conditions bearing
the same ratio to equal magnitudes.
2. That it be convenient for practical or theoretical pur-
poses.
3. That it be of the same kind as the quantity measured.
7. Two systems of measuring angles are in use, the sexa-
gesimal and the circular.
The sexagesimal system is used in most practical applica-
tions. The right angle, the unit of measure in geometry,
though it is invariable, as a measure is too large for con-
venience. Accordingly it is divided into 90 equal parts,
called degrees. The degree is divided into 60 minutes^ and
the minute into 60 seconds. Degrees, minutes, seconds, are
indicated by the marks ^ ' ", as 36^ 20' 15".
The division of a right angle into hundredths, with subdivisions into
hundredths, would be more convenient. The French have proposed such
MEASUREMENT OF ANGLES.
a centesimal system, dividing the right angle into 100 grades, the grade
into 100 minutes, and the minute into 100 seconds, marked ^^ ^\ as 50s
70^ 28^\ The great labor involved in changing mathematical tables,
instruments, and records of observation to the new system has prevented
its adoption.
8. The circular system is important in theoretical con-
siderations. It is based on the fact that for a given angle
the ratio of the length of its arc to the length of the radius
of that arc is constant, i,e. for a fixed
angle the ratio arc : radius is the same
no matter what the length of the
radius. In the figure, for the angle ^,
OA OB 00
AA' BB' 00'
That this ratio of arc to radius for a fixed angle is constant
follows from the established geometrical principles :
1. The circumference of any circle is 2 tt times its radius.
2. Angles at the centre are in the same ratio as their arcs.
The Radian. It follows that an angle whose arc is equal
in length to the radius is a constant angle for all circles,
since in four right angles, or the perigon, there are always
27r such angles. Thi% constant angle^
whose arc is equal in length to the radius,
is taken as the unit angle of circular
measure, and is called the radian. From
the definition we have
4 right angles = 360® = 2 tt radians,
2 right angles = 180® = tt radians,
1 right angle = 90® = - radians.
Fio. 6.
TT is a numerical quantity, 3.14159+, and not an angle. When we
speak of 180° as tt, 90° as ^, etc., we always mean v radians^ ^ radians, etc.
6 PLANE TRIGONOMETRY.
9. To change from one system of measurement to the
other we use the relation,
2 TT radians = 360®.
1QAO
. •. 1 radian = =^ = 57^2958- ;
TT
i.e. the radian is 57®.3, approadmately.
Ex. 1. Express in radians 75"" W.
75° 30' = 75°.5 ; 1 radian = 57^3.
.% 75** 30' = ^ = 1.317 radians.
67.3
2. Express in degree measure 3.6 radians.
1 radian = 57*^.3.
.-. 3.6 radians = 3.6 x 57^3 = 206° 16' 48".
EXAMPLES.
1. Construct, approximately, the following angles ; 50®, — 20°, 90°,
179°, -135°, 400°, -380^ 1140°, ^radians, ^radians, -^radians,
4 3 6
3 TT radians, — -^ I'^dians, — ^ radians. Of which quadrant is each
angle? ^ ^
2. What is the measure of :
(a) f of a right angle, when 30° is the unit of measure ?
(6) an acre, when a square whose side is 10 rds. is the unit ?
(c) m miles, when y yards is the unit ?
3. What is the unit of measure, when the measure of 2} miles is 50?
4. The Michigan Central K.R. is 535 miles long, and the Ann Arbor
R.R. is 292 miles long. Express the length of the first in terms of the
second as a unit.
5. What will be the measure of the radian when the right angle is
taken for the unit ? Of the right angle when the radian is the unit ?
6. In which quadrant is 45°? 10°? -60°? 145°? 1145°? -725°?
Express each in right angles ; in radians.
7. Express in sexagesimal measure
2[, Z., 1, 6.28, i, I^, - i2[, radians.
3' 12' ' TT 3 3 '
EXAMPLES, 7
8. Express in each system an interior angle of a regular hexagon ;
an exterior angle.
9, Find the distance in miles between two places on the earth's
equator which are 11° 15' apart. (The earth's radius is about 3963 miles.)
10. Find the length of an arc which subtends an angle of 4 radians
at the centre of a circle of radius 12 ft. 3 in.
11. An arc 15 yds. long contains 3 radians. Find the radius of the
circle.
12. Show that the hour and minute hands of a watch turn through
angles of 30' and 6° respectively per minute ; also find in degrees and in
radians the angle turned through by the minute hand in 3 hrs. 20 mins.
13. Find the number of seconds in an arc of 1 mile on the equator ;
also the length in miles of an arc of 1' (1 knot).
14. Find to three decimal places the radius of a circle in which the
arc of 71° 36' 3".6 is 15 in. long.
15. Find the ratio of - to 5°.
6
16. What is the shortest distance measured on the earth's surface
from the equator to Ann Arbor, latitude + 42° 16' 48" ?
17. The difference of two angles is 10°, and the circular measure of
their sum is 2. Find the circular measure of each angle.
18. A water wheel of radius 6 ft. makes 30 revolutions per minute.
Find the number of miles per hour travelled by a point on the rim.
4
/ I
/ '
^v
CHAPTER II.
THE TRiaONOMETRIC FUNCTIONS.
10. Trigonometry, as the word indicates, was originally
concerned with the measurement of triangles. It now
includes the analytical treatment of certain functions of
angles^ as well as the solution of triangles by means of cer-
tain relations between the functions of the angleg of those
triangles.
U. Function. If one quantity depends upon another for
its value, the first is called a function of the second. It
always follows that the second quantity is also a function of
the first ; and, in general, functions are so related that if one
is constant the other is constant, and if either varies in value,
the other varies. This relation may be extended to any
number of mutually dependent quantities.
Illustration. If a train moves at a rate of 30 miles per
hour, the distance travelled is a function of the rate and
time, the time is a function of the rate and distance, and the
rate is a function of the time and distance.
Again, the circumference of a circle is a function of the
radius, and the radius of the circumference, for so long as
either is constant the other is constant, and if either changes
in value, the other changes, since circumference and radius
are connected by the relation 0=2 irR.
Once more, in the' right triangle
NOP^ the ratio of any two sides is
a function of the angle a, because
all the right triangles of which a is
one angle are similar, Le. the ratio
8
THE TRIGONOMETRIC FUNCTIONS.
9
of two corresponding sides is constant so long as a is con-
stant, and varies if a varies.
Thus, the ratios
JVP N'F' N"P"
OP OP'
OP"
and
ON 0N> ON"
;> etc.,
NP N'P' N"P"'
depend on a for their values, i.e. are functions of a.
12. The trigonometric functions. In trigonometry six
functions of angles are usually employed, called the trigono-
metric functions.
By definition these functions are the six ratios between the
sides of the triangle of reference of the given angle. The
triangle of reference is formed ly drawing from some point in
the initial line^ or the initial line produced^ a perpendicular to
that line meeting the terminal line of the angle.
X N
—X
—X
Fig. 8.
Let a be an angle of any quadrant. Each triangle of
reference of a, NOP^ is formed by drawing a perpendicular
to OJf, or OX produced, meeting the terminal line OR in P.
10 PLANE TRIGONOMETRY.
If a is greater than 360°, its triangle of reference would
not differ from one of the above triangles.
It is perhaps worthy of notice that the triangle of reference might be
defined to be the triangle formed by drawing a perpendicular to either
side of the angle, or that side produced, meet-
ing the other side or the other side produced.
In the figure, NOP is in all cases the triangle
of reference of a. The principles of the fol-
lowing pages are the same no matter which
of the triangles is considered the triangle of
FiQ. u. reference. It will, however, be as well, and
perhaps clearer, to use the triangle defined
under Fig. 8, and we shall always draw the triangle as there described.
13. The trigonometric functions of a (Fig. 8) are called
the sine^ co%ine^ tangent^ cotangent^ secant^ and cosecant of a.
These are abbreviated in writing to sin a, cos a, tan a, cot a,
sec a, CSC a, and are defined as follows :
sin a = 2?3^ = M, whence y ^rsina;
nyp. r ^
CM a = i:^ = —9 whence x^^rcosa;
nyp. r
taiia = ^^ = ~» whence y = xtana;
cot tt = = —J whence x=i y cotct;
sec a = ^^^ = —9 whence r = x8eca;
base 0?
CSC a = ^^ = -9 whence r^yc%c a.
perp. y ^
1 — cos a and 1 — sin a, called versed-sine a and coversed-sine a, respec-
tively, are sometimes used.
Ex. 1. Write the trigonometric functions of j8, NPO (Fig. 8), and .•
compare with those of a above.
The meaning of the prefix co in cosine, cotangent, and cosecant
appears from the relations of Ex. 1. For the sine of an angle equals the
cosine, i.e. the complement-siney of the complement of that angle ; the tangent
THE TBIGONOMETRIC FUNCTIONS.
11
of an angle equals the cotangent of its complementary angle, and the secant
of an angle equals the cosecant of its complement'
ary angle.
2. Express each side of triangle ABC in
terms of another side, and some function of an
angle in all possible ways, as a = & tan Ay etc. Fio. 10.
14. Constancy of the trigonometric functions. It is impor-
tant to notice why these ratios are functions of the angle^ i.e.
are the same for equal angles and different for unequal
angles. This is shown by the principles of similar triangles.
Fig. 11.
In each figure show that in all possible triangles of refer-
ence for a the ratios are the same, but in the triangles of
reference for a and a', respectively, the ratios are different.
The student must notice that sin a is a single symbol. It is the name
of a number, or fraction, belonging to the angle a ; and if it be at any
time convenient, we may denote sin a by a single letter, such as o, or x.
Also, sin^a is an abbreviation for (sin a) 2, Le. for (sin a) x (sin a).
Such abbreviations are used because they are convenient. Lock, Ele-
mentary Trigonometry.
15. Fundamental relations. From the definitions of Art. 13
the following reciprocal relations are apparent :
sina =
cosa =
CSC a
1
seca^
tana = -4— >
cot a
Also from the definitions,
cos a
C8Ca=:
8eca =
cota =
cota =
sinci
1
CO8 a
1
tan a
COS a
sin a
12 PLANE TRIGONOMETRY.
From the right triangle NOP, page 9,
j,2 + a^ = r2;
whence (1)
^ + ^-1
^ + ^-1'
•
(2)
(3)
1 + '^='*.
y* y*
From (1) 8iii*a+co8*a=lj nn «= VI — coi? a ; eosass?
(2) ta]i^a+l=sec^a| tow «= Vwc^ a— 1 ; «e<?a=s?
(S) l-f-eot*a=C8C*a; cot a=^^/c8e^ a-^1; esca=i?
The foregoing definitions and fundamental relations are of
the highest importance, and must be mastered at once. The
student of trigonometry is helpless without perfect familiarity
with them.
These relations are true for all values of ce, positive or negative, but
the signs of the functions are not in all cases positive, as appeal's from
the fact that in the triangles of reference in Fig. 8 x and y are sometimes
negative. The equations sin a = ± VI — cos^ a, tan a = ± Vsec^ a — 1,
cot a = ± Vcsc^ a — 1, have the double sign ±, Which sign is to be used
in a given case depends on the quadrant in which a Ues.
16. The relations of Art. 15 enable us to express any
function in terms of any other, or when one function is
given, to find all the others.
Ex. 1. To express the other functions in terms of tangent :
r.;« ^1 1 tan a . ^^x « 1' .
sm a = = = — ; cota= ;
CSC a VI + cot2 a VI + tan^ a ^^ana
1 1
^^^^^TITZ- /. . =' seca = Vl + tan2a;
sec a Vl + tana « ^
tani = tan«; csc« = ^^ + ^^^ >
tana
THE TRIGONOMETRIC FUNCTIONS.
18
In like manner determine the relations to complete the following
table :
sma
cosce
tana
cot a
seca
CSC a
sma
cos a
tana
tan a
VI + tan* a
1
VI + tan* a
tana
1
tana
VI + tan*a
VI 4- tan* g
tana
cot a
A
^.
sec a
CSC a
2. Given sin a = f ; find the other functions.
co8a=vT^^ = jV7; tana =-*-= — = f\/7;
jVr V7
1 /- 14/- 14
cot a = -^=- = i V7 ; sec a = = -=- = 4 V7 ; esc a = i = ^«
^V7 Ja/7 V7 t 3
3. Given tan ^ + cot ^ = 2 ; find sin ^.
1
tan^ +
tan^
= 2, tan* ^ - 2 tan ^ + 1 = 0, tan ^ = 1.
... sin <^= , ^°^ - = }V2.
VI + tan* ^
Or, expressing in terms of sine directly, 5HL$+ S21$^ = 2,
cos^ sm^ I
sin* ^ + cos* ^ = 2 sin ^ cos ^ sin* ^ — 2 sin ^ cos ^ + cov ^ = ;
whence sin ^ — cos ^ = 0, sin ^ = cos ^. .*. sin ^ = ^VS.
4. Prove sec* x — sec* x = tan* x + tan* a?.
sec** — sec*x = sec* a; (sec* x — 1) = (1 + tan* x) tan* ar = tan*x + tan*x.
5. Prove sin* y + cos* y = 1 — 3 sin* y cos* y.
sin* y + cos* y = (sin* y + cos* y) (sin* y — sin* y cos* y + cos* y)
s= (sin* y + cos* y)* — 3 sin* y cos* y = 1 — 3 sin* y cos* y.
14 PLANE TRIGONOMETRY.
,6. Prove -i5^^ + -52ii_=8ec«c8c^ + l.
1 — cot z 1 — tau z
sinz C08Z
tau z cot z cos z sin z
+ i 7- = V
1 — cot z 1 — tan z • __ cos 2 ^^ _ sing
sin 2 cos z
sin* 2; _^ cos* z
cos 2 (sin 2 — cos z) sin 2 (cos z — sin 2)
_ sin' 2 — cos' 2 _ sin* 2 4- sin 2 cos 2 4- cos* 2
sin 2 cos z (sin 2 — cos 2) sin 2 cos 2
^l+8iD»C08»^_^ +l=8ec*C8C^ + l.
Sin 2 cos 2 sm 2 cos 2
In solving problems like 3, 4, 5, and 6 above, it is usually safe, if no
other step suggests itself, to express all other functions of one member
in terms of sine and cosine. The resulting expression may then be re-
duced by the principles of algebra to the expression in the other member
of the equation. For further suggestions as to the solution of trigono-
metric equations and identities see page 66.
\ ^ EXAMPLES.
y 1. Find the values of all the functions of a, if sin « = | ; if tan ce = f ;
if sec a = 2 ; if cos a = J V3 ; if cot a = | ; if esc a = v^.
2. Compute the functions of each acute angle in the right triangles
whose sides are : (1) 3, 4, 5 ; (2) 8, 15, 17 ; (3) 480, 31, 481 ; (4) a,h,c\
>,ex 2a:y a:* + y^
(5) — ?-, . ^ ^ , X + y.
x-y x-y
3. n cos a = JV, find the value of ^^"""^^^""
cos a— cot a
y 4. If 2 cos a = 2 — sin a, find tan a.
5. If sec* a esc* a — 4 = 0, find cot a.
6. Solve for sin p in 13 sin j8 + 5 cos* P = 11.
Prove
-^ 7. sin* ^ ~ cos* ^ = 1 — 2 cos* ^.
V 8. (sin a + cos a) (sin a — cos a) = 2 sin* a — 1.
'' V ^ 9. (sec a + tan a) (sec a — tan a) = 1.
'y^ 10. cos* p (sec* )8 ~ 2 sin* j8) = cos* p + sin* p.
^ ^ J. cos v
11. tan t; 4- sec r =
/
1 — sin V
12 sin w _ 1 4- cos w
1 ~ cos w sin w
13. (sectf + l)(l-costf) = tan*tfcosA
/
FUNCTIONS OF CERTAIN ANGLES. 15
14. sin* t — sin^ t = cos* t — cos^ U
1 — sin p sin p
16. (tan A + cot il)^ = sec^il csc« A.
17. sec^x — sin* x = tan^x + cos^ar.
In the triangle ABC, right angled at C,
18. Given cos A = ^^ BC = 45, find tan B, and AB.
•2 _ ♦,2
' ,19. If cos A = ^, " ^ . and AB = m^ + n^, find ^ C and 5a
fn* 4- n*
, 20. If -4 C = m + n, BC zzm—n, find sin il, cos 5.
21. In examples 18, 19, 20, above, prove sin^il + cos^il = 1;
' 1 + tan^il =8ec2^.
17. Functions of certain angles. The trigonometric func-
tions are numerical quantities which may be determined for
any angle. In general these values are taken from tables
prepared for the purpose, but the principles already studied
enable us to calculate the functions of the following angles.
18. Functions of 0^. If a be a very small angle, the
value of y is very small, and
decreases as a diminishes. , ,t7 ■ [« ^
Clearly, when a approaches fig 12 ^
0^ as a limit, y likewise ap-
proaches 0, and X approaches r, so that when a = 0®,
y = 0, and x = r.
.'. «t«0» = ^ = 0, cot 0" = —^ = CO,
r tan
co,0° = -=l, «ecO° = -^ = l,
r cos 0°
to„0°=2^ = 0, c»cO° = -r^==cc.
X sm U
In the figure of Art. 18, by diminishing a it is clear that \?e can make
y as small as we please, and by making a small enough, we can make the
value of y less than any assignable quantity, however small, so that sin a ap-
proaches as a limit 0. This is what we mean when we say sin 0° = 0.
In like manner, it is evident that, by sufficiently diminishing a we can
make cot a greater than any assignable quantity. This we express by
saying cot QP = co.
16
PLANE TRIGONOMETRY.
19. Functions of 30**. Let NOP be the triangle of refer-
,R ence for an angle of 30^. Make
triangle NOP' = NOP. Then
POP' is an equilateral triangle
(why?), and ON bisects PF.
Hence
9%n
FiQ. 13.
30<> = ^ = f = J.
r 2y 2
Also a; = Vr2 - y2 = V3y2 = y VS.
c.«tf 30^ = 2,
r 2y *
•etf 30° = f VS,
«a« 80° = ^ = -^ = 4= « i^. <»«30°=V3.
X yV3 V8 *
20. Functions of 45°. Let NOP be the triangle of refer>
ence. If angle NOP = 45°, OPN= 46°.
Fio. U.
,♦. jf = x, and r = Va:* + y* = V2 a? = a; V2.
Then «iw 46° = « = -^ = J V2,
«iw46° = ^ =
CO* 46° = - = -^ = i V2,
«an45° = ^ = - = l.
a; a;
Find cot 46°, «ec 46°, cse 46°.
FUNCTIONS OP CERTAIN ANGLES.
17
2L Functions of 60^. The functions 'of 60^ may be com-
puted by means of the figure, or
they may be written from the func-
tions Off the complement, or 30®.
Let the student in both ways ^how
that
sin 60° = J V3, cos 60® = J,
Compute also the other functions of 60®.
22. Functions of 90'
x2r
Fio. 16.
. If a be an angle very near 90®,
the value of x is very small, and de-
creases as a increases toward 90®.
Clearly when a approaches 90® as a
limit, X approaches 0, and t( ap-
proaches r, so that when
a = 90®, a;=0, y = r.
.-. «m90®=l, c?o8 90®=0, eaw90®=Qo.
Compute the other functions. Also find the functions of
90® from those of its complement, 0®.
23. It is of great convenience to the student to remember
the functions of these angles. They are easily found hy
recalling the relative values of the sides of the triangles of
reference for the respective angles^ or the values of the other
functions may readily be computed by means of the funda-
mental relations, if the values of the sine and cosine are
remembered, as follows :
a
0»
80°
45»
60°
90°
sine
cosine
iVi
ivI
iVs
JV2
Jvl
iVi
1V5
18 PLANE TRIGONOMETRY.
ORAL WORK.
1. YOiich is greater, sin 45** or i sin 90** ? sin 60° or 2 sin 30'' ?
2. From the functions of 60^ find those of 30°; from the functions of
90°, those of 0°. Why are the functions of 45° equal to the co-functions
of 45°?
3. Given sin A =:\, find cos A ; tan A»
4. Show that sin ^ esc ^ = 1 ; cos C sec C = 1 ; cot x tan x = 1,
6. Show that sec« - tan^ e = c8c^0- cot^ $ = sin« $ + cos^ 0.
6. Show that tan 30° tan 60° = cot 60° cot 30° = tan 45°.
7. Show that tan 60° sin^ 45° = cos 30° sin 90°.
8. Show that cos a tan a = sin a; sin j3 cot j3 = cos j3.
9. Show that l-*ang30° ^ ^^ g^o = i cos 0°.
l + tan2 30° '
10. Show that (tan y + cot y) sin y cos y = 1.
* EXAMPLES.
^ *^ 1. Show that sin 30° cos 60° + cos 30° sin 60° = sin 90°.
Y^^ ' 2. Show that cos 60° cos 30° + sin 60° sin 30° = cos 30°.
3. Show that sin 45° cos 0° - cos 45° sin 0° = cos 45°.
4. Show that cos2 45° - sina45° = cos 90°.
5. Show that tan 45° + tan 0° ^ ^^^ ^^o.
/ 1 ~ tan 45° tan 0°
If il = 60°, verify
a o^r, A jl — c os A
6. 8m_ = ^
7. tan4 = A^
2 ^1 +
— cos^
/ 2 ^1 + coSil
8. coSi4 = 2cos24~l = l-2sin«:i.
2 2
If a = 0°, )8 = 30°, y = 45°, 8 = 60°, c = 90°, find the values of
9. sin p + cos 8.
/ 10. cosjS + tan8.
11. sin j3 cos 8 + cos j3 sin 8 — sin e.
12. (sin P + sin f ) (cos « + cos 8) — 4 sin a (cos y + sin c) .
VARIATIONS IN THE FUNCTIONS.
19
24. Variations in the trigonometric functions.
Signs. Thus far no account has been taken of the signs of
the functions. By the definitions it appears that these de-
pend on the signs of re, y, and r. Now r is always positive,
and from the figures it is seen that x is positive in the fii*st
Ml. +
C90, +
(r-)
(r+)
Cot. 4^
Sin. +
C09. +
Tan,+
CM, +
See, +
C!MX +
%-)
Co».-h
See. -»-
Fig. 17.
and fourth quadrants, and y is positive in the first and
second. Hence
For an angle in the first quadrant all functions are positive^
since re, y, r are positive.
In the second quadrant x alone is negative^ so that those
functions whose ratios involve 2?, viz. cosine^ tangent^ co-
tangent^ secant^ are negative; the others, sine and cosecant^
are positive.
In the third qvxidrant x and y are both negative^ so that
those functions involving r, viz. sine^ cosine^ secant^ cosecant^
are negative; the others, tangent and cotangent^ rtb positive.
In the fourth quadrant y is negative^ so that sine^ tangent^
cotangent^ cosecant are negative^ and cosine and secant^ positive.
Valves. In the triangle of reference of any angle, the
hypotenuse r is never less than x or y. Then if r be taken of
any fixed length, as the angle varies, the base and perpen-
dicular of the triangle of reference may each vary in length
X u
from to r. Hence the ratios - and - can never be greater
r r
than 1, nor if x and y are negative, less than —1; and -^ -
X y
20
PLANE TRIGONOMETRY.
cannot have values between + 1 and — 1. But the ratios
^ and - may vary without limit, i.e. from + oo to — oo.
X y
Therefore the possible values of the functions of an angle
are :
sine and cosine between + 1 and — 1,
i.e. nnt and cosine cannot be numerically greater than 1;
tangent and cotangent between + oo and — oo,
i.e. tangent and cotangent may have any real valite;
secant and cosecant between + oo and + 1, and — 1 and — oo,
i.e. secant and cosecant may have any real values^ except
values between + 1 and — 1.
These limits are indicated in the following figures. The
student should carefully verify.
Ate iy>to
Bin, M**-!
Cm »0°«10
Tan, W-ioD
Cm IW^a^i
Tan UO-70
Ml. 41
Cm. -0
2kra. —flo
F
* t *
• 9S 6 £
sm o-±o o /♦« -1 -0
Cm^.
Tom o'^-tO
Bin, -1
Cm. -0
+1
+0
QQ O K
•i-o. -t-i. -t-0^ 0'
O -0. +1, -0-^380"
-1
+
S/II8T0--1
(M« 170^^70
Tom STO'£ao
Tan, +«>| -00
If'
270"
FiQ. 18.
25. In tracing the changes in the values of the functions as
a changes from 0® to 360°, consider the revolving line r as
of fixed length. Then x and y may have any length between
and r.
Sine, At 0°, sin a = ^ = - = 0. As a increases through
r r
the first quadrant, y increases from to r, whence - increases
from to 1. In passing to 180® sin a decreases from 1 to 0,
VARIATIONS IN THE FUNCTIONS. 21
since y decreases from r to 0. As a passes through 180°, y
changes sign, and in the third quadrant decreases to nega-
tive r, so that sin a decreases from to — 1. In the fourth
quadrant y increases from negative r to 0, and hence sin a
increases from — 1 to 0.
Co%ine depends on changing values of x. Show that,
as a increases from 0° to 360°, cos a varies in the four
quadrants as follows: 1 to 0, to — 1, — 1 to 0, to 1.
Tangeid depends on changing values of both y and x.
At 0°, y = 0, a; = r, at 180°, y = 0, a; = - r,
at 90°, 2? = 0, y = r, at 270°, a; = 0, y = - r.
Hence tan 0° = -2^ = - = 0. As a passes to 90°, y increases
X r
to r, and x decreases to 0, so that tan a increases from to oo.
As a passes through 90°, x changes sign, so that tan a
changes from positive to negative by passing through oo.
In the second quadrant x decreases to negative r, y to 0, and
tan a passes from — oo to 0. As a passes through 180°,
tana changes from minus to plus by passing through 0,
because at 180° y changes to minus. In the third quadrant
tana passes from to oo, changing sign at 270° by passing
through 00, because at 270° x changes to plus. In the fourth
quadrant tan a piasses from — oo to 0.
Cotangent. In like manner show that cot a passes through
the values oo to 0, to — oo, oo to 0, to — oo, as a passes
from 0° to 360°.
Secant depends on x for its value. Noting the change
in X as under cosine, we see that secant passes from 1 to oo,
— 00 to — 1, — 1 to — 00, 00 to 1.
Cosecant passes through the values oo to 1, 1 to oo,
— 00 to — 1, — 1 to — 00 .
The student should trace the changes in each function
fully, as has been done for sine and tangent, giving the
reasons at each step.
22
PLANE TRIGONOMETRY.
a
0° to 90°
90° to 180°
180° to 270°
270° to 360*»
sin
to 1
1 to
-0 to -1
- 1 to -Q
cos
1 to
- to - 1
-1 to -0
to 1
tan
to 00
- 00 to -
to 00
-00 to -0
cot
00 to
- to -00
00 to
- to -00
sec
1 to 00
— 00 to — 1
— 1 to —00
00 to 1
CSC
00 to 1
1 to 00
— 00 to — 1
— 1 to —00
* 26. Graphic representation of functions. These variations
are clearly brought out by graphic representations of the
functions. Two cases will be considered : I, when a is a
constant angle ; II, when a is a variable angle.
I. When a is a constant angle.
The trigonometric functions are ratios, pure numbers.
By so choosing the triangle of reference that the denomi-
nator of the ratio is a side of unit length, the side forming
the numerator of that ratio will be a geometrical representa-
tion of the value of that function, e.g. if in Fig. 19 r = 1,
then sin a = ?f = |.= y. This may be done by making a a
r 1
central angle in a circle of radius 1, and drawing triangles
of reference as follows :
FiQ. 19.
GRAPHIC REPRESENTATION OF FUNCllONS. 23
In all the figures AOP = a, and
SP SP T>r>
OB OB ^j. \j ,^' * ;^ ,. /•
. BP AD AD .J.
tan a = = = = AD^
OB OA 1
. OA HO EC Tpr,
OP OD OB ^n
OP 00 00 f,n
''"' = BP = OE = -=^^-
It appears then that, by taking a radius 1,
sine is represented by the perpendicular to the initial line,
drawn from that line to the terminus of the arc sub-
tending the given angle;
cosine is represented by the line from the vertex of the
angle to the foot of the sine;
tangent is represented by the geometrical tangent drawn
from the origin of the arc to the terminal line, produced
if necessary;
cotangent is represented by the geometrical tangent drawn
from a point 90° from the origin of the arc to the
terminal line, produced if necessary;
secant is represented by the terminal line, or the terminal
line produced, from the origin to its intersection with
the tangent line;
cosecant is represented by the terminal line, or the terminal
line produced, from the origin to its intersection with
the cotangent line.
24
PLANE TRIGONOMETRY.
The9e lines are not the functions^ but in triangles drawn
as explained their lengths are equal to the numerical values
of the functions, and in this sense the lines may be said to
represent the functions. It will be noticed also that their
directions indicate the signs of the functions. Let the
student by means of these representations verify the results
/ of Arts. 24 and 25.
II. WTien a is a variable angle.
Take XX* and TT' as axes of reference, and let angle
units be measured along the 2;-axis, and values of the func-
tions parallel to the y-axis, as in Art. 3. We may write
corresponding values of the angle and the functions thus :
a-(f, 30°, 46^ 60», W, 120**, 136^ 150°, 180°, 210°, 226°,
Bina=0, J, 1V2,JV3,1, iV3, Jv^, J, 0, - J, - Jv^,
«= 240°, 270°, 300°, 316°, 330°, 360°, -30°, -46°, -60°, -90°, etc.,
Bina=-J>/3, -1, -jV3, -jV2, -J, 0, -J, -jV2, -i>/a, -1, etc.
These values will be sufi&cient to determine the form of the
curve representing the function. By taking angles between
those above, and computing
the values of the function, as
given in mathematical tables,
the form of the curve can be
determined to any required
degree of accuracy. Reduc-
ing the above fractions to
decimals, it will be convenient
to make the y-units large in
comparison with the a;-units.
In the figure one a?-unit repre-
sents 15°, and one y-unit 0.25.
Measuring the angle values along the a;-axis, and from these
points of division measuring the corresponding values of sin a
parallel to the y-axis, as in Art. 3, we have, approximately.
Curves of Sine and Cosecant,
Sine
Cosecant
Fio.20.
GRAPHIC REPRESENTATION OF FUNCTIONS. 25
C>Xi = 30° = 2 units, 0X^ = ^5'' « 3 units,
Xi Fi = J =2 units, Xg Fg = 0. 71 = 2. 84 units,
OX^ = 60° =4 units, etc.,
-y8F8 = 0.86 = 3.44 units, etc.
We have now only to draw through the points F^, Fg, Fg,
etc., thus determined, a continuous curve, and we have the
sine-curve or sinusoid.
The dotted curve in the figure is the cosecant curve. Let
the student compute values, as above, and draw the curve.
In like manner draw the cosine and secant curves, as
follows :
I 1
Curves of Cosine and SecarU,
Cosine
Secant ""
Fig. 21.
Tangent curve. Compute values for the angle a and for
tan a, as before :
a = 0°, 30°, 46°, 60°, 90°, 120°, 186°, 160°, 180°, 210°, 226°, 240°, 270%
tan a = 0, jV3, 1, VS, ±«, -V3, -1, -jVS, 0, JV3, 1, V8, ±oo,
a = - 30°, - 46°, - 60°, - 90°, etc.,
tan a = — J V3, — 1, — V3, ± «, etc.
Then lay off the values of a and of tan « along the a?, and
parallel to the y-axis, respectively. It will be noted that,
26
PLANE TRIGONOMETRY.
as a approaches 90^, tan a increases to oo, and when a passes
90^, tan a is negative. Hence the value is measured parallel
Curves of Tangent and Cotangent
Tangent
Cotangent
Fio. 22.
to the y-axis downward, thus giving a discontinuous curve,
as in the figure.
* 27. The following principles are illustrated by the curves :
1. The sine and cosine are continuous for varying values
of the angle, and lie within the limits + 1 and — 1. Sine
changes sign as the angle passes through 180°, 360®, •••,
n 180**, while cosine changes sign as the angle passes through
90*=; 270*^, — , (2w + l) 90°. Tangent and cotangent are
discontinuous, the one as the angle approaches 90°, 270°, ...,
(2 w+1) 90°, the other as the angle approaches 180°, 360°, —,
71180°!, and each changes sign as the angle passes through
these vetoes. The limiting values of tangent and cotangent
are + oo and — oo.
2. A line parallel to the y-axis cuts any of the curves in
but one point, showing that for any value of a there is but
one value of any function of a. But a line parallel to the
a;-axis cuts any of the curves in an indefinite number of
points, if at all, showing that for any value of the function
there are an indefinite number of values, if any, of a.
GRAPHIC REPRESENTATION OF FUNCTIONS. 27
8. The curves afford an excellent illustration of the varia-
tions in sign and value of the functions, as a varies from
to 360^, as discussed in Art. 25. Let the student trace these
changes.
4. From the curves it is evident that the functions are
periodic^ i.e. each increase of the angle through 360® in the
case of the sine and cosine, or through 180° in the case of
the tangent and cotangent, produces a portion of the curve
like that produced by the first variation of the angle within
those limits.
6. The difference in rapidity of change of the functions
at different values of a is important, and reference will be
made to this in computations of triangles. (See Art. 64,
Case III.) A glance at the curves shows that sine is chang-
ing in value rapidly at 0®, 180°, etc., while near 90°, 270°,
etc., the rate of change is slow. But cosine has a slow rate
of change at 0°, 180°, etc., and a rapid rate at 90°, 270°, etc.
Tangent and cotangent change rapidly throughout.
Ex. Let the student discuss secant and cosecant curves.
ORAL WORK.
1. Express in radians ISO'*, 120°, 45°; in degrees, ^ radians, 2ir,
fir, iir.
2. If } of a right angle be the unit, what is the measure of } of a
right angle? of 90°? of 135°?
3. Which is greater, cos 30° or 4 cos 60°? tan^ or cot J? sin ? or cos- ?
^ ' ' 6 3 4 4
4. Express sin a in terms of sec a ; of tan a ; tan a in terms of cos a ;
of sec a,
5. Given sin a = {, find tan a. If tan a = 1, find sin a, esc a, cot a ;
also tan 2 a, sin 2 a, cos 2 a.
6. If cos a = 4, find sin ^, tan ^<
' 2 2
7. In what quadrant is angle f, if both sin t and cos t are minus ? if
sinf is plus and cos< minus? if tan< and cot< are both minus? if sint
and CSC f are of the same sign ? Why ?
8. Of the numbers 3, f, — 5, — 4> ^> "" ^> °°» ^> which may be a value
of sin;) ? of Bocp ? of tan p ? Why ?
\
28 PLANE TRIGONOMETRY.
EXAMPLES.
1. If 8in26''40' = 0.44880, find, correct to 0.00001, the cosine and
tangent.
2. If tan a = V3, and cot p = | V3, find sin a cos /3 — cos a sin p,
3 j,^^!^^^ sin 30^ cot 300^ cos 60O tan 00°
sin 90° cos 0°
Prove the identities :
4. tan^(l -cot2>4) + coti4(l -tan2^) = 0.
5. (sin A + sec i4 )2 + (cos A + esc .4)2 = (1 + sec ^ esc Ay.
6. sin^ X cos a; esc re — cos' x esc x sin^ x + cos^ x sec a; sin :e = sin* x cos x
+ cos'z sinx.
7. tan^ w + cot^ w = sec^ to esc* to — 2.
8. sec* V + cos* u = 2 + tan* i; sin* v,
9. cos* r + 1 = 2 cos« r sec « + sin*«.
10. CSC* t — sec* ^ = cos* t esc* ^ — sin* t sec* f.
11. The sine of an angle is ; find the other functions.
12. If tan il + sin .4 = m, tan A — sin -4 = n, prove to* — n* = ^Vmi,
Solve for one function of the angle involved the equations :
13. sintf + 2costf = 1. 16. 2sin*x + cosx - 1 = 0.
, ^ cos a 3 ^^' 8ec*x — 7 tan x — 9 = 0.
14. = -.
tan a 2 18. 3 esc y + 10 cot y - 35 = 0.
15. V3csc*^ = 4cotft 19. sin*i;-}cost;-l = 0.
20. a sec* to + b tan to + c — a = 0.
21. If ?HLii=v^, *!£ii = v^, find .4 and B.
sm B tan B
22. Find to five decimal places the arc which subtends the angle of
1° at the centre of a circle whose radius is 4000 miles.
23. If CSC il = }\/3, find the other functions, when A lies between
~ and w.
2
24. In each of two triangles the angles are in G. P. The least angle
of one of them is three times the least angle of the other, and the sum of
the greatest angles is 240^ Find the circular measure of each of the
angles.
CHAPTER III.
FXJITCTIOITS OF ANY ANGLE — INVERSE FUNCTIONS.
28. B}*^ an examination of the figure of Art. 24 it is seen
that all the fundamental relations between the functions hold
true for any value of a. The table of Art. 16 expresses the
functions of a, whatever be its magnitude, in terms of each
of the other functions of that angle if the ± sign be prefixed
to the radicals.
The definitions of the trigonometric functions (Art. 12)
apply to angles of any size and sign, but it is always possible
to express the functions of any angle in terms of the func-
tions of a positive acute angle.
The functions of any angle ^, greater than 360°, are the
same as those of ^ ± n • 360°, since and ±n' 360° have
the same triangle of reference. Thus the functions of 390°,
or of 750°, are the same as the functions of 390°— 360°, or
of 750°— 2-360°, i.e. of 30°, as is at once seen by drawing a
figure. So also the functions of — 315°, or of — 675° are
the same as those of - 315° + 360°, or of - 675° + 2-360°,
i.e. of 45°.
For functions,of angles less than 360° the relations of this
chapter are important.
29. To find the relations of the functions of — 0, 90° ± ^,
180° ± 0, and 270° ±0to the functions of 0^ being any angle.
Four sets of figures are drawn, I for an acute angle, II
for obtuse. III for an angle of the third quadrant, and
IV for an angle of the fourth quadrant.
In every case generate the angles forming the compound
angles separately, i.e. turn the revolving line first through
29
80 PLANE TBIGONOMETBY.
W (») (0)
-e i80°±e
^:5 3KF
^^
j:^T
FUNCTIONS OF ANY ANGLE.
W
82 PLANE TRIGONOMETRY.
O*', 90°, 180°, or 270°, and then from this position through
0^ or — 0^ as the case may be. Form the triangles of refer-
ence for (a) the angle 0, (b) - 0, ((?) 180° ± 0, (rf) 90° ± 0,
(6)270°±^.
The triangles of reference (a), (6), ((?), (c?), and (e), in
each of the four sets of figures, I, II, III, IV, are similar,
being mutually equiangular, since all have a right angle and
one acute angle equal each to each. Hence the sides x, y^ r
of the triangles (d) are homologous to x\ y\ r^ of the cor-
responding triangles (6) and ((?), but to y\ a;', r', of the
corresponding triangles (c?) and (e). For the sides x of
triangle (a) and aj' of the triangles (6) and ((?) are opposite
equal angles, and hence are homologous, but the sides y^ are
opposite this same angle in triangles (rf) and (e), and there-
fore sides y' of (df) and («) are homologous to x of (a).
Attending to the signs of x and x', y and y in the similar
triangles (a) and (6),
sin(-^) = ^ = -^ =-sin^.
a?' X
r
r
cos (— ^)=-r = - = cos ^,
tan (- ^) = ^= - ^ = - tand.
^ ^ aj' a;
Also in the similar triangles (a) and ((?),
sin (180° - ^) = ^ = I = sin ^,
cos (180° - ^) = ^ = - ? = - cos d,
tan (180° - ^) = ^ = - ^ = - tand.
X X
In like manner show that
sin(180° + ^) = -sin^,
cos(180° + ^) = -cos^,
tan (180° + 0)= tan 0.
FUNCTIONS OF ANY ANGLE. 83
Again, in the similar triangles (a) and (d"),
8in(90° + d) = ^ = - =cosd,
cos (90° + 0)=^ = -^=-ava0,
tan(90° + ^)=^ = --=-cot^. •
X y
Show that
sin (90*^ - ^) = cos 0,
cos (90^ - ^) = sin 0,
tan (90^-^) = cot ^.
.; ' ' Finally, from the similar triangles (a) and (e), show that
sin (270°±^)=-cos^,
cos (270° ± ^) = ± sin 0,
tan (270° ± ^) = T cot 0.
From the reciprocal relations the student can at once
write the corresponding relations for secant, cosecant, and
cotangent.
30. Since in each of the four cases x\ y^ of triangles
(6) and (js) are homologous to x, y of triangle (a), while
x\ y* of the triangles (d) and (g) are homologous to y, x
of triangle (a), we may express the relations of the last
article thus :
The functions of -[qm . n correspond to the same functions
of 0, while those of oyno . n correspond to the co-functions
of 0j due attention being paid to the signs.
The student can readily determine the sign in any given
case, whether be acute or obtuse, by considering in what
quadrant the compound angle, 90° ± 0^ 180° ± 0^ etc., would
84 PLANE TRIGONOMETRY.
lie if were an acute angle, and prefixing to the correspond-
ing functions of the signs of the respective functions for
an angle in that quadrant. Thus 90° + ^, if ^ be acute, is
an angle of the second quadrant, so that sine and cosecant
are plus, the other functions minus. It will be seen that
sin (90° + ^) = +COS 0, cos (90° + ^) = - sin 0, etc., and this
will be true whatever be the magnitude of 0. It will assist
in fixing in the memory these important relations to notice
that when in the compound angle is measured from the
y-axis, as in 90° ± 0^ 270° ± 0^ the functions of one angle
correspond to the co-functions of the other, but when in the
compound angle is measured from the 2;-axis, as in ± d,
180° ± ^, then the functions of one angle correspond to the
same functions of the other.
These relations, as has been noted in Art. 28, can be
extended to angles greater than 360°, and it may be stated
generally that
function ^ = ± function (2 w • 90° ± ^),
function ^ = ± co-function [(2 n + V) 90° ± ^].
Computation tables contain angles less than 90° only. The chief
utility of the above relations will be the reduction of functions of angles
greater than 90° to functions of acute angles. Thus, to find tan 130° 20',
look in the tables for cot 40° 20', or for tan 49° 40'. Why?
Ex. 1. What anglea less than 360° have the same numerical cosine
as 20°?
cos 20° = - cos (180° ± 20°) = cos (360° - 20°).
.-. 200°, 160°, 340° have the same cosine numerically as 20°.
2. Find the functions of 135° ; of 210°.
sin 135° = sin (90° + 45°) = cos 45° = J V^,
cos 135° = cos (180° - 45°) = - cos 45° = - i V2, etc.
sin 210° = sin (180° + 30°) = - sin 30° = - J.
Let the student give the other functions for each angle.
INVERSE FUNCTIONS. 85
ORAL WORK.
1. Determine the sine and tangent of each of the following angles :
30°, 120°, - 30°, - 60°, J IT, 2} ir, - 135°, - ir.
2. Which is the greater, sin 30° or sin (-30°)? tan 135=» or tan 45° ?
cos 60° or cos( - 60°) ? sin 22° 30' or cos 67° 30' ?
3. What positive angle has the same tangent as -? the same sine
as 50°? ^
4. If tantf = -1, findsinft
5. Find sin 510°, cos(- 60°), tan 150°
6. Reduce in two ways to functions of a positive acute angle, cos 122°
tanl40°30',sin(-60°).
7. Find all positive values of x, less than 360°, satisfying the fol-
lowing equations : cos x = cos 45°, sin 2 x = sin 10°, tan 3 a: = tan 60°,
sin a? = sin 30°, tan x = tan 135°.
8. What angles are determined when (a) sine and cosine are + ?
(6) cotangent and sine are — ? (c) sine + and cosine — ? (<f) cosine —
and cotangent + ?
INVERSE FUNCTIONS.
31. That a is the sine of an angle 6 may be expressed in
two ways, viz., sin ^ = ^, or, inversely, = sin""^ a, the latter
being read, 6 equals an angle whose sine is a, or, more briefly,
is the anti-sine of a.
The notation sin*^ a, cos"^ a, tan*^ a, etc., is not a fortunate one, but
is so generally accepted that a change is not probable. The symbol may
have been suggested from the fact that it ax = b, then x = a'^ 6, whence,
by analogy, if sin 6 = a, 6 = sin"^ a. But the likeness is an analogy only,
for there is no similarity in meaning. Sin~^ a is an angle 6, where sin = a,
and is entirely different from (sin a)-^ = -: — . In Europe the symbols
arc sin a, arc cos a, etc., are employed.
32. Principal value. We have found that in sin ^ = a,
for any value of ^, a can have but one value ; but in
= sin"-^ a, for any value of a there are an indefinite number
of values of (Art. 27, 2).
Thus, when sin ^ = a, if a = J, ^ may be 80°, 150°, 890°,
610°, - 330°, etc., or, in general, wtt +(- 1)«80°.
In the solution of problems involving inverse functions.
86
PLANE TRIGONOMETKY,
the numerically least of these angles, called the principal
value^ is always used ; i.e. we understand that sin~i a, tan""-^ a,
are angles between + 90® and — 90®, while the limits of
cos-^a are 0® and 180®.
Thus, sin-i^ = 30®, sin-i(- J)=:- 30®, cos-^J^GO®,
cos--i(-^)=120®,
ORAL WORK.
How many degrees in each of the following angles? How man}
radians ?
1. COS'
2
2. tan-il?
3. cot-i(-V3)?
4. 8in-i(-jV2)?
5. cos-i(-i>/2)?
7. tan-iV5?
8. cos-iQ?
9. sin-il?
10. tan-iQ?
11. tan-i(-l)?
12. sin-i(-l)?
Find the values of the functions :
13. sin(tan-i jVa).
14. tan(cos"^ 1).
15. tan(cot"^[— «]).
16. cos(tan~ioo).
17. sin(am-ijV2).
18. tau(tan~^x).
3. K tf = csc-i a, prove = cos
19. cos(sin-iO).
20. 8in(cos-i[- 1]).
21. cos(cot-i>/3).
22. tan(sin-i[- 1]).
23. 8in(tan-i[-l]).
Ex. 1. Construct cot~i |.
Construct the right triangle xyr, so that 2 = 4,
y = 3, whence angle xr = cot"^ |.
2. Find co8(tan~^ -j^).
Let = tan"i -j^, whence
tan 6 = ^y and cos = ^.
.'. cos tf = cos(tan-i -fg) = J^.
a
C8C$ = a ; .'. sin tf = ->
a
and
cos
0=^1^, = ^^^. or 6i = co8-»:i^iHI
EXAMPLES- 87
EXAMPLES.
1. Construct sin*i}, tan-^^, cos~i(-- }).
/i/*^ 2. Find tan(sin-lA)»si"(tan-lA)•
/ . 3. K tf = sin-i a, prove B = tan-^ »
4. Show that sm"^ a = 90® — cos"^ a.
#
5. Prove tan-i V3 + cot-iV3 = ^.
6. Prove tan-if sin ^^ = cos-i Jv^.
/
7. What angles, less than 360°, have the same tangent numerically
as 10°?
8. Given tan 143° 22' = - 0.74357 ; find, correct to 0.00001, sine and
cosine.
9. If cot2(90° + i8) + csc(90° - )8) - 1 = 0, find tan j3.
10. Find all positive values of a:, less than 360°, when sin a:=8in 22° 30';
when tan 2 ar = tan 60°.
11. When is sin x = " ^ ^ possible, and when impossible ?
12. Verify sin'i i + cos-i^ + tan-i>/3 = sin-i ^.
13. What values of x will satisfy sin-i(a:2 - x)- 30°?
14. If tan^ 6 - sec« a = 1, prove sec B + tan« tf esc ^ = (3 + tan« a)*.
15. Prove sin -4 (1 + tan ^ ) + cos ^ (1 + cot A) = sec il + esc A.
16. Solve the simultaneous equations :
sin-i(2 a: + 3 y) = 30° and 3ar + 2y = 2.
17. Verify (a) tan 60° = ^^
(J) cos 60° =
~ cos 120° .
cos 120°*
1 - tan^ 30° ,
1 + tan2 30°'
(c) 2 sina 60° = 1 - cos 120°.
18. Show that the cosine of the complement of ^ equals the sine of
the supplement of 2-
88 PLANE TRIGONOMETRY.
REVIEW.
Before leaving a problem the student should review and master all
principles involved.
1. Construct cos"^ ^ ; 8in-i( — j) ; tan"i 2.
2. Find cos (sin-i f ) ; tan (cos"^ [ — i] ) •
1/ 3. Prove cot"^ a = cos"^ — ^^ »
4. Given a = cot-i f , find tan a + sin (90** + a),
5. Find tan (sin-ij + cos-i^).
6. State the fundamental relations between the trigonometric func-
tions in terms of the inverse functions. Thus,
sin*^a = csc*^-, sin"^a = cos"^Vl — a^, etc.
a
7. Find all the angles, less than 360**, whose cosine equals sin 120^
8. Given cot'^ 2.8449, find the sine and cosine of the angle, correct
tb 0.0001.
9. K tan2 (180° - tf) - sec (180'' + tf) = 5, find cos 0.
10. K sin 61 = I, find J-^B!|±co8?|
Y^ ' tan^tf-cos^^
11. Is sin 2? — 2 cos a; + 3 sin 2? — 6 = a possible equation ?
12. Verify (a) sin 60*' = ^ ^^^ ^Q° .
^ ^ ^ l + tan230°
(b) 2 cos^ 60' = 1 + cos 120*».
(c) cos 60° - cos 90° = 2 cos« 30° - 2 cos« 46®.
13. If sin a: = ^ ^^^ "^, — ^^— -, find sec x and tan x.
14. Prove 1 + sing - cos^ 1 + sin^-f cosg^ ^^^g^
^ 1 + sing + cosg 1 + sing - cosg
15. Prove
cos 45° + cos 135° + cos 30° + cos 150° - cos 210° + cos 270° = sin 60°.
- 16. If tan = prove that
sin g(l + tan 6) + cos g(l + cot 6) - sec g = ?•
17. Solve sin^x + sin^ (x + 90°) + sin^ (x + 180°) = 1.
EXAMPLES. 89
18. Given cos^ a = m sin a — n, find sin a.
19. If sin2i8=— 2--r,findj3.
^ 20. Given tan 238° = 1.6, find sin 148°.
21. Prove tan-i m + cot'i m = 90^
22. Find sin (sin"^ jo + co8~^/>) .
23. Solve cot2 ^ (2 esc ^ - 3) + 3 (esc tf - 1) = 0.
24. Prove sin^ a sec^ )8 + tan^ j3 cos^ a = sin^ a + tan^ )3.
25. Prove cos« V + sin« F=l-3sinaF+3 sin* F.
26. What values of A satisfy sin 2 il = cos 3 /I ?
27. ntanC = ^ ^-"^^ and tan Z> ==VLil22L2^ find tan D in terms
-^ m 'l+cosC
Ol 171.
28. If sin a: — cos ar + 4 cos^a: = 2, find tan x ; sec ar. '^^^ i j:
]^ 2 cos* ar
29. Does the value of sec a?, derived from sec* a: = — — , give a
possible value of a:? ""
30. Prove
[cot (90° - ^ ) - tan (90'' + il )] [sin (180° - il) sin (90° + ^)] = 1.
31. Prove (1 + sin -4)* [cot il + 2 sec il (1 — esc -4)] + esc A cos« ^ = 0.
32. Given sin a: = m sin y, and tan x = n tan y, find cos x and cos y.
33. Given cot201° = 2.6, find cos 111°.
34. Find the value of
* cos-i J + sin-i J V2 + csc-i( - 1) + tan'i 1-2 cot-^ V3.
^35. Solve 2cos*^ + llsintf -7 = 0.
36. Prove
cos* B + cos* (J5 + 90°) + cos* (J5 + 180?) + cos* (J5. + 270°) = 2.
CHAPTER IV.
COMPUTATION TABLES.
33. Natural functions. It has been noted that the trigo-
nometric functions of angles are numherB^ but the values
were found for only a few angles, viz. 0°, 30°, 45*^, 60°,
90°, etc. In computations, however, it is necessary to know
the values of the functions of any angle, and tables have
been prepared giving the numerical values of the functions
of all angles between 0° and 90° to every minute. In
these tables the functions of any given angle, and con-
versely the angle corresponding to any given function, can
be found to any required degree of accuracy ; e,g. by look-
ing in the tables we find sin 24° 26'= 0.41363, and also
1 .6415 = tan 58° 39'. These numbers are called the natural
functions^ as distinguished from their logarithms, which are
called the logarithmic functionB of the angles.
Ex. 1. Find from the tables of natural functions :
sin 35° 14'; cos 54" 46'; tan 78° 29'; cos 112° 58'; sin 135°.
2. Find the angles less than 180° corresponding to :
sin-i 0.37865 ; cos'i 0.37865 ; tan-i 0.58870 ; cos'i 0.00291 ; sin-i 0.99999.
34. Logarithms. The arithmetical processes of multi-
plication, division, involution, and evolution, are greatly
abridged by the use of tables of logarithms of numbers
and of the trigonometric ratios, which are numbers. The
principles involved are illustrated in the following table :
Write in parallel columns a geometrical progression having
'the ratio 2, and an arithmetical progression having the dif-
ference 1, as follows :
40
LOGARITHMS.
41
It will be perceived that the numbers in
the second column are the indices of the
powers of 2 producing the corresponding
numbers in the first column, thus : 2^ = 64,
2" = 2048, 218 = 262144, etc. The use of
such a table will be illustrated by examples.
Ex. 1. Multiply 8192 by 128.
From the table, 8192 = 2i«, 128 = 2^ Then by
actual multiplication, 8192 x 128 = 1048576, or by the
law of indices, 2i» x 2^ = 2» = 1048576 (from table).
Notice that the simple operation of addition is sub-
stituted for multiplication by adding the numbers in
the second column opposite the given factors in the
first column. This sum corresponds to the number
in the first column which is the required product.
2. Divide 16384 by 512.
16384 -T- 512 = 32, which corresponds to the result
obtained by use of the table, or 2" -j- 2« = 2« = 32.
The operation of subtraction takes the place of
division.
3. Find v^262144.
v^262l44 = v^ = 2^ = 2« = 8.
In the table, 262144 is opposite 18. 18 -r- 6 = 3,
which is opposite 8, the required root ; i.e, simple division takes the
place of the tedious process of evolution.
' 6. Find \^^2768.
G. p.
A. p.
1
2
1
4
2
8
3
16
4
32
5
64
6
128
7
256
8
512
9
1024
10
2048
11
4096
12
8192
13
16384
14.
32768
15
65536
16
131072
17
262144
18
524288
19
1048576
20
4. Cube 64.
5. Multiply 256 by 4096.
7. Divide 1048576 by 32768.
35. The above table can be made as complete as desired
by continually inserting between successive numbers in the
first column the geometrical mean, and between the opposite
numbers in the second, the arithmetical mean, but in prac-
tice logarithms are computed by other methods. The num-
bers in the second column are called the logarithms of the
numbers opposite in the first column. 2 is called the base of
this system, so that the logarithm of a number is the exponent
by which the base is affected to produce the number.
42 PLANE TRIGONOMETRY.
Thus, the logarithm of 612 to the base 2 is 9, since
29 = 512.
Logarithms were invented by a Scotchman, John Napier, early in the
seventeenth century, but his method of constructing tables was different
from the above. See Encyc, Brit., art. " Logarithms,^^ for an exceedingly
interesting account. De Morgan says that by the aid of logarithms the
labor of computing has been reduced for the mathematician to about
one-tenth part of the previous expense of time and labor, while Laplace
has said that John Napier, by the invention of logarithms, lengthened
the life of the astronomer by one-half.
Columns similar to those above might be formed with any
other number as base. For practical purposes, however, 10
is always taken as the base of the system, called the common
system, in distinction from the natural system, of which the
base is 2.71828 •••, the value of the exponential series (^Higher
Algebra^. The natural system is used in theoretical discus-
sions. It follows that common logarithms are indices, positive
or negative, of the powers of 10.
Thus, 108 = 1000 ; i.e, log 1000 = 3 ;
10-2 = ^=0.01; i.g. log0.01 = -2.
36. Characteristic and mantissa. Clearly most numbers
are not integral powers of 10. Thus 300 is more than the
second and less than the third power of 10, so that
log 300 = 2 plus a decimal.
Evidently the logarithms of numbers generally consist of
an integral and a decimal part, called respectively the charac-
teristic and the marvtissa of the logarithms.
37. Characteristic law. The characteristic of the loga-
rithm of a number is independent of the digits composing
the number, but depends on the position of the decimal
point, and is found by counting the number of places the first
significant figure in the number is removed from the units'
place, being positive or negative according as the first significant
LOGARITHMS. 48
figure is at the left or the right of units^ place. This follows
from the fact that common logarithms are indices of powers
of 10, and that 10**, n being a positive integer, contains n-\-l
places, while lO'** contains w — 1 zeros at the right of units'
place. Thus in 146.043 the first significant figure is two
places at the left of units' place ; the characteristic of log
146.048 is therefore 2. In 0.00379 the first significant digit
is three places at the right of units' place, and the charac-
teristic of log 0.00379 is - 3.
To avoid the use of negative characteristics, such charac-
teristics are increased by 10, and — 10 is written after the
logarithm. Thus, instead of log 0.00811 = 3.90902, write
7.90902 — 10. In practice the — 10 is generally not written,
but it must always be remembered and accounted for in the
result.
£x. Determine the characteristic of the logarithm of :
1; 46; 0.009; 14796.4; 280.001; 10« x 76; 0.525; 1.03; 0.000426.
38. Mantissa law. The mantissa of the logarithm of a
number is independent of the position of the decimal point,
but depends on the digits composing the number, is always
positive^ and is found in the tables.
For, moving the decimal point multiplies or divides a
number by an integral power of 10, i.e. adds to or subtracts
from the logarithm an integer, and hence does not affect the
mantissa. Thus,
log 225.67 = log 225.67,
log 2256.7 = log 225. 67 X 101 = log 225.67 + 1,
log 22567.0 = log 225.67 x 102 = log 225.67 + 2,
log 22.567 = log 225.67 x lO-i = log 225.67 + (- 1),
log 0.22567 = log 225.67 x 10-8 = log 225.67 4- (- 3),
so that the mantissse of the logarithms of all numbers com-
posed of the digits 22567 in that order are the same, .35347.
Moving the decimal point affects the characteristic only.
The student must remember that the mantissa is always positive.
44 PLANE TRIGONOMETRY.
Log 0.0022567 is never written -3 +.35347, but 3.35347, the minus
8ign being written above to indicate that the characteristic alone is nega-
tive. In computations negative characteristics are avoided by adding
and subtracting 10, as has been explained.
39. We may now define the logarithm of a number as the
index of the power to which a fixed number^ called the base^
mu8t be raised to produce the given number.
Thus, a' = 6, and x = logjb (where \ogJb is read logarithm
of b to the base a) are equivalent expressions. The relation
between base, logarithm, and number is always
(base)^** = number.
To illustrate: log28=3 is the same as 2^=8; logg81=4and
3*= 81 are equivalent expressions ; and so are logiQlOOO = 3
and 103 _ 1000, and logio0.001= -3 and 10-8= q.OOI.
Find the value of :
log^ei; log5l25; log8243; log«(a)*; log27 3; log^l.
40. From the definition it follows that the laws of indices
apply to logarithms, and we have :
I. The logarithm of a product equals the sum of the loga-
rithms of the factors,
II. The logarithm of a quotient equals the logarithm of the
dividend minus the logarithm of the divisor,
III. The logarithm of a power equals the index of the
power times the logarithm of the number.
IV. The logarithm of a root equals the logarithm of the
number divided by the index of the root.
For if a* = w and a^ = w,
then w X m = a^"'"'', .•. log nm = x + y = log n + log m;
and n-*-w = a*"*', .'.log— =a: — y = logw — logm;
m
also n^= Qa^y= a***, .*. log n*" =rx = r x log n ;
finally, Vn = Vo* = a**, .'. logVw = ? = - log n.
r r
LOGARITHMS. 45
EXAMPLES.
Given log 2 = 0.30103, log 3 = 0.47712, log 5 = 0.69897, find :
1. log 4. 4. log 9. 7. logl5». 10. logVjf.
2. log 6. 5. log 25. 8. logf. f
3. log 10. 6. log>/3. 9. log 15x9. ^- ^^g^/;
92 X 5«
2*xl0
USE OF TABLES.
41. To find the logarithm of a number.
First, Find the characteristic, as in Art. 37.
Second. Find the mantissa in the tables, thus :
(a) When the number consists of not more than four
figures.
In the column N" of the tables find the first three figures,
and in the row iV the fourth figure of the number. The
mantissa of the logarithm will be found in the row opposite
the first three figures and in the column of the fourth figure.
Illustration. Find log 42.38.
The characteristic is 1. (Why ?)
In the table in column N" find the figures 423, and on the
same page in row iV the figure 8. The last three figures of
the mantissa, 716, lie at the interseetion of column 8 and
row 423. To make the tables more compact the first two
figures of the mantissa, 62, are printed in column only.
Then log42.38 = 1.62716.
Find log 0.8734 = 1.94121,
log 3.5 = log 3.500 = 0.54407,-
log 36350 =4.56050.
(6) When the number consists of more than four figures.
Find the mantissa of the logarithm of the number com-
posed of the first four figures as above. To correct for the
remaining figures we interpolate by means of the principle of
proportional parts, according to which it is assumed that^ for
differences small as compared with the numbers, the differences
46 PLANE TRIGONOMETRY.
between several numbers are proportional to the differences be^
tween their logarithms.
The theorem is only approximately correct, but its use
leads to results accurate enough for ordinary computations.
Ex. 1. To find log 89.4562.
As above, mantissa of log 894500 = 0.95158,
mantissa of log 894600 = 0.95163,
.-. log 894600 - log 894500 = O.OOOOb^called the tabular difference.
Let log 894562 - log 894500 = x hundred-thousandths.
Now, by the principle of proportional parts,
log 894562 - log 894500 ^ 894562 - 894500
log 894600 - log 894500 894600 - 894500'
or - = :^, whence ar = .62 of 5 = 3.1
5 100
.-. log 89.4582 = 1.95158 + 0.00003 = 1.95161,
all figures after the fifth place being rejected in five-place tables. If,
however, the sixth place be 5 or more, it is the practice to add 1 to the
figure in the fifth place. Thus, H x = 0.0000456, we should call it
0.00005, and add 5 to the mantissa.
2. Find log 537.0643.
To interpolate we have x : 9 = 643 : 1000, i.e. x = 5.787;
.-. log 537.0643 = 2.72997 + 0.00006.
3. Find log 0.0168342 = 2.22619.
4. Find log 39842.7 = 4.59816.
42. To find the number corresponding to a given logarithm.
The characteristic of the logarithm determines the posi-
tion of the decimal point (Art. 37).
(a) If the mantissa is in the tables, the required number
is found at once.
Ex. 1. Find log"^ 1.94621 (read, the number whose logarithm is
1.94621).
The mantissa is found in the tables at the intersection of row 883 and
column 5.
.-. log-n .94621 = 88.35,
the characteristic 1 showing that there are two integral places.
LOGARITHMS. 47
(h) If the exact mantissa of the given logarithm is not in
the tables, the first four figures of the corresponding num-
ber are found, and to these are annexed figures found by
interpolating by means of the principle of proportional
parts, as follows :
Find the two successive mantissae between which the given
mantissa lies. Then, by the principle of proportional parts,
the amount to be added^ to the four figures already found is
such a part of 1 as the difference between the successive
mantissse is of the difference between the smaller of them
and the given mantissa.
2. Find log-U.43764.
Mantissa of log 2740 = 0.43775
of log2739 = 0.43759
Differences 1 16
Mantissa of log required number = 0.43764
of log2739 = 0.43759
Differences x 5
By p. p. a? : 1 = 5 : 16 and x = ^ = 0.3125.
Annexing these figures, log-* 1.43764 = 27.393+.
3. Find log-i T.48762.
The differences in logarithms are 14, 6.
.'. x=-^ = .428+,
14
and log-i 1.48762 = 0.30744+.
4. Find log 891.59; log 0.023; logi; log 0.1867; log V2.
5. Find log-i 2.21042 ; log'i 0.55115 ; log-i 1.89003.
43. Logarithms of trigonometric functions. These might
be found by first taking from the tables the natural func-
tions of the given angle, and then the logarithms of these
numbers. It is more expeditious, however, to use tables
showing directly the logarithms of the functions of angles
less than 90° to every minute. Functions of angles greater
than 90° are reduced to functions of angles less than 90° by
<./?' S^-^S'^
48 PLANE TRIGONOMETRY.
the formulsB of Art. 29. To make the work correct for
seconds, or any fractional part of a minute, interpolation
is necessary by the principle of proportional parts, thus :
Ex. 1. Find log sin 28° 32' 21".
In the table of logavithms of trigonometric functions, find 28° at the
top of the page, and in the minute column at the left find 32'. Then
-under log sin column find log sin 28° 32' = 9.67913 - 10
log sin 28° 33 = 9.67936 - 10
Differences 1' 23
By p. p. a: : 23 = 21" : 60", t.e. x = — x 23 = 8.06.
•^ ^ '^ 60
.-. log sin 28° 32' 21" = 9.67913 + 0.00008 - 10
= 9.67921 - 10.
Whenever functions of angles are less than unity, i,e, are decimals
f(as sine and cosine always are, except when equal to unity, and as tan-
:gent is for angles less than 45°), the characteristic of the logarithm will
ibe negative, and, accordingly, 10 is always added in the tables, and it
must be remembered that J.0 is to be subtracted. Thus, in the example
above, the characteristic of the logarithm is not 9, but I, and the log-
;arithm is not 9.67913, as written in the tables, but 9.67913 — 10.
2. Find log cos 67° 27' 50''.
In the table of logarithms at the foot of the page, find 67°, and in the
iminute column at the right, 27'. Then computing the difference as
;above, x = 25.
But it must be noted that cosine decreases as the angle increases
itoward 90°. Hence, log cos 67° 27' 50" is less than log cos 67° 27', t.c.
ithe difference 25 must be subtracted, so that
log cos 67° 27' 50" = 9.58375 - 0.00025 - 10
= 9.58350 - 10.
44. To find the angle when the logarithm is given, find the
.^successive logarithms between which the given logarithm
ilies, compute by the principle of proportional parts the
seconds, and add them to the less of the two angles corre-
sponding to the successive logarithms. This will not neces-
sarily be the angle corresponding to the less of the two
logarithms ; for, as has been seen, the number, and, therefore,
the logarithm, may decrease as the angle increases.
LOGARITHMS. 49
Ex. 1. Find the angle whose log tan is 9.88091.
log tan 37° 14' = 9.88079 - 10
log tan 37*» 15' = 9.88105 - 10
Differences 60" 26
log tan 37° 14' = 9.88079 - 10 ■
log tan angle required = 9.88091 — 10
Differences a:" 12
.-. X : 60 = 12 : 26, or a:" = ij x 60" = 28", approximately, and the
angle is 37° 14' 28".
2. Find the angle whose log cos = 9.82348.
We find a: = A X 60" = 26", and the angle is 48° 14' 26".
3. Show that log cos 25° 31' 20" = 9.95541 ;
log sin 110° 25' 20" = 9.97181 ;
log tan 49° 52' 10" = 0.07417.
4. Show that the angle whose log tan is 9.92501 is 40° 4' 39" ; whose
log sin is 9.88365 is 49° 54' 18" ; whose log cos is 9.50828 is 71° 11' 49".
45. Cologarithms. In examples involving multiplications
and divisions it is more convenient, if n is any divisor, to
add log - than to subtract log n. The logarithm of - is
n n
called the cologarithm of n. Since
log - = log 1 — log n = — log w,
it follows that colog w = — log «, i.e. logw subtracted from
zero. To avoid negative results, add and subtract 10.
Ex. 1. Find colog 2963.
log 1 = 10.00000 - 10
log 2963= 3.47173
.-. colog 2963 = 6.52827 - 10
2. Find colog tan 16° 17'.
log 1 = 10.00000 - 10
log tan 16° 17' = 9.46554 - 10
.-. colog tan 16° 17' = 0.53446
50 PLANE TRIGONOMETRY.
By means of the definitions of the trigonometric functions, the parts
of a right triangle may be computed if any two parts, one of them being
a side^ are given. Thus,
■B given a and A in the rt. triangle ABC.
Then c =^ a -t- sin A, b = a -t- tan Ay
and B = 90°-A,
Again, if a and b are given, then
tan A = -, c = a -h sin ^, and B = 90° — -4-
b
3. Given c = 25.643, B = 37** 25' 20", compute the other parts.
^ = 90° - 37° 25' 20" = 52° 34' 40".
a = c cos B, & = a tan B.
log c = 1.40897 log a = 1.30889
log cos B = 9.89992 log tan J5 = 9.88376
log a = 1.30889 log b = 1.19265
.-. a = 20.365. .-. b = 15.583.
Check: c^ = a^ -^ b^ = 20.3652 + 15.5832 = 657.57 = 25.6432.
4. Given b = 0.356, B = 63° 28' 40", compute the other parts.
A = 26° 31' 20".
a =
sin B tan B
log b = 9.55145 log b = 9.55145
colog sin B = 0.04829 colog tan B = 9.69816
log c = 9.59974 log a = 9.24961
c = 0.3979 a = 0.1777
Check: c^ - a^ = 0.1583 - 0.03157 = 0.12673 = bK
EXAMPLES.
Compute the other parts :
^ 1. Given a = 9.325, A = 43° 22' 35".
.. ' ^2. Given c = 240.32, a = 174.6.
-^ • >- 3. Given jB = 76° 14' 23", a = 147.53.
y 4. Given a = 2789.42, b = 4632.19.
'^ 5. Given c = 0.0213, il = 23° 14".
S. Given 6 = 2, c = 3.
CHAPTER V.
APPLICATIONS.
46. Many problems in measurements of heights and dis-
tances may be solved by applying the preceding principles.
By means of instruments certain distances and angles may
be measured, and from the data thus determined other
distances and angles compnted. The most common instru-
ments are the chain, the transit, and the compasi.
The chain is used to measure distances. Two kinds are in
use, the engineer's chain and the Q-unter*! chain. They each
contain 100 links, each link in the engineer's chain being
12 inches long, and in the Gunter's 7.92 inches.
The trarmt is the instrument moat used to measure hori-
zontal angles, and with certain attachments to measure verti-
cal angles. The figure shows the form of the instrument.
62
PLANE TRIGONOMETRY.
The mariner''» compass is used to determine the directions,
or bearings, of objects at sea. Each quadrant is divided
into 8 parts, making the 32 points of the compass, so that
each point contains 11° 15'.
^yx^'^ip>
|>8
^ s
FiQ. 27.
Fig. 28.
47. The angle between the horizontal plane and the line
of vision from the eye to the object is called the angle of
elevation^ or of depression^ according
as the object is above or below the
EievaUoiry"-.^ obscrvcr.
It is evident that the elevation
angle of B^ as seen from A^ is equal
to the depression angle of A^ as seen from jB, so that in the
solution of examples the two angles are interchangeable.
PROBLEMS.
48. Some of the more common problems met with in
practice are illustrated by the following:
To find the height of an object
when the foot is accessible.
The distance jB(7, and the eleva-
tion angle B are measured, and x
is determined from the relation b
X=BOt^nB. Fig. 29.
/
APPLICATIONS.
53
Ex. 1. The elevation angle of a cliff measured from a point 300 ft.
from its base is found to be 30®. How high is the cliff?
Then
BC = 300, B = 30°.
a: = 300 . tan 30° = 300 • i V3 = 100 V3.
2. From a point 175 ft. from the foot of a tree the elevation of tho
top is found to be 27° 19'. Find the height of the tree.
The problem may be solved by the use of natural functions, or of
logarithms. The work should be arranged for the solution before the
tables are opened. Let the student complete.
5C = 175. jB = 27°19'.
Then x = BC tan B, Or by natural functions,
logjBC= BC = 175
log tan jB = tan iB = 0.5165
log a: =
.-. X = 90.39.
To find the height of an object
when the foot is inaccessible.
Measure BB\ and ^'.
.-. X = 90.3875.
Then x =
BO BB'-^-B'O
cot cot
But B^O = x cot ^', whence substituting,
^^ BB^
cot - cot 0''
which is best solved by the use of the natural functions of
and 0'.
3. Measured from a certain point at its base the elevation of the
peak of a mountain is 60°. At a distance of one mile directly from this
point the elevation is 3Q°. Find the height of the mountain.
BB' = 5280 ft., e = 30°, & = 60°.
X = IL±^?§5. But y = arcot 60°.
cot 30° ^
» m X — •
5280
cot 30° - cot 60°
= 4572.48 ft.
54
PLANE TRIGONOMETRY.
In surveying it is often necessary to make measurements
across a stream or other obstacle, too wide to be spanned by
a single chain.
To find the distance from O to a
point B on the opposite side of a
stream.
At C measure a right angle, and
take CA a convenient distance.
Measure angle A^ then
FiQ. 31. ~ 5(7= (7^. tan A.
4. Find CB when angle A = 47° 16', and CA = 250 ft.
5. From a point due south of a kite its elevation is found to be
42° 30'; from a point 20 yds. due west
from this point the elevation is 36° 24'.
How high is the kite above the ground ?
^J5 = x.cot42°30',
^C = ar.cot36°24',
AC^-AB^ = BC^ = 400. ^
.-. x^ (cota 36° 24' - cot^ 42° 30') = 400,
whence
EXAMPLES.
4-^\
Fia. 32.
1. What is the altitude of the sun when a tree 71.5 ft. high casts
a shadow 37.75 ft. long ?
2. What is the height of a balloon directly over Ann Arbor when
its elevation at Ypsilanti, 8 miles away, is 10° 15' ?
3. The Washington monument is 555 ft. high. How far apart are
two observers who, from points due east, see the top of the monument
at elevations of 23° 20' and 47° 30', respectively?
4. A mountain peak is observed from the base and top of a tower
200 ft. high. The elevation angles being 25° 30' and 23° 15', respec-
tively, compute the height of the mountain above the base of the tower.
5. From a point in the street between two buildings the elevation
angles of the tops of the buildings are 30° and 60°. On moving across
APPLICATIONS.
55
/
the street 20 ft. toward the first building the elevation angles are found
to be each 45^ Find the width of the street and the height of each
building.
6. From the peak of a mountain two towns ar6 observed due south.
The first is seen at a depression of 48® 40', and the second, 8 miles farther
away and in the same horizontal plane, at a depression of 20° 50'. What
is the height of the mountain above the plane ?
7. A building 145 ft. long is obseiTed from a point directly in front
of one corner. The length of the building subtends .tan-i 3, and the
height tan-i 2. Find the height.
8. An inaccessible object is observed to lie due N.E. After the ob-
server has moved 8.£. 2 miles, the object lies N.N.E. Find the distance
of the object from each point of observation.
9. Assuming the earth to be a sphere with a radius of 3963 miles,
find the height of a lighthouse just visible from a point 15 miles distant
at sea.
10. The angle of elevation of a tower 120 ft. high due north of an
observer was 35° ; what will be its angle of elevation from a point due
west from the first point of observation 250 ft. ? Also the distance of
the observer from the base of the tower in each position ?
11. A railway 5 miles long has a uniform grade of 2° 30' ; find the rise
per mile. What is the grade when the road rises 70 ft. in one mile ?
(The grade depends on the tangent of the angle.)
12. The foot of a ladder is in the street at a point 30 ft. from the
line of a building, and just reaches a window 22J ft. above the ground.
By turning the ladder over it just reaches a window 36 ft. above the
ground on the other side of the street. Find the breadth of the street.
13. From a point 200 ft. from the base of the Forefathers* monument
at Plymouth, the base and summit of the statue of Faith are at an eleva-
tion of 12° 40' 48" and 22° 2' 53", respectively; find the height of the
statue and of the pedestal on which it stands.
14. At a distance of 100 ft. measured in a horizontal plane from the
foot of a tower, a flagstaff standing on the top of the tower subtends an
angle of 8°, while the tower subtends an angle of 42° 20'. Find the
length of the flagstaff.
15. The length of a string attached to a kite is 300 ft. The kite's
elevation is 56° 6'. Find the height of the kite.
16. From two rocks at sea level, 50 ft. apart, the top of a cliff is ob-
served in the same vei'tical plane with the rocks. The angles of eleva-
tion of the cliff from the two rocks are 24° 40' and 32° 30'. What is the
height of the cliff above the sea ?
CHAPTER VI.
OXSNERAL FORMUIt^l— TRiaONOMBTRIC EQUATIONS
AND IDENTITIES.
49. Thus far functions of single angles only have been
considered. Relations will now be developed to express
functions of angles which are sums, differences, multiples,
or sub-multiples of single angles in terms of the functions
of the single angles from which they are formed.
First it will be shown that,
sin (a±p) = 8lnaco8p±co8asinp,
cos (a ± p)= cosacosp T sinasinp
tan (a ± p) = *!_.
1 q=tanatanp
The following cases must be considered :
1. a, /8, a + /8 acute angles.
2. a, )8, acute, but a + 13 an obtuse angle.
3. Either a, or )8, or both, of any magnitude, positive or
negative.
The figures apply to cases 1 and 2.
P)
•Q
Let the terminal line revolve through the angle a, and
then through the angle /3, to the position OB^ so that angle
56
X
GENERAL FORMULA. 67
XOB=: oi + 13. Through any point P in OB draw perpen-
diculars to the sides of a, DP and OP, and through O draw
a perpendicular and a parallel to OX, MO and IfO.
Then the angle QOA = a (why?), and OJ^P is the triangle
of reference for angle QOP = 90^ + a.
CNP is sometimes treated as the triangle of reference for angle CPN.
The fallacy of this appears when we develop cos (a + j8), in which PC
would be treated as both plus and minus.
Now sin(« + /3)=sinX05 = ^ = ^+^
or expressing in trigonometric ratios,
^MO 00^, NP OP
00' OP op' op
= sin a cos )8 + sin (90** + a) sin /8.
Hence, since sin (90*^ + a) = cos a, we have
sin (a + p^=z sin a cos/8 + cos a sin /8-__^ """
In like manner
cos(a + ^) = cosXOB = — = — + -^
or expressing in trigonometric ratios,
^OM 00 ON OP
00 ' op'^ op' op
= COS a COS 13 + cos (90** + a) sin /3.
And since cos(90*' + «)= — sin a, we have
cos (^u + I3^=s cosa cos/8 — sin a sin/8.
It will be noted that the wording of the demonstration ap-
plies to both figures, the only difference being that when a+jS-
is obtuse OD is negative. ON is negative in each figure.
50. In the case, when a, or /8, or both, are of any magni-
tude, positive or negative, figures may be constructed as-
before described by drawing through any point in the terminat
line of 13 a perpendicular to each side of a, and through the foot
of the perpendicular on the terminal line of a a perpendicular
and a parallel to the initial line of a. Noting negative lines.
58 PLANE TRIGONOMETRY.
the demonstrations already given will be found to apply for
all values of a and /3.
To make the proof complete by this method would require an unlim-
ited number of figures, e.g. we might take a obtuse, both a and p obtuse,
either or both greater than 180% or than 360% or negative angles, etc.
Instead of this, however, the generality of the proposition
is more readily shown algebraically, as follows :
Let a' = 90^ + « be any obtuse angle, and a, /3, acute
angles.
Then
sin («' + /3) = sin (90** + a + )8) = cos (a + /8)
= cosacos/3— sinasin/3
= sin (90^ + a) cos /3 + cos (90° + a) sm /8(why?)
= sin a' cos 13 + cos a' sin fi.
In like manner, considering any obtuse angle /8' = 90*^ + /8,
it can be shown that
sin (a' + /8') = sin a' cos /8' + cos a' sin^S'.
Show that cos (a' + /8') = cos a' cos /8' — sin a' sin /8'.
By further substitutions, e.ff. a" = 90*^ ± «', /8" = 90° ± /8',
etc., it is clear that the above relations hold for all values,
positive or negative, of the angles a and /3.
Since a and 13 may have any values, we may put — /8 for /8,
and sin (a + [ — /8])
s= sin (a — /8) = sin a cos(— /8) + cosasin ( — /8)
5= sin a cos )8 — cos a sin /8 ( why ?) .
Also cos (a — /3)= cos a cos(— /8) — sin a sin (— /8)
= cos a cos /8 + sin a sin /3.
Finally,
tan C ±S^^ sin (« ± /8) _ sin « cos /3 ± cosce sin /3
cos (a ± /8) cos a cos /8 :f sin a sin )8
sin a cos /3 , cos a sin /3
COS a cos iS cos a cos iS tan (x ± tan B
cos « cos 13 sin a sin /8 l^tanatan/S
cos tt cos 13 cos a cos )3
EXAMPLES. 59
ORAL WORK.
By the above f onnulse develop :
1. sin(2A+SB). 7. sin 90« = sin (45<» + 45*>).
2. cos (90° -5). a cos 90°.
3. tan (45° + ^). 9. tan 90°.
4. sin2 ^ = sin (A + A). 10. sin(90° + jS + y).
5. cos 2 ft 11. cos (270° - m - n).
6. tan (180° + C). 12. tan (90° + m + n).
Ex. 1. Find sin 76°.
sin 75° = sin (45° + 30°) = sin 45° cos 30° + cos 45° sin 30*
= -L.v2^.J_.UL±^=0.9659.
V2 2 v^ 2 2V5
2. Find tan 15°.
tan 45° - tan 30°
1--L
= ^ =:5^i^= 2 --v/3 = 0.2679.
1 + J_ V3 + 1
V3
3. Prove 2ii^-S2ilii=2.
Sin A cos A
Combining si^^^ cos^— co8 3^ sin^ _ sin(3^ — vi)
sin A cos A sin A cos A
s= 8"^2i4 _ sin (A + il) __ sin ^ cos i4 + cos ^4 sin A _ g
sin ^ cos A sin ^1 cos^ sin A cos ^1
4. Prove tan-* a + tan-^ b = tan-^ ° "^ »
1 — ao
Let a = tan-^o, fi = tan-^J, y = tan-^ Z' "*" *
1 — ab
Hence, tan a^^a, tan B = bj tan y = "^ *
1—00
Then a + /3 = y, and hence tan (a + /3) = tan y.
Expanding, tan « + tan ^ ^ ^^^
^ ^ l-tanatanj8 '
Substituting, ^L±\ = :^±A.
1 — ao 1 — ao
60 PLANE TRIGONOMETRY.
r6
EXAMPLES.
1. Find COB 15% tan 75**.
2. Prove cot (a i /8) = cotacot^Tl .
^ '^^ cot /8i cot a
3. Prove geometrically sin (a + j3) = sin a cos /3 + cos a sin j9,
and cos (a + j3) = cos a cos /3 — sin a sin ^
given (a) a acute, /3 obtuse ;
(6) a, j3, obtuse ;
(c) a, /3, either, or both, negative angles.
4. Prove geometrically tan (a + iS) = ^^^ "*" ^^ ^^ '
1 — tan OL tan p
Verify the formula by assigning values to a and j3, and finding the
values of the functions from the tables of natural tangents.
5. Prove cos (a + /8) cos (a — /3) = cos' a — sin'jS.
6. Show that tan a + tan j8 == 5^5J[«±^
cos a cos p
7. Given tan a = J, tan /8 = f, find sin (a + /8)
a Given sin 280° = «, find sin 170^
9. If a = 67° 22', /8 = 128° 40', by use of the tables of natural func-
tions verify the formulae on page 56.
10. Prove tan-i^5^^JtJ^ = tan- V« + tan-i>/a.
1 — Vai
U. Prove tan-i?^^ + tan-i2Ajl£^^^j^.i^
12. Prove sec-* — ^ = sin-^ ^
Va« - a:' a
13. If a + /3 = 0), prove cos'oe + cos'jS — 2 cos a cos j9 cos » = sin*c0.
14. Solve i sin = 1 -. cos 0.
y^ 15. Prove sin {A + -B) cos A — cos (-4 + 5) sin il = sin B,
16. Prove cos(il +B)cos(i4--B) + sin(il +-B)sin(-4--B) = cos2A
17. Prove sin (2 a - /8) cos (a - 2 j8)
- cos (2a - )8)8in(a - 2/8) = sin(a + /8).
1& Prove sin(n'— l)acos(n+l)a+cos(n-l)asin(n+l)a = 8in2na.
~. 19. Prove sin(135° - tf) + cos(135° + tf ) = 0.
ADDITION— SUBTRACTION FORMULA. 61
20. Provel-tan«atan«i8 = 528!fzJI^
21. Prove ^"^ + ^^" ^ = tan « tan ^.
cot a + cot /3
22. tan« (? - «U 1^4^^««2i«.
\4 / 1 + 2 sm a cos a
51. The following formiilse are very important and should
be carefully memorized. They enable us to change sums
and differences to products, i.e, to displace terms by factors.
sin e + sin ♦ = 2 sin^i^cos^^
sine - 8iii+ = 2 CM^±^9ln^,
cose + COS+ = 2C08-^^€08-^^>
cose - C08+ = - 2 sin -i-*sin -=^«
Since sin (« + /8) = sin a cos jS + cos a sin )8,
and sin (« — /8) = sin a cos /8 — cos a sin )8,
then sin (^a + fi^-^- sin (a — )8) = 2 sin a cos )8, (1)
and sin (a + /8) — sin (a — )8) = 2 cos a sin /8. (2)
Also since cos (« + /8) = cos a cos /8 — sin « sin /8,
and cos (a "13^ = cos a cos fi + sina sin )8,
then cos C^ + 13^ + cos (a — /8) = 2 cos a cos )8, (3)
and cos (« + /8) — cos (a — /8) = — 2 sin « sin /8. (4)
Put a + l3 =
and a-— ^ = <f>
2a = ^ + 0, anda = ^i:-^,
2/8 = ^-^, and /3 = ^^.
Substituting in (1), (2), (3), (4), we have the above
formulae.
62 PLAICE TRIGONOMETRY.
EXAMPLES.
1. Prove "»^^ + ^^"; = tan^.
CO82 + COS0 2
By formuke of last article the first member becomes
2 sm -e cos ^
2 2 ^ se
2cos--ecos^ *
2 2
2 Prove sin « + 2 sin 3 « + sin 5 a __ sin3«
sin 3 a + 2 sin 5 a + sin 7 a sin 5 a
(sin tt + sin 5 tt) + 2 sin 3 cc __ 2 sin 3 « cos 2 « + 2 sin 8 «
(sin 8 a -h sin 7 a) + 2 sin 5 a 2 sin 5 a cos 2 a + 2 sin 5 a
_ (cos 2 « + 1) sin 8 « _ sin 8 a
(cos 2 a + 1) sin 5 ce sin 5 a
cos(4i4 -25)+ 008(45-2-4) \ ^ ^
o»:^ 4^ -25 + 45-2^ ^^, 4^ -25 -45 + 2.4
If sm cos L__
^ 2 2
a^^.4il-25 + 45-2^_, 4^-25-45 + 2"T
SJ cos rr cos ^
2 2
;=eHLii^ = tan(^+5).
cos(^+5) ^ ^
4. Prove sin 50° - sin 70^ + sin 10° = 0.
2 cos ^° ^ "^^^ sin ^''"•'^^ = 2 cos 60° sin (- 10°) = - sin 10°
5. Prove '^'^«°*»^«-°°^^''°"»^'*+'^«'^^"«=cot6«cot5tt.
sin 4 a sin 3 a— sin 2 a sin 5 a+sin4a sin 7 a
By (3) and (4), p. 61,
cos 5 « + cos « — cos 9 ft — cos 5 « + cos 11 « + cos 9 a
cos a — cos 7 a — cos 8 a + cos 7 a + cos 8 a — cos 11 a
^costt + cosll«^ 2co86«cos5« ^^^g^^^g^
cos a — cos 11a 2 sin 6 a sin 5 a
ORAL WORK.
By the formulae of Art. 51 transform :
6. cos 5 a + cos a. 8. 2 sin 3 cos 0.
7. cos a — cos 5 a. 9. sin 2 a — sin 4 a.
FUNCTIONS OF THE DOUBLE ANGLE. 68
10. cos 9 cos 2 0. 16. cos (30°+ 2<^) sin (30°- <^).
11. sintf + sin^. 17. sin(2r + ») + sm (2r-»).
12. sin 75° sin 15°. 18. cos (2 /S ^ a) - cos 3 a.
13. cos7i>--cos2;). >c ». sin 36° + sin 54°.
14. co8(2p + 8,)8in(2p-35). ^ ««, ««» + cos 20-.
15. sin^-sin^. ^ 8m30» + co8m
Prove: 22. !JE«±«]l| = tan«±^cot^.
Sin a — sm /3 2 2
23 cos « + cos /3 ^ cotgL+^cot^M^
cos j8 — cos a 2 2
^^ COS a: + cos y 2
^^ COS X — cos y 2
.^ 26. cos 55° + sin 25° = sin 85°.
Simplify: 27. 8ini? + sin2^ + 8in3^
^ "^ C08 5+C08 2jB +C08 3iJ
23 sin C- sin 4 C + sin 7 C - sin 10 C
cos C — COS 4 C + cos 7 C — cos 10 C
52. Functions of an angle in terms of those of tKe half angle.
If in sin (« + /8) = sin a cos /3 + cos a sin /3^ a = fi^
then sin (a + a) = sin 2 a = 2 sin acos a*.
In like manner
cos (a + a) = COS 2 a = cos* a - sin^a
= 2cos«a-l
= l-2sin*a;
2 tan a
and tan 2 a
1-tanSa
64 PLANE TRIGONOMETRY.
ORAL WORK.
Ex. Express Ib terms of functions of half the given angles :
1. sin 4 m. 4. cos x, 6. sin (2p — q),
2. co83/>. . . iS ^« cos (30° + 2 «^).
3. tan 5 1. ^ 8. sin {x + y),
9. From thefimctious of 30° find those of 60°; from the functions of
45°, those of 90°.
53i Functions of an angle in terms of those of twice the angle.
By Art. 62, co9«=:l -2sin2?= 2cos2^- 1.
/. 2 sin*^ = 1 — cos a, and 2 cos^^ = 1 + cos «•
. a
8in;r
.•. taii- = =±\r
— cosa
2 ^^„a ^l + cosa
cos-
Explain the significance of the ± sign before the. radicals.
Express in terms of the double angle the functions of
120®; SO*'; 90*^, with proper signs prefixed.
Ex. 1. Express in terms of functions of twice the given angles each
of the functions in Examples 1-8 above.
•
2. From the f auctions of 45° find those of 22° 30^ ; from the functions
of 36°, those of 18° (see tables of natural functions).
3. Find the corresponding functions of twice and of half each of the
following angles, and verify results by the tables of natural functions :
Given sin 26° 42' = 0.4493,
tan 62° 24' = 1.9128,
cos 21° 34' = 0.9300.
4. Prove tan-i Vt*^^^ = ^ 5-2 tan-i x = tan-^ -^^
^1 + cosx 2 l-a;«
(^^y^ 6. Ji A^ B, C toe angles of a triangle, prove
ARC
sin il 4 sin C + sin jB = 4 cos —^ cos — cos ^.
7. K cos^a + cos« 2 a + cos^ 3 a = 1, then
cos a cos 2 a cos 3 a = 0.
8. Proye cot il — cot 2 il = esc 2 il.
9. Prove
fS
n
\
^ .i^'^,
f'-)
>-:i
tanf^-*^ fl-tan*"
V4 2/ 2
(M) '
10.
tan
tana
/
/•
c
V
/ )
tan (a + ^)
= 1-
1 + tani
2 J
2 sin ^
sin (2 a + ^) + sin ^
11. If V = tan-i2 g T^' + ^1 ~ ^' . prove :ca = sin 2y.
12. Prove tan^i 21,+ ^^" 1 + tan^i ^^ , = -tan-V
a;
13. Ify = sin-i
JT
^ prove X = tan ^.
14. Prove cos^ a + cos' j8 - 1 = cos (a + j8) cos (a - j8).
15. Prove V(cos a - cos py + (sin a - sin py = 2 sin ^^^-
16. Prove sin-i -J-^ = tan-i a/- = ^ cos"* ^^-^•
^a + a; 'a 2 a + a;
17. Prove cos« B - cos^ <^ = sin (<^ + 9) sin (<^ - S).
la Provetanil+tan(il + 120°)+tan(il -120«)=8taii3il
19. Prove tan a — tan ^ = tan ^ sec a.
2 2
20. 8tan-ia = tan-i5iLr«?.
l-3a2
21. cos* 3 A (tan« 3 il - tan« ^) = 8 sin« J cos 2 il.
22. 1 + cos2(i4 - 5) cos 25 = cos* il +cos*(il -2B).
23. cot«f2: + ?\ = 2c8c2^-jecl
V4 2/ 2csc20 + sec0
•■'./'
66 PLANE TKIGONOMETRY.
TRIGONOMETRIC EQUATIONS AND IDENTITIES.
54. Identities. It was shown in Chapter I that
sin»d + 008^^=1
is true for all values of 0^ and in Chapter YI, that
sin (a + )8)= sin a cos fi + cos a sin )8
is true for all values oi a and fi. It may be shown that
sin 2 A . A
r t;— 7= tan -4.
1 + cos 2 J.
is true for all values of J., thus :
sin 2 J. _ 2 sin A cos A (by trigonometric transf orma-
l + cos2A"l + 2 cos* J. - 1 tion)
= ^^^^ (by algebraic transformation^
cos A
= tan A (by trigonometric transformation).
Such expressions are called trigonometric identities. They
are true for all values of the angles involved,
55. Equations. The expression
2 cos* a — 3 cos a + 1 =
is true for but two values of cos a, viz. cos a= ^ and 1, i.e.
the expression is true for a = 0®, 60®, 300®, and for no other
positive angles less than 360®. Such expressions are called
trigonometric equations. They are true only for particular
v(ilu£S of the angles involved.
56. Method of attack. The transformations necessary at
any step in the proof of identities, or the solution of equa-
tions, are either trigonometric^ or algebraic; i.e. in prov-
ing an identity, or solving an equation, the student must
choose at each step to apply either some principles of algebra,
or some trigonometric relations. If at any step no algebraic
operation seems advantageous, then usually the expression
METHOD OF ATTACK. 67
should be simplified by endeavoring to state the different
functionB involved in terms of a Bingle function of the angle,
or if there are multiple angles^ to reduce all to functions of a
single angle.
f Algebraic
Transformations
Trigonometric, f Single function
to change to a 1 Single angle
No other transformations are needed, and the student will
be greatly assisted by remembering that the ready solution
of a trigonometric problem consists in wisely choosing at
each step between the possible algebraic and trigonometric
transformations. Problems involving trigonometric func-
tions will in general be simplified by expressing them entirely
in terms of sine and cosine.
EXAMPLES.
1. Prove 8in3^_cos3^^2
T> 1 V sin 3 ^ cos 3 il sin 3 ^ cos ^ — cos 3 ^ sin il
By algebra, — : — -r- = : — -z -z
sin ^ cos ^ sin A cos A
V . . ,, sin (3 -4 — i4 ) sin 2 ^
by trigonometry, = — A— -r^ = -, — j
•^ ° "^ sinilcosil smAcosA
2 sin i4 cos A
sin A cos A
Or, by trigonometry,
= 2.
sin 3^ . cos3i4 _ 3sinii —isin^A _ 4cos»^ ~3coSil
sin^ cosil sin J. cos^
by algebra, = 3 - 4sin«^ -icoa^A + 3
= 6 - 4(sinM + cos2^)= 2.
8ec80-l tan 8^
2. Prove
sec4^-l tan 2^
No algebraic operation simplifies. Two trigonometric changes are
needed. 1. To change the functions to a single function, sine or cosine.
2. To change the angles to a single angle, 8^, 4^, or 2^1.
68 PLANE TRIGONOMETRY.
By trigonometry and algebra,
1 -cos8^ sinSg
cos 8 ^ __ cos 8 B ^
1 -cos4g "" sm2g '
cos 4 COB 2 ^
by algebra, co84g(l - cos8g) ^ 8in8gcos2^ .
1 — COS 4 tf sin 2 ^
by trigonometry,
cos4^(l~l +28in«4^) _ 2sin4^cos4gco82g ,
l-l + 28in82d 8in2fl
byalgebra, 5ilLi| = 2 cos 2 tf ;
sin 2 9
and sin40=:28in20co820,
which is a trigonometric identity.
3. Solve 2 cos^ ^ + 3 sin = 0.
By trigonometry, 2(1 - sin« tf) + 3 sin tf = 0,
a quadratic equation in sin B,
By algebra, 2sin30 - 3sin^ - 2 = 0,
and (sin tf - 2)(2 sin tf + 1) = 0.
.% sin ^ = 2, or — J. Verify.
The value 2 must be rejected. Why ?
.*. B = 210% and 330° are the only positive values less than 860° that
satisfy the equation.
4. Solve sec — tan ^ = 2.
Here tan ^ = — 0.75, .*. from the tables of natural functions,
B = 143° 7' 48", or 323° V 48".
Find sec B, and verify.
5. Solve 28in0sin30-8mS20 = O.
By trigonometry, cos 2B — cos 4 ^ — sin* 20 = 0,
also oos20 - 008*20 + 8in*20 - sin*20 = 0.
By algebra, cos 2 B(\ - cos 2 0) = 0.
.•. cos 20 = or 1,
and 2 = 90°, 270°, 0°, or 360°,
whence = 45°, 135°, 0°, or 180°. Verify.
^
o
"-,
(^
d-
o
.a
TRIGONOMETRIC EQUATIONS. 69
Or, by trigonometry,
2 Bin 0(3 sin - 4 sin« 0) - 4 sin^ co8> = ;
by trigonometry and algebra,
esin^d - 88in4 tf - 48in2fl + 48in*fl = 0;
by algebra, 2 sin^ 0-4 sin^ = 0,
and 2 sin^ 0(1 - 2 sin^ 0) = 0.
.*. sin = 0, or ± Vj,
and = (F, 180°, 45°, 135°, 225°, or 315°.
The last two values do not appear in the first solution, because only
angles less than 360° are considered, and the solution there gave values
of 2 0, which in the last two cases would be 450° and 630°.
X
Solve \\y 6. tan = cot 0.
..SJ^ aLo^2^ ^ ^- 8in«0 + CO80 = l.
/ 7 T^- Prnv^ 10. 2cot2i4 =cot-4 -tan-4
8. 2cos20-2Bin0=l.
9. sin 2 cos = sin 0.
11. cos 2a; + cos 2y = 2 cos(x + y)cos (a: — y).
12. (cos a + sin a)^ = 1 + sin 2 a.
57. Simultaneous trigonometric equations.
13. Solve cos(a: + y)+cos(a:-y) = 2,
sin ? + sin ? = 0.
2 2
By trigonometry,
cos a; cos y — sin ar sin y + cos x cos y + sin a; sin y = 2^
so that
also,
and
Substituting,
cos a: cos y = 1 ;
^l-cosa: ^^ l-cosy ^^^
and
2 • ^ 2
.% cos X = COS y.
cos' a: = 1,
COS 0? = db 1.
.% X = 0°, or 180%
y = a: = 0°, or 180°. Verify.
/
- c^
5 tnI
^^
70 PLANE TRIGONOMETBT
14. Solve for R and F. ^nJ
W " Fain t - iSoost s 0^
1F+ Fooet - /Jam t =s 0.
To eUminate F,
IF" cos i — Fsin t cos t — ^ cos*t = 0,
IFsin t + Fcos itani" R sin* t = 0.
Adding, 1F(8in t + cos t) - /{(sin* t + oos* t) = 0.
.'. R = 1F(8in t + cos 0. ^4^
Sabstitating, IF — Fsin t — Tr(sin t + cos t)cos t =
. « IF — 1F(8in t + cos cos t
sint
If TF= 3 tons, and t = 22^3(y, compute Fand R.
/? = 3(0.3827 + 0.9239) = 3.9198. >;;;^
p ^ 3 - 3(0.3827 + 0.9239)0.9239 « _ j g24
0.3827
Solve: <^
15. 472cot0-263cot<^ = 49O, 3O7cot0-379cot^ = O.
16. sin 2 X + 1 = cos x + 2 sin x.
17. co8«fl + 8ind = l.
la If 2A(cosS0-8in3^-2a8in0co80 + 2ft8in0cos0 = O, provo
tf=:jtan-i-i^.
a —
Prove :
19. tany =(1 + secy) tan ^
20. 2 cot-i X = csc-i i^-^
2x
21. 8in(^ + 45°) + sin (^ + 135°) = \/§ cos ^.
22 cos p 4- cos 3 p 1 ^
cos 3tf + co6 5t; 2cos2tf — 8ec2r
23. cos 3 X — sin 3 X = (cos x + sin x) (1 — 2 sin 2 x).
Solve:
24. sin 2 ^ + sin = cos 2 + cos $.
25. 4co8(tf+60°)-V2=V6-4co8(tf+30°).
26. cot 20 = tan 0-1.
27. cos0 + cos20 + cos30 = O.
\'
V.
)
1
N
v-^
\
"•-rs
K.
Vj
TRIGONOMETRIC EQUATIONS. 71
28. sin 2 a; + v^ cos 2 x = 1.
29. 3 tan«p + 8 cos^p = 7.
30. Determine for what relative valaes of P and W the following
equation is true:
CO82^-^COS^-i=0.
2 IF 2 2
-or TMT
31. Compute N from the equation iV+ -J^ cos a — ~- sin a — IF cos a = 0,
3 3
when W = 2000 pounds and a satisfies the equation 2 sin a = 1 + cos a.
32. sin — tan ^(cos + sin 9) = cos 0, sin $ — tan ^ cos = 1.
Prove:
33. cot(« + 15°)-tan(/-15°)= ^^^^^^
28in2< + l
34. sin-* J — sin"* ^ = sin"* J}.
35. ^^J^+^)=^l\±^^
\4 2/ ^1— sin 01
36. 2 sin-* } = cos"* J.
37. If siuii is a geometric mean between sinB and cosB, prove
cos 2 ^ = 2 sin(45 - B) cos (45 + B).
38. Prove 8in(a + j8 + y) = sin a cos j3 cos y + cos a sin j3 cos y
+ cos a cos j3 sin y — sin a sin fi sin y.
Also find cos(a + jS + y).
39. Prove tan(« + ff + y) = ^ V^ ^""'^ ^o*" ^"^ V ^" " !''" ^^'^ ^ '
1— tanatanjS— tanj3tany-*tanytance
lift, Pi and y are angles of a triangle, prove
40. tan a + tan fi + tan y = tan a. tan fi tan y.
41. C0t^+C0t^+ COt^rrCOt^COtccot?*
2 2 2 2 2 2
If a + j8 + y = 90**, prove
42. tan a tan fi + tan j3 tan y + tan y tan a = 1.
Prove :
43. sin na = 2 sin (n — 1) a cos a — sin (n — 2)ce.
44. cos na = 2 cos (n — 1) a cos a — cos (n — 2)a.
A- 4. tan fn — 1) a 4- tan a
45. tan na = - f — ^ " ^ ^^ ^
1 — tan (n — 1) a tan a
CHAPTER VIT.
TRIANOIiEa.
58. In geometry it has been shown that a triaiigle is
determined, except in the ambiguous case, if there are given
any three independent parts, as follows :
I. Two angles and a side.
II. Two sides and an angle,
(a) the angle being included by the given sides,
(6) the angle being opposite one of the given sides (am-
biguous case).
III. Three sides.
The angles of a triangle are not three independent parts, since they
are connected by the relation A + B + C = 180°.
The three angles of a triangle will be designated J., jB, (7,
the sides opposite, a, 5, c.
But the principles of geometry do not enable us to compute
the unknown parts. This is accomplished by the following
laws of trigonometry:
sin A sin B sin
I. Law of Sines^
a
11. Law of Tangents, tan|(^-^) ^a-^ ^^^
^ tanJ(J. + ^) a + b
III. Law of Cosines, cos -4. = — ^^r^ , etc.
2 00
59. Law of Sines. In any triangle the sides are propor-
tional to the sines of the angles opposite.
Let ABO be any triangle, p the perpendicular from B
on 6. In I (Fig. 84), O is an acute, in II, an obtuse, in III,
72
LAW OF SINES — OF TANGENTS.
78
a right angle. The demonstration applies to each triangle,
but in II, sin^CB=sin2)CB (why?); in III, sin (7=1
(why?).
B
b D C A b C D A
I. II.
Fia. 34.
6
III.
Now
sin -4.=
P
P
/. I? = {? sin A.
or
sin (7= ^t .*. » = a sin (7.
a ^
Equating values of j?, ^ {? sin J. = a sin (7,
sin A sin
a c
By dropping a perpendicular from A, or (7, on a ^ ^, show
that
whence
sin B sin C sin ^ sin B
— — - = ^, or = — — ,
he a
sin J. _ sin ^ _ sin
_ - ,
a e
60. Law of Tangents. The tangent of half the difference
of two angles of a triangle is to the tangent of half their sum^
as the difference of the sides opposite is to their sum.
BA^ii'Ck ^ sin A
QinB
By composition and division,
g — 6 _ sin ^ — sin ^ _ 2 cos ^ (J. + -B) sin ^(A — B)
a + 6 " sin^ + sin^" 2 sin J(^ + ^)co8 J(J. - B)
^ tan^(^-^) ,
tanJ(^ + -B)'
^j. tanK^-^) ^«"S
' tanj(^-|-^) a + b'
74
PLANE TRIGONOMETRY.
61. Law of Cosines. The cosine of any angle of a triangle
is equal to the quotient of the sum of the squares of the adjacent
sides less the square of the opposite side^ divided by twice the
product of the adjacent sides.
In each figure a^^p^-\-I)C^
(in Fig. 84, II, 2) (7 is negative; in III, zero)
= c2 - ^2)2 + J2_ 26 . ^2) + ^i>2
= J2_|_^_26.^2).
But
AD=e cos
A
cos J.=
2 be
2 be
•
cos
A;
Prove that
d
(ioaB =
fioaCs
a' + c^-
2ao
a2 + 62_
2ah
62. Though these formulsB may be used for the solution
of the triangle, they are not adapted to the use of loga-
rithms (why?). Hence we derive the following:
Since cos ^ = 2 cos^ ^-1 = 1-2 sin^- ,
2 2
we have
2 cos^:^ == 1 + cos A, and 2 sin^^ = 1 - cos A.
LAW OF COSINES. 75
From the latter
2 2bc 2be
2 be 2 be
Let a+6+(?=2«,thena+6— {?=a + 6+c— 2i?=2«— 2tf;
*•«• a + J-<?=2(«-.c),
In like manner, a -- 6 + c = 2(« — J).
"a-»-6 + (?=2(«-a).
Substituting, 2 sin2^ = 2(<^- ft) '2rg-<?-)
2 2be
2 ^ be
Show that sin-=?
also sin-^== ?
From 2 cos?— = 1 + cos J,
show that cos - = ■yj"^"-'^),
also cos — = ?
and cos 1^ = ?
AJso derive the formulae
V >v.^ tanf = ?
tan~ = ?
2
76 PLANE TBIGONOMETRY.
63. Area of the triangle. In the figures of Art. 59 the
area of the triangle ABC=^ A = \'pf>.
But ^scsinA .'. A = J6(?8inil. (i)
Again, by law of sines, * = — :
Substituting, A =
sin (7
g^sinjlsinJ?
2sin(7
c^sinJ.sinjB
2sin(^+^)
(why?). (ii)
A A
Finally, since sin J. = 2 sin — cos ~, we have from (i)
A 1 iL a ' A A T ^ /«(« — d)(s — h)(s — c)
A = * 6<? • 2 sm— ■ cos — = oc\-^ ^ — - ^^ ^
^ 2 2 ^ be ' be
or A = V« (« — a)(« — 6)(« — e). (iii)
Find A ; (1) Given a = 10, 5 = 12, C = 45*».
(2) Given a = 4, J = 6, c = 6.
(3) Given a = 2, B = 45% C = GO*.
SOLUTION OF TRIANGLES.
64. For the solution of triangles we have the following
formulae, which should be carefully memorized :
J gja A __ 9in B _ gin C
II. tiui|(^-B) = ^^i|taii|(^ + B).
* a + 6 *
III. ««- = V^^ g ^' w ««_ = ^-L^
2 ' »(»-a) ' , .
IV. A = I dc 8to ^ = ^ «j^°^^yg = V«(* -aK»- b) (5 - g).
SOLUTION OF TRIANGLES. 77
Which of the above formulae shall be used in the solution
of a given triangle must be determined by examining the
parts known, as will appear in Art. 69. It is always pos-
sible to express each of the unknown parts in terms of three
known parts.
In solving triangles such as Case I, Art. 58, the law of
sines applies; for, if the given side is not opposite either
given angle, the third angle of the triangle is found from
the relation A + B + 0=: 180®, and then three of the four
quantities in — — = ^^ — being known, the solution gives
a
the fourth.
In Case II (6) the law of sines applies, but in II (a) two
only of the four quantities in — — = ^^ — are known.
a
Therefore, we resort to the formula
tanJ(^-5) = 5L::^tanJ(^ + 5),
in which all the factors of the second member are known.
In Case III, tan4 = \^ ""^^^*"7^^ is clearly applicable,
and is preferred to the formulae for sin— and cos — ; for,
first, it is more accurate since tangent varies in magnitude
from to 00, while sine and cosine lie between and 1.
(See Art. 27, 6.)
Let the student satisfy himself on this point by finding, correct to
seconds, the angle whose logarithmic sine is 9.99992, and whose loga-
rithmic tangent is 1.71668. Does the first determine the angle ? Does
the second?
And, second, it is more convenient, since in the complete
solution of the triangle by sin — 8ix logarithms must be taken
A A
from the table, by cos — seven^ and by tan — but four.
The right triangle may be solved as a special case by the
law of sines, since sin (7=1.
78
PLAKE TRIGONOMETRY.
65. Ambiguous case. In geometry it was proved that a
triangle having two sides and an angle opposite one of them
of given magnitude is not always determined. The marks
of the undetermined or ambiguous triangle are :
1. The parts given are two sides and an angle opposite one.
2. The given angle is acute,
3. The side opposite this angle is less than the other given
side.
When these marks are all present, the number of solutions
must be tested in one of two ways :
(a) From the figure it is apparent that there will be no
solution when the side opposite is less than the perpendicular
p ; one solution when side a equals p ; and two solutions when.
a is greater than p.
A b c
No Solution,
A b C A b C C'
One Solution, Two Solutions,
Fig. 35.
And since sin ii = — » it follows that there will be no solt^
c ^
tion^ one solution^ two solutions^ according as sin A =-,
< c
(6) A good test is found in solving by means of loga-
rithms ; and there will be no solutions, one solution, two solu*
tions, according as log sin proves to be impossible, zero,
possible, i,e, as log sin Q is positive, zero, or negative. This
results from the fact that sine cannot be greater than unity,
whence log sine must have a negative characteristic, or be
zero.
G6. In computations time and accuracy assume more than
usual importance. Time will be saved by an orderly arrange-
ment of the formulae for the complete solution, before open-
ing the book of logarithms, thus :
SOLUTION OF TRIANGLES. 79
Given -4., By a. Solve completely.
(7=180<^-(A+5), 6 = 554!]^, c:=^Lmf^ A^labsinO.
am A am A ^
ISC'* log a = log a =
A -\-B = log sin 5 = log sia C =
••• C = colog sin A = colog sin A =
log 6 = log c =
.'. 6= .'. c =
log a = log (« — 6) =
log 6 = log (« — c) =
log sin C = colog 8 =
colog 2 = colog (« — a) =
logA= 2 )
.-. A = log tan — =
.-. A =
67. Accuracy must be secured by checks on the work at
every step ; e.g. in adding columns of logarithms, first add,
up, and then check by adding down. Too much care can-
not be given to verification in the simple operations of
addition, subtraction, multiplication, and division. A final
check should be made by using other formulae involving the
parts in a different way, as in the check above. As far as.
possible the parts originally given should be used through- '
out in the solution, so that an error in computing one part:
may not affect later computations.
6& The formulae should always be solved for the unknown
part before using^ and it should be noted whether the solu-
tion gives one value, or more than one, for each part ; e.g.^
the same value of sin 5 belongs to two supplementary angles,,
one or both of which may be possible, as in the ambiguous,
case.
69. Write formulae for the complete solution of the fol-^
lowing triangles, showing whether you find no solution, one
solution, two or more solutions, in each case, with reasons for
your conclusion :
80
PLANE TRIGONOMETRY.
a
h
c
A
B
C
1.
8P 26' 28"
44** 11' 20"
540 22' 12"
2.
78.54
63° 18' 20"
41** 30' 18"
3.
135.82
26.89
63*^ 28' 30"
4.
0.75
0.85
0.95
"•
5.
243
562
36** 15' 40"
6.
38.75
25.92
63° 50' 10"
7.
0.058
78^5'
83^46'
8.
2986
1493
30°
9.
48
50
26n5'
MODEL SOLUTIONS.
1. Given a = 0.785, b = 0.85, c = 0.633. Solve completely.
Check: A + B + C =z 180°. A = V«(« - a)(«- 6)(» - c).
a = 0.735
h = 0.85
c = 0.633
2)2.268
s = 1.134
s-a = 0.349
5-6=0.284
« - c = 0.501
Check:
A = 61° 53' 38"
J?= 72° 46' 4"
C= 45° 20' 20"
180° 0' 2"
log(«-6)= 9.45332
log(«-c)= 9.69984
oology = 9.94539
colog («-«)= 0.45717
2)19.55572
logtanJ^= 9.77786
^A =30° 56' 49"
A = 61° 53' 38"
log(»-a)= 9.54283
log(«-6)= 9.45332
colog«= 9.94539
colog (« - c) = 0.30016
2 )19.24170
logtanJC= 9.62085
J C = 22° 40' 10"
C = 45° 20' 20"
log(»-a)= 9.54283
log (« - c) = 9.69984
colog 9= 9.94539
colog(»-6)= 0.54668
2 )19.73474
log tan i 5= 9.86737
J5 = 36°23'2"
B = 72° 46' 4"
log«= 0.05461
log («-«)= 9.54283
log(«-6)= 9.45332
log (« - c) = 9.69984
2)18.75060
logA= 9.37530
A = 0.2373
(1) Given a = 30, 6 = 40, c = 50.
(2) Given a = 2159, b = 1431.6, c = 914.8.
(3) Given a = 78.54, 6 = 32.56, c = 48.9.
SOLUTION OF TRIANGLES.
81
2. Given A = 57° 23' 12", C= 68° 15' 30", c = 832.56. Solve completely.
a —
csinii
sin C
5 = 180O-(.l +C)
= 54° 21' 18".
j^ciBin^ ^ = j5c8in^.
sin C
Check: taxi^A
-4
logc = 2.92042
log sin A = 9.92548
colog sin C = 0.03204
log a = 2.87794
a= 754.98
Check: a= 754.98
6= 728.38
c= 832.56
log c = 2.92042
log sin B = 9.90990
colog sin C = 0.03204
log b = 2.86236
6= 728.38
9-a= 402.98
« - 6 = 429.58
8-c= 325.40
2 )2315.92
5 = 1157.96
« (« — a)
log 6= 2.86236
logc= 2.92042
log sin ^ = 9.92548
log 2 A = 5.70826
A = 515811 = 255405.5
2
log(«-6)= 2.63304
log(«-c)= 2.51242
colog «= 6.93634
colog(«-a)= 7.39471
2 )19.47651
logtani^= 9.73826
i il = 28° 41' 38'
A = 57° 23' 16'
Solve :
(1) Given a = 215.73, B = 92° 15', C = 28° 14'.
(2) Given b = 0.827, A = 78° 14' 20", B = 63° 42' 30".
(3) Given b = 7.54, c = 6.93, if = 54° 28' 40".
3. Given a = 25.384, c = 52.925, B = 28° 32' 20". Solve completely.
(Why not nse the same formulffi as in Example 1, or 2?)
tan
C-A = £^tan.^ + ^
2 c -^a 2
180° - 5 = C + ^ = 151° 27' 40".
.-. i(C + A)=: 75° 43' 50".
, c sin ^ J. , ^^ .^ o
sinC
Check: b =
asinB
sin^ '
c= 52.925 log (c- a) = 1.43998 .-. i(C-A)= 54° 7'38"
a= 25.384 colog (c+ a) =8.10619 l(C+.4)= 75°43'50"
c+a= 78.309 log tan KC+^)= 0.59460 ^^^^^^^ C=129°51'28"
c-a= 27.541 log tan i(C-i4) =0.14077 subtracting, ^4= 21°36'12''
log a =1.40456
log c= 1.72366
log sin 5=9.67921
colog sin C =0.11484
log ft =1.51771
b= 32.939
Check: log a =1.40456
log sin 5=9.67921
colog sin i4 =0.43395
log 6= 1.51772
log c= 1.72366
log sin 5=9.67921
log 2 A =2.80743
^^641^^ 320.92
82 PLANE TRIGONOMETRY.
Solve : (1) Given a = 0.325, c = 0.426, B = 48° 50' 10".
(2) Given h = 4291, c = 3194, A = 73° 24' 50".
(3) Given b = 5.38, c = 12.45, A = 62° 14' 40".
4. Ambiguous cases. Since the required angle is found
in terms of its sine, and since sin a =* sin (180° — a), it fol-
lows that there may be two values of a, one in the first, and
the other in the second quadrant, their sum being 180®. In
the following examples the student should note that all the
marks of the ambiguous case are present. The solutions will
show the treatment of the ambiguous triangle having no
solution, one solution, two solutions.
(a) Given 6 = 70, (? = 40, (7=47° 32' 10". Solve. Why
ambiguous ?
o;« » 6 sin (7 log b = 1.84610
sin JL> = . '^
logsin (7= 9.86788
colog c = 8.39794
log sin B = 0.11092
.*. 5 is impossible, and there is no solution. Why?
Show the same by sin (7 > -•
(J) Given a = 1.5, c = 1.7, A = 61° 66' 38". Solve.
. /Y csinA loff <? = 0. 23045
sin ^ . °
a log sin A = 9.94564
colog a = 9.82391
logsin (7= 0.00000
(7=90°
* and there is one solution. Why? Show the same by
sin -4. = -. Solve for the remaining parts and check the
work.
X
c^
N
SOLUTION OF TRIANGLES. 88
((?) Given a = 0.236, b = 0.189, B = 36° 28' 20". Solve.
„. J a sin 5 ^ bsinO
smjS
log a = 9.37107 log 6 = 9.27646 9.27646
logsin5=: 9.77411 log sin (7= 9.99772 or 9.28774
colog b = 0.72354 colog sin B = 0.22589 0.22589
log sin A = 9.86872 log c = 9.50007 or 8.79009
A =: 47° 89' 25" c = 0.31628 or 0.06167
or 132° 20' 85".
.-. (7= 95° 52' 15" or 11° 11' 5".
Solve for A, and check. Show the same by sin 5 < —
a
Solve :
(1) Given 6 = 216.4, (? = 593.2, 5= 98° 15'.
(2) Given a = 22, 6 = 75, -B = 82° 20'.
(8) Given a = 0.353, c = 0.295, A = 46° 15' 20"..
(4) Given a = 293.445, b = 460, A = 40° 42'.
(6) Given b = 631.03, (? = 629.20, jB = 84° 28' 16".
Solve completely, given :
a .ft \ c
1. 50 60
2. 10 11
3. 4 6 6
4. 10
5. 40 51
l^ 6. 852.25 513.27 482.68
^ ^ 7. 0.573 0.394
^ 8. 107.087
9. v^
10. 197.63 246.35
11. 4090 8850 3811
3795
234.7 185.4
14. 26.234 22.6925
15. 273 136
I
c
/
^
^ ^13.
A
B
C
78° 27' 47'*
03° 35'
109° 28' 16"
38° 56' 54"
49° 28' 32"
112° 4'
56° 15'
48° SS'
117°
45°
34° 27'
73° 15' 15"
42°18'30'
84° 36'
49° 8' 24"
72° 25' 13"
84 PLANE TRIGONOMETRY.
APPLICATION&
70. Measarements of heights and distances often lead to
the solution of oblique triangles. With this exception, the
methods of Chapter V apply, as will be illustrated in the
following problems.
The bearing of a line is the angle it makes with a north
and south line, as determined by the magnetic needle of the
mariner's compass. If the bearing does not correspond to
any of the points of the compass, it is usual to express it
thus: N. 40** W., meaning that the line bears from N. 40®
toward W.
EXAMPLES.
1. When the altitude of the sun is 48^, a pole standing on a slope
inclined to the horizon at an angle of 15° casts a shadow directly down
the slope 44.3 ft. How high is the pole?
2. A tree standing on a mountain side rising at an angle of 18° 3(K
breaks 82 ft. from the foot. The top strikes down the slope of the moun-
tain 28 ft. from the foot of the tree. Find the height of the tree.
3. From one corner of a triangular lot the other corners are found to
be 120 ft. £* by N., and 150 ft. S. by W. Find the area of the lot, and
the length of the fence required to enclose it.
4. A surveyor observed two inaccessible headlands, A and B. A was
W. by N. and B, N.E. He went 20 miles N., when they were.S.W. and
S. by E. How far was A from B ?
5. The bearings of two objects from a ship were N. by W. and N.E.
by N. After sailing E. 11 miles, they were in the same line W.N. W.
Find the distance between them.
6. From the top and bottom of a vertical column the elevation angles
of the summit of a tower 225 ft. high and standing on the same hori-
zontal plane are 45° and 55°. Find the height of the column.
7. An observer in a balloon 1 mile high observes the depression angle
of an object on the ground to be 35° 20'. After ascending vertically and
uniformly for 10 mins., he observes the depression angle of the same object
to be 55° 40'. Find the rate of ascent of the balloon in miles per hour.
8. A statue 10 ft. high standing on a column subtends, at a point
100 ft. from the base of the column and in the same horizontal plane, the
same angle as that subtended by a man 6 ft. high, standing at the foot
of the column. Find the height of the column.
9. From a balloon at an elevation of 4 miles the dip of the horizon
is 2° 83' 40". Required the earth's radius.
TRIANGLES — APPLICATIONS. 85
10. Two ships sail from Boston, one S.E. 50 miles, the other N.E. by
£. 60 miles. Find the bearing and distance of the second ship from the
first.
11. The sides of a valley are two parallel ridges sloping at an angle of
dO°. A man walks 200 yds. up one slope and observes the angle of eleva-
tion of the other ridge to be 15°. Show that the height of the observed
ridge is 273.2 yds.
12. To determine the height of a mountain, a north and south base
line 1000 yds. long is measured ; from one end of the base line the sum-
mit bears E. 10° N., and is at an altitude of 13° 14'. From the other end
it bears E. 46° SO' N. Find the height of the mountain.
13. The shadow of a cloud at noon is cast on a spot 1600 ft. due west
of an observer. At the same instant he finds that the cloud is at an ele-
vation of 23° in a direction W. 14° S. Find the height of the cloud and
the altitude of the sun.
14. From the base of a mountain the elevation of its summit is 54° 20'.
From a point 3000 ft. toward the summit up a plane rising at an angle
of 25° 30' the elevation angle is 68° 42'. Find the height of the mountain.
15. From two observations on the same
meridian, and 92° 14' apart, the zenith
angles of the moon are observed to be
44° 54' 21" and 48° 42' 57". CaUing the
earth's radius 3956.2 miles, find the dis-
tance to the moon. 1 ^^-^C^'ZenUh angle
16. The distances from a point to three
objects are 1130, 1850, 1456, and the angles
subtended by the distances between the three objects are respectively
102° 10', 142°, and 115° 50'. Find the distances between the three objects.
17. From a ship A running N.E. 6 mi. an hour direct to a port dis-
tant 35 miles, another ship B is seen steering toward the same port, its
bearing from A being E.S.E., and distance 12 miles. After keeping on
their courses IJ hrs., B is seen to bear from A due E. Find B's course
and rate of sailing.
18. From the mast of a ship 64 ft. high the light of a lighthouse is
just visible when 30 miles distant. Find the height of the lighthouse,
the earth's radius being 3956.2 miles.
19. From a ship two lighthouses are observed due N.E. After sailing
20 miles E. by S., the lighthouses bear N.N.W. and N. by E. Find the
distance between the lighthouses.
20. A lighthouse is seen N. 20° E. from a vessel sailing S. 25° E. A
mile further on it appears due N. Determine its distance at the last
observation.
EXAMPLES FOR REVIEW.
In connection with each problem the student should review
all principles involved. The following list of problems will then
furnish a thorough review of the book. In solving equations,
find all values of the unknown angle less than SCO"* that satisfy
the equation.
1. If tan a = ^, tan j3 = }, show that tan ()3 ~ 2 a) = |^.
2. Prove tan a + cot a = 2 esc 2 a.
3. From the identities sin^4 + cos^— = 1, and 2 sin •— cos-r- = sin il,
2 2 2 2
prove 2 sin -- = ± V 1 + siu -4 db V 1 — sin -4,
and 2 cos — = db VI + sin A qp VI — sin^.
4. Remove the ambiguous signs in £x. 3 when il is in tnm an angle
of each quadrant.
5. A wall 20 feet high bears S. SQ"" 5' E. ; find the width of its shadow
on a horizontal plane when the sun is due S. and at an altitude of 60^
6. Solve sin X + sin 2 X + sin 3 X = 1 + cos X + cos2 «•
7. Prove tan-i i + tan-i i = |.
8. K il = 60% B = 45% C = SO^*, evaluate
tan A + tan B + tan C
tan A tan B + tan B tan C + tan C tan A
9 Prove ^^^ (^ + B) cos C _ 1 — tan A tan B
cos (il + C) cos B 1 — tan A tan C
10. Solve completely the triangle whose known parts are h = 2.35,
c = 1.96, C = 38^ 45'.4.
11. Fmd the functions of 18°, 36^ 54°, 72°.
Let«-18°. Then 2x = 36°, 3z = 54°, and 2x + 3x-90°,
12. If cot a = -, find the value of
sin a + cos a + tan a + cot a + sec a + esc a.
86
EXAMPLES FOR REVIEW. 87
13. Prove "°.^«"».2g-«!°^^''!°^«=l + 4cog«co8i8.
Sin 2 a Sin )3 — sm 2 p Bin a
14. From a ship sailing due N., two lighthouses bear N.E. and
N.N.E., respectively; after sailing 20 miles they are observed to bear
due £. Find the distance between the lighthouses.
15. Solve 1 — 2 sin X = sin 3 X.
16. Prove sin-K/-^ = tan-^V?-
^a + 6 ^b
17. If cos tf — sin tf = V2 sin ^, then cos tf + sin tf = "\/2 cos A
18. Solve completely the triangle ABC, given a = 0.256, b = 0.387,
C = 102° 20'.5.
2 cos 2 a — 1
19. Prove tan (30° + a) tan (B0°- a) =
2 cos 2 a + 1
20. Solve tan (45° - 6) + tan (45° + ^) = 4. ,
21. Prove sin^ a cos* P — cos* a sin* j8 = sin* a — sin* j8.
22. Prove cos* a cos* j8 - sin* a sin* j8 = cos* a — sin* j8.
23. A man standing due S. of a water tower 150 feet high finds its
elevation to be 72° 30' ; he walks due W. to A street, where the elevation
is 44° 50' ; proceeding in the same direction one block to B street, he finds
the elevation to be 22° 30'. What is the length of the block between A
and B streets ?
24. Prove tan-* i + tan-i 1 + tan-i i + tan-i g = |.
25. U P = 60°, Q - 45°, R = 30°, evaluate
sin P cos Q + tan P cos Q
sin P cos P + cot P cot 72
26. If cos (90° + a) = - i evaluate 3cos 2 a + 4 sin 2 a.
27. If sin B + sin C = m, cos B + cos C = n, show that tan — 5— = —
2 n
28. Show that sin 2 p can never be greater than 2 sin j3.
29. Prove sin-* J + sin-^ ^ = tan-* JJ.
30. Solve cot-^x + sin-»| V^ = ^.
5 4
31. Solve sin-* x + sin-* (1 — a;) = cos-* x.
32. A man standing between two towers, 200 feet from the base of
the higher, which is 90 feet high, observes their elevations to be the same ;
70 feet nearer the shorter tower he finds the elevation of one is twice that
of the other. Find the height of the shorter tower, and his original
distance from it.
88 PLANE TRIGONOMETRY.
33. Solve cos 3 j3 + 8 cos* j3 = 0.
34. Solve cot m - tan (ISO** + m) = sec m + sec (OO** - m).
35. Solve i-=^-5?:2J = 2 cos 2 ^
1 + tan <
36. Prove cotil + cot^ = "°(^ t ^^ '
sin A sin 2^
37. Prove cot P - cot Q = - ^^P ^^ 7 9 '
sin P sin Q
38. In the triangle ABC prove
a = 6 cos C + c cos 5,
6 = c cos il + a cos C,
c = a cos 5 + ft cos -4.
39. Solve completely the triangle, given
a = 027.56, b = 648.25, c = 738.42.
40. Prove cos^ a - sin (30° + a) sin (30° - a) = |.
^1 "D- ^ A-. Q i. cos 2 a: — cos 4 x
41. Prove tan3xtanx = — •
cos 2 X + cos 4 X
42. Simplify cos (270° + a) + sin (180° + «) + cos (90° + a),
43. Simplify tan (270° -$)- tan (90° + ^ + tan (270° + 0).
44. Solve cos 3 ^ - cos 2 ^ + cos ^ = 0.
45. Solve cos il + cos 3 -4 + cos 5 il + cos 7 -4 =0.
46. The topmast of a yacht from a point on the deck subtends the
same angle a, that the part below it does. Show that if the topmast be
a feet high, the length of the part below it is a cos 2 a.
47. A horizontal line ^4^ is measured 400 yards long. From a point
ID. AB SL balloon ascends vertically till its elevation angles at A and B
are 64° 15' and 48° 20', respectively. Find the height of the balloon.
sin a a n
48. If cos ^ = n sin a, and cot ^ = - — ^, prove cos P = .
tan^ Vl+n^cos^a
49. Find cos 3 a, when tan 2 a = — }.
50. Solve completely the triangle, given a = 0.296, B = 28°47'.3,
C = 84° 25'.
51. Evaluate sin 300° + cos 240° + tan 2250.
52. Evaluate sec ^ - esc ^ + tan Is:.
O o u
EXAMPLES FOR REVIEW. 89
53. If ^^^_ f&nacmy-n\nPtmy
cos a cos y — cos fi siu y
and t^^^8in«siDY-8i"^co8Y,
COS a sin y — cos p cos y
show that tan(0 + ^) = tan (a + p),
54. If tan iee** 16' 38" = - ^, find the sine and cosine of 233° T 49".
CSC a — cot a sec ce — tan a
55. Prove
sec a + tan a esc a + cot a
56. Prove co8(« - 3/3)-co8(f« -/?) = 2 8in(« - /9).
sm2a + sin2p
57. Prove sin 80° = sin 40° + sin 20°.
58. Prove cos 20° = cos 40° + cos 80°.
59. Prove 4 tan-i - - tan'i — = E.
5 239 4
60. From the deck of a ship a rock bears N.N.W. After the ship
has sailed 10 miles E.N.E., the rock bears due W. Find its distance*
from the ship at each observation.
61. Find the length of an arc of 80° in a circle of 4 feet radius.
62. Given tan ^ = |, tan ^ = A, evaluate sin(^ + ^) + cos(^ — ^).
63. If tan ^ = 2 tan ^, show that sin(^ -f ^) = 3 sin(^ — ^).
64. Prove cos(a+)8)cos(a-j8)+8in(a+i8)sin(a-i8)=^^I^5^.,
l4-tan*/S
65. Solve 4 cos 2 ^ + 3 cos ^ = 1.
66. Solve 3 sin a = 2 sin (60° - a).
67. Prove (sin a — esc a)' — (tan a — cot ay +(cos a — sec a)*='l.
68. Prove 2 (sin® a + cos* a) + 1 = 3(sin* a + cos* a).
69. Prove esc 2 )8 + cot 4 j8 = cot j8 — esc 4 j8.
70. K tanj3=A, cos2g=5|Z, thencsc-2_=L2 = 5vlg.
71 Solve completely the triangle, given
a = 0.0654, 5 = 0.092, 5 = 38°40'.4.
72. Solve completely the triangle, given
6 = 10, c = 26, 5 = 22°37'.
73. A railway train is travelling along a curve of \ mile radius at the
rate of 25 miles per hour. Through what angle (in circular measure)
will it turn in half a minute ?
90 PLANE TRIGONOMETRY.
74. Express the following angles in circular measure :
630, 4°3(y, 6° 12' 36".
75. Express the following angles in sexagesimal measure:
«" 3jr 17 IT
6' 8* 64
76. 11 Af By C are angles of a triangle, prove
cos il + cos 5 + cos C = 1 + 4 sin — sin — sin ^
77. Prove sin 2 x + sin 2 y + sin 2 2 = 4 sin :c sin y sin 2, when x, y^ z
are the angles of a triangle.
78. Prove sec a = 1 + tan a tan ^*
79. Prove sin* (a + jS) - sin2(a - )3) = sin 2 a sin 2 j8.
80. Prove cos«(a + /3)- sin2(a - j8) = cos 2 a cos 2 j8.
81. Prove ^? ^^? + ^\" ^JP = 2cos9;).
sm 10 j3 + 8m8/>
82. Consider with reference to their ambiguity the triangles whose
known parts are :
(a) a = 2743, h = 6452, B = 43° 15'
(6) a = 0.3854, c = 0.2942, C = 38°20r
(c) 6= 5, c = 53, 5 = 15^22'
((/) a = 20, 6 = 00, il = 63° 28'.5.
83. From a ship at sea a lighthouse is observed to bear S.E. After
the ship sailed N.E. 6 miles the bearing of the lighthouse is S. 27° 30' E.
Find the distance of the lighthouse at each time of observation.
84. Prove «ai^)i«aM±^.2co8((? + «.
sm 2 u -\- sin J ^p
85. Prove cos 15° - sin 15° = — •
y/2
86. Show that cos (a + j8) cos (a - )8) = cos« a - sin* p
= cos*)8-sin*a.
87. Show that tan (a + 45°) tan (a - 45°) = 2 8in*tt-l ,
^ ^ ^ ^ 2co82«-l
88. Solve sin (x + y) sin (x — y) = }, cos (x + y) cos (x - y) = 0.
89. Prove ^ + T" " '^" = tang.
1 + sm a + COS a 2
EXAMPLES FOR REVIEW. 91
90. Prove tan2tf + sees tf = 52S4±«n|.
co$ ^ — sin 9
91. If tan ^ = -9 then a cos 2 ^ + 6 sin 2 ^ = a.
a
92. Prove sin-*— + cot-i 3 = ?•
93. Solve cos A + cos 7 -4 = cos 4 -4.
94. Two sides of a triangle, including an acute angle, are & and 7
the area is 14 ; find the other side.
95. Show that ^.^^"^^"^.'"''o^'r.^^ = *an 2 ft
sin5^-3sin30 + 4sin^
96. A regular pyramid stands on a square base one side of which is
173.6 feet. This side makes an angle of 67° with one edge. What is
the height of the pyramid?
97. From points directly opposite on the banks of a river 600 yards
wide the mast of a ship lying between theqi is observed to be at an eleva-
tion of 10** 28'.4 and 12° 14'.5, respectively. Find the height of the mast.
98. Show that (sin 60° - sin 45°) (cos 30° + cos 45°) = sin^SO^
99. Find ar if sin-i X + sin-i ^ = ?.
2 4
100. Trace the changes in sign and value of sin a + cos a as a
changes from 0° to 360°.
CHAPTER VIII.
MZBCSLZiANBOnS PROPOSITIONS.
71. The circle inscribed in a given triangle is often called
the ineircle of the triangle, its centre the incentre^ and its
radius is denoted by r. The incentre is the point of inter-
section of the three bisectors of the angles of the triangle
(geometry).
The circle circumscribed about a triangle is called the
circumdrcle^ its centre the circumcentre^ and its radius M,
The circumcentre is the point of intersection of perpendicu-
lars erected at the middle points of the three sides of the
triangle (geometry).
Incirde.
Cirenmcircle. Escribed circle opposite A.
PlO. 37.
The circle which touches any side of a triangle and the
other two sides produced is called the escribed circle; its
radius is denoted by r^, r^, or r^, according as the escribed
circle is opposite angle -4, 5, or C.
Again, the altitudes from the vertices of a triangle meet
in a point called the orthoeentre of the triangle.
Finally, the medians of a triangle meet in a point called
the centroid^ which is two-thirds of the length of the median
from the vertex of the angle from which that median is
drawn (geometry).
Certain properties of the above will now be considered.
02
c
MISCELLANEOUS PEOPOSITIONS.
7Z To find the radius of the incircle.
Let A, A', A," A'" represent
the areas of triangles ABO,
COB, A 00, BOA, respectively.
Then
A = A' + A" + A'"
93
And since A = V«(« — a)(« — 6)(« — (?),
(Art. 63)
8
Cob. To express the angles in terms of r and the sides,
divide each member of the above equation by « — a.
Then
s
-I,=J (^"/)(^ Z£)=.tan^A (Art. 62)
In like manner tan i -B = ^ , ; tan i (7 = — ^«
'^ « — 6 " 8 —
73. To find the radius of the circumcirde.
Fio. 39.
In the figure ABO is the given triangle, and A'O sl diam-
eter of the circumcirde. Then, angle A = A', or 180®— A'.
.•. sin J. = sin -4.'.
Since A* BO is a right angle,
BO
sin A^ =
a
AIO 2R
• . ./I/ — f r-
94
PLANE TRIGONOMETRY.
CoE, 1. As above, 25 = -: — 7 = -: — =r = -: — :^ which is
»
sin A sill B sin (7*
another proof of the ^^ law of sines.
a
Cob. 2. From 5=
, we have
E
2sin^
abe
2 be sin ^ 4 A
= --—, where A= area ABO.
74i To find the radii of the escribed circles.
Now
Represent areas ABC^ BOA^
AOO, BOO, by A, A', A'', A"',
respectively. Then r^ is the
altitude of each of the triangles
^ BOA, AOO, BOO.
A = A' + A" — A'"
= J^a(<? + 6 - a)= r„(« - a>
A
•*• * « ^
In like manner, r^ =
« — a
A
1'» ^^ =
« — c
75. The ortfaocentre.
Denote the perpendiculars on the sides
a, 6, {?, by AP^ BPf^ OP^ and let it be p^
required to find the distances from their
intersection to the sides of the tri-
angle, and also to the vertices. ^
OPf, = APf, tan OA 0.
But APt^'CcosA, and (740 = 90^-0:
••. 0PA = <?cosilcot(7=-: — -cosAcosO.
sin (7
= 2Rco8A cos 0. (Art. 73, Cor. 1)
MISCELLANEOUS PROPOSITIONS. 95
In like manner, OPc = 2Bco8 £ cos A,
OPa^2Eco8Cco8£.
Again, the distances from the orthocentre to the vertices
are,
cosCAO sin (7
= 2RcobA.
Also, 0B = 2R €08 B,
and 00 =^ 2 R 608 0.
76. Centroid and medians.
The lengths of the medians may be computed as follows :
In the figure the medians to the ^
sides a, J, <?, are AMa^ BM^^ OM^^
meeting in the centroid 0.
Now, by the law of cosines, from ^ /^^-""^ ^\ ^""^ ^^^ g
the triangle -B-Mi (7,
BM^^ =^ ^2 + jf^ (72 _ 2 a . Jf, (7 . c OS (7
= a^ + — ■ — a6 cos (7.
4
But, cos O = — ~— -r ^
zoo
. »if2_^2 . J' a2^.62_^ _ 2aa+2cg-6g
whence, BMf, = ^■V2a^-\-2(^-b^ = ^^a^ + €^+ 2aeco8B
a^ + ^-V^ J,
since — — = cos-o.
2ae
In like manner,
OM, = ^^/2b^ + 2a^-(^=^i^/b^ + a^ + 2baco8 0,
and AM^=lW2€^ + 2V^-a^=^Vc^+b^ + 2cbco8A.
96 PLANE TRIGONOMETRY,
EXAMPLES.
t
1. In the triangle, a = 25, ( = 35, c = 45, find 22, r, r..
2. Given a = 0.354, h = 0.548, C = 28"" 34' 20V, find the distances to
C and By from the circumcentre, the incentre, the oentroid, and the
orthooentre.
3. In the ambiguous triangle show that the circumcircles of the two
triangles, when there are two solutions, are equal.
4. Prove that 1 + 1 + 1 = -.
r. r^ r« r
5. In any triangle prove A ^y/rr^ri^r^
6. Prove that the product of the distances of the incentre from the
vertices of the triangle is 4 r^R.
7. Prove that the area of all triangles of given perimeter that can be
circumscribed about a given circle is constant.
8. Prove that the area of the triangle ABC is Jf2r(8in il + sin £ + sin C).
CHAPTER IX.
SERIES— DE MOIVRE'S THEOREM— HYPEBBOLIC
FUNCTIONS.
77. First consider some series by means of which loga-
rithms of numbers and the natural functions of angles may
be computed. For this purpose the following series is
important :
6 = 1 H 1 1-
• ••■
(
It may be derived as follows :
By the binomial theorem,
1 , 1 , nxCnx—l') 1 , nxCiix—l')(nx-'2) 1 ,
n \2 n^ [3 rfi
X
(^-D 4-S(^-!)
. • a.
= 1+^ + [2-+ 2 ^
and if n increase without limit,
This is called the exponential series^ and is represented by
6*, so that
e* = 1 + » + — +—+ •.. + ^+ ....
It is shown in higher algebra that this equation holds for
all values of x ; whence, if a; = 1,
97
98 PLANE TRIGONOMETRY.
This value of e is taken as the base of the natural or
Naperian system of logarithms.
Thifl yalue «, howeyer, is not the base of the system of logarithms
computed by Napier, but its reciprocal instead. The natural logarithm
is used in the theoretical treatment of logarithms, and, as will presently
appear, it is customary to compute the common logarithm by first
finding the natural, and then multiplying it by a constant multiplier
called the modulus. Art. 82 ; t.6. in the Naperian system the modulus
is taken as 1, and the base is computed. In the common system the
base 10 is chosen and the modulus computed.
78l From the exponential series the value of e may be
computed to any required degree of accuracy.
1+1 + 1-2.5
i = 0.1666666666
i = 0.0416666666
,4 = 0.0083333333
i = 0.0013888888
i = 0.0001984126
1 = 0.000Q248016
1 = 0.0000027557
,47 = 0.0000002765
|10
Adding, e = 2.7182818, correct to 7 decimal places.
SERIES. 99
79. To expand a* in ascending powers of jr.
Let «* = 6', then z = log^ a* = a; • log^ a. (Arts. 35, 40)
Substituting
Now put 1 + a for a, and
(l + a)' = l + a;.log.(l + a)
[2 ^ [3 ^ V
But by the binomial theorem,
[2 [3
Equating coefficients of a; in the second members of the
above equations,
log.(l + a)=a-| + |-f + ...;
or writing « f or a,
%.(l + a:) = a:-| + |-| + ....
In this form the series is of little practical use, since it
converges very slowly, and only when x is between + 1 and
— 1 (higher algebra).
Put — X for a;, and
log,(l-a:)=-a;-----~ ;
.-. loge(l + a;) - log,(l - a;)
..og.l±|-2(. + f + f+...)
SS1730A
100
PLANE TRIGONOMETRY.
Finally, put
2n + l
for Xy and
W 4- 1
loge — ^ = log, (w + 1) - log, n
n
.Mog.(»+l)=lo,.»+2{^-Kl(^)V|(^)*+...j,
a series which is rapidly convergent.
>
80. From this series a table of logarithms to the base e
may be computed.
To find log, 2 put w = 1. Then, since log, 1 = 0, the series
becomes
log,2 = log.l + 2{| + ^ + ^ + ^ + ^_
11-3" 13. 3»
}-«•
693147.
The computations may be arranged thus :
3
9
9
9
9
9
9
2.00000000
.66666667 = .66666667
.07407407-*- 3 = .02469136
.00823046 -»- 5 = .00164609
.00091449 -f- 7 = .00013064
.00010161+ 9 = .00001129
.00001129 -i- 11 = .00000103
.00000125 -f- 13 = .00000009
.69314717
whence log; 2 = 0.693147, correct to 6 decimal places.
To find log, 3, put » = 2, and
log.3 = log,2 + 2(l + ^ + ^ + ^ + ^ + ...).
SERIEa
6
2.00000000
25
.40000000
.40000000
25
.01600000 + 3 =
.00633338
25
.00640000 + 5 =
.00012800
25
.00025600 -J- 7 =
.00000366
.00000102 + 9 =
.00000011
.40546510
log. 2 =
.69314717
101
.% log, 3 = 1.098612,
correct to 6 decimal places.
log, 4 = 2 X log, 2, log, 6 = log, 3 + log^ 2, etc. (Why ?)
The logarithms of prime numbers may be computed as above
by giving proper values to w.
8L Having computed the logarithms of numbers to base «,
the logarithms to any other base may be computed by means
of the following relation :
Let loga n=:x; then a^=n.
Also, logft w = y ; then b^ = w.
Hence, log^ (a^ = log« (J^i
and .'. x^i/logah.
It follows that loga n = log^ n • log^ h ;
whence log^ n = logfa n • - •
logo ft
This factor ; is called the modulvs of the system of
logarithms to base 5. Using it as a multiplier, logarithms
of numbers to base b are computed at once from the loga-
rithms of the same numbers to any other base a.
102
PLANE TRIGONOMETRY.
82. To compute the common logarithms.
Common logarithms are computed from the Naperian by
use of the modulus
log. 10
9 ».^.
logio n = log, n
loge 10
By Art. 80, log^lO can be, found, and
•- — — - as .434294, the moduliis of the common system.
loffe 10
Ex. Compute the common logarithms of :
2, 3, 4, 6, 6, 10, 15, 216, 3375.
COMPLEX NUMBERS.
83. In algebra it is shown that the general expression for
complex numbers is a + W, where a represents all the real
terms of the expression, b the coefficients of all the imagi-
nary terms, and % is so defined that i^ = — 1 ; whence
t = V— 1, i* = — 1, t^ = — 2, f* = 1, etc.
The laws of operation in algebra are found to apply to
complex numbers. Moreover, it is further shown that if
two complex numbers are equal, the real terms are equal,
and the imaginary terms are equal ; i,e, if
a + bi = c + di^
then a=sb and c = d.
Finally, the complex niunber
may be graphically represented as
follows :
'0
r
Flo. 43.
The real number is measured
along OX^ a units ; the imaginary
parallel to OF, 6 units. The line
r is a graphic representation of
a +
DE MOIVBE'S THEOREM. 103
Since a = r cos 6 and J = r sin 6^
.". a + if = r (cos 6 + i sin ^).
The properties of complex numbers are best developed by
using this trigonometric form. If r be taken as unity, then
cos 6 + % sin 6 represents any complex number.
84. De Home's Theorem. To prove that, for any value
of w,
(608 + i sin 0)** = cos nO + i sinnO.
I. When w is a positive integer.
By multiplication,
(cos a -♦- i sin a) (cos /8 + i sin ff)
= cos a cos fi — sin a sin, /8 + f (sin a cos /8 + cos a sin /8)
= cos(a + )8) + tsin(a + /8).
In like manner,
(cos a + i sin a) (cos /8 + f sin )8)(cos 7 + 1 sin 7)
= cos(a + )8 + 7) + i sin (a + /3 + 7) 5
and finally,
(cos a+i sin a) (cos P+i sin /8)(cos 7+i sin 7) ••• to n factors
= cos(a + )8 + 7 + •••) + « sin(« + /8 + 7+ ••*)•
Now let a = /8 = 7 = •••, and the above becomes
(cos a + i sin «)*» = cos na + i sin na.
II. When w is a negative integer.
Let w = — m ; then
(cos a + f sin «)*» = (cos a + i sin «)""•
(cos a + i sin a)*" cos ma + i sin wa
cos ma — f sin ma
(cos Twa + i sin m«)(cos ma — t sin ma)
cos ?w« — i sin ?w«
cos^ ???« + sin^ ma
= cos ma — f sin ma = cos (— m) a + 1 sin (— wi) a.
104
PLANE TRIGONOMETRY.
Substituting n for — m, the equation becomes
(cos a + i sin a)" = cos na + i sin na.
III. When w is a fraction, positive or negative.
p
Let n = — , ^ and q being any integers.
Now
cos - + 1 sin - = cos a • - + i sin 7 • - = cos a + 1 sm a (dv I >
Then
(COS - + f sin - J = (cos a + i sin «)?•
Raising each member to the power p^
Tcos a + % sm «)« = cos - -♦- i sm - = cos ~ a + 1 sin ^ a.
\ 9 qJ q q
COMPUTATIONS OF NATURAL FUNCTIONS.
85. The radian measure of an
acute angle is greater than its sine
and less than its tangent, i.e.
sin a<a< tan a.
Let a be the circular, or radian,
measure of any acute angle A OP.
Then, in the figure,
area of sector OAF < area of triangle OAT^
i.e. iOA*2LTcAP<^OA'AT.
.-. arc.4P<^T:
Now, since N^P < arc AP,
NP ^ SLTcAP AT
OP OP
OP
But
whence
bxgAP
OP
= circular measure of AOP = a ;
8ina<a<tana.
NATURAL FUNCTIONS. 105
86. Since sina<a<tan€e,
i<-A-< ^
Sin a cos a
Hence, however small a may be, -: — lies between 1 and
^ sm a
When a approaches 0, cos a approaches unity.
cos a
Therefore, by diminishing a sufficiently, we may make
-, — differ from unity by an amount less than any assign-
sma
able quantity. This we express by saying that when a
approaches 0, -: — approaches unity as a limit, i.e. — — = 1,
sin a sm a
approximately. Multiplying by co«a(= 1, nearly), we have
^ = 1, approximately. Whence, if a approaches 0,
tan a
tan az=sina=za^ approximately.
87. Sine and cosine series,
cos na + { sin na = (cos a + i sin «)•, (De Moivre's Theorem).
Expanding the second member by the binomial formula,
it becomes,
, r^Cw— l)Cw — 2) - . •9*3
^ — ^^ ^Av ^cos" ' a . p sm' a
li
Substituting the values of »*, t*, «*, etc., we have
. • • « w (n — 1 ) -_» • *
cos na + 1 sin »i« = cos" «6 i-— — ^-cos" 'asin'a
If
^ n(n-lXn-2Xn-3) ^^^,_ „ ^^, „ _ ...
li
+ ifn cos*-' g sin « - " (" ~ |X» - ^) cos*-* a sin* «+...)
lOa FLANE TKIGONOMETRY.
Equating the real and imaginary parts in tlie two members,
^ nTw - V)(n - 2)(w - 3) -.4 ^ . 4 ^
[4
and m na ss n co*"*"^ a ^ "" ^^ """"^ cofi""* a mV a + —.
Ex.1. Find cos 8a; sin 3 a.
In the above put n = 8| and cos 3 a = cos' a — 3 cos a sin^a
= 4 cos* a — 3 cos a;
= 3 sin a — 48in'a.
also "Mvv ^ '\- sinr a = 3 cos^ a sin a — sin' a
2. Find sin 4a; cos 4 a; sin 5 a; cos 5 a.
It will be noticed that in the series for cosna and Wnna the terms
are alternately positive and negative, and that the series continues till
there is a zero factor in the numerator.
88l If now in the above series we let na == d, then
-(--1)
a\a J
cos 6 = cos" a r^ 008""^ a sin^ a
a\a J\a J\a J . . .
^ — ^— 1-; ^ cos* * a sin* a— •••
0(0^ a) -_2 /sinaV
= cos" a ^^— - — ^cos*. ^ al
2 V a /
11
(4 \ « y
If now remain constant, and a decrease without limit,
then will n become indefinitely great, and -. — - and every
a
NATURAL FUNCTIONS. 107
power thereof, and cos a and every powet of cos« will
approach unity as a limit, so that
Similarly, **^* = *"^"*'^""^"^':-
By algebra it is shown that these series are convergent for
all values of 0. By their use we can compute values of sin
and C09 to any required degree of accuracy.
Show from the above that tan 0=^0 + -^ + -— - + -..
Ex. 1. Compute the value of sin V, correct to 5 places.
In sin $ = $ ^ — + _ — — -f •.., make $ the radian
[3+16 [7 +
measmp of 1» = -|L = 0,01745 +.
180
Then, 6 = 0.01745 +
.? = 0.0000008.
Li
.-. sin tf = 0.01745 +.
The terms of the series after the first do not affect the fifth place, so
that the value is given by the first term, an illustration of the fact that,
if a is small, sin a=: a, approximately. Compare the value of tan 1°.
2. Show that sin 10° = 0.17365 ; cos 10° = 0.98481 ; sin 15° = 0.25882 ;
cos 60° = 0.50000.
3. Find the sine and cosme of 18° 30' ; 22° 15' ; 67° 45'.
It is unnecessary to compute the functions beyond 30°, for since
sin (30° + tf) + pin (30° - tf) = cos tf (why ?),
.-. sin (30° + ^) = cos tf - sin (30° - $}.
So,also, cos (30° + ^) = cos (30° - ^) - sill tf .
Giving proper values the functions of any angle from 30° to 45° are
determined at once from the functions of angles less than 30°.
Thus, sin 31° = cos 1° - sin 29° ;
cos 31° = cos 29° - sin 1°.
4. Find sine and cosine of 40° ; of ^50°.
108 PLANE TRIGONOMETRY.
I The following are sometimes useful in applied
mathematics :
Ex. 1. To find the Bum of a series of sines of angles in A. F^ such as
sin a + sin (a + )8)+ sin (a + 2)8)+ ... + sin (a-^-ln- l]/3).
2sinasin^= cosf a — ^J — cosf a + q)»
2sin(a + )8)sin| = cos(a + |)-co8^a+^),
2sin(a + 2)8)sin| = co8fa+^)-co8fa+^],
28m (a + [n - l]/3) Bm^ = ooB(a +^Jt^p\ - cos f « +?iLli^y
Adding
2{sin « + sin (a + )3)+ sin (a + 2/3)+ ... + sin (a-\-[n- l])3)}sin|
= co8(«-|)-ooe(« + 2iL^V)
= 2 sin (a +2^/3) sin ^^. .
.•. sin a + sin (« + fi}+ sin (« + 2^)+ ... + sin (« + [n - 1]^)
sin (« + r^p) sin5^
Similariy it can he shown that
cosa + co8(a + /?)+ co8(a + 2)3) + — + cos(a + [n - 1])8)
cos (a + 2^)3) sin|i3
sin^
2
HYPERBOLIC FUNCTIONS. 109
90. Th^ series e^==l-|-a;-|--~ + ~H |-±l + ...i8 proved
in higher algebra to be true for all values of x^ real or
imaginary. Then if a; = iff^
c** = 1 + t^+—- + -— + -. + -— + ...
[2 |3 [r
.-. e«* = <?(?« ^ + i «m e (Art. 87).
In like manner, e*'* ==co80 — i sin tf .
Adding, co« ^ =
e<« -I- e'^
subtracting, sin =
2»
HYPERBOLIC FUNCTIONS.
9L Since sin ^ = — tt-, , and cos == ^ are
2t 2
true for all values of ^, let 6 = i^.
Then, sin (i0) = ^^^^^-^ = i ^——^ = i sinh 0,
and cos (i0) « ^ "^^ =a cosh ^,
so that tan (f ^) = ^Hllg} » \l^ = ; tanh ^,
COS {%0) cosh ^
where tinh 0^ co%h 0^ tanh 0^ are called the hyperbolic sine^
cosine^ and tangent of 0. The hyperbolic cotangent, secant,
and cosecant of are obtained from the hjrperbolio sine,
cosine, and tangent, just as the corresponding circular func-
tions, cotangent, secant, and cosecant, are obtained from
tangent, cosine, and sin6. The hyperbolic functions have
the same geometric relations to the rectangular hyper-
110 PLANE TRIGONOMETRY.
bola that the circular functions have to the circle, hence the
napie hyperbolic functions.
stnh6 = ^ J , .'. C8ch6= ^ ° A
eogh6 = — \^ ■, .-. gech0 = -T- -A
tanh6 = ^^-^"^ . ... cothe = 4-±-^.
92. From the relations of Art. 91 it appears that to any
relation between the circular functions there corresponds a
relation between the hyperbolic functions.
Since cos? (id^ + sin^ (i^) = 1,
cosh^ e + i^ sinh^ 5 = 1,
or cosh' d — sinh' 5 = 1.
This may also be derived thus :
cosy - sinh* e =(^'s^'-(^lifrj
c» + 2 + e-» - e» + 2 - e-» .
4 ^'
Also since
sin (ia + i)8) = sin (ia) cos (i)8) + cos (ia) sin (t)8),
.•. i sinh (a + ^) = i sinh a cosh y9 + cosh a • i sinh )8,
and sinh (a + y9) = sinh a cosh )8 + cosh a sinh )8.
Let the student verify this relation from the exponential
values of sinh and cosh.
EXAMPLES.
Prove
1. cosh (a + )3) = cosh a cosh + sinh a sinh p.
2. cosh ( «+ )3) — cosh (a — )8) = 2 sinh a sinh j8.
3. co8h2tf=l + 28inh2tf = 2cosh»tf-l.
4. sinh 2 a = 2 sinh a cosh a.
EXAMPLES. Ill
^ « ^ ^ ^
I
6. sinli 3 = 3 sinh tf + 4 sinh* $.
7. sinhtf + sinh ff>= 2 sinh "^ ^ cosh ~ r ,
' .....
8. sinh a+sinh (a + )8) + sinh (a + 2 )3) + — + sinh (« + [n - 1])3)
sinh ia + ^^/S) sinh|)3
sinh?
2
^ ^ 1 + tanh tanh 6
10. sinh^* X = cosh"! VT+^ = tanh-^— -
11. cosh (a + P) cosh (a - /8) = cosh' a + sinh^ /J = cosh* ft + sinh* a.
12. 2 cosh na cosh a = cosh (n + 1) <% + cosh (n — 1) a.
13. cosh a = J (c« + e-*) = 1 + |L + ^ + ....
14. -sinh a = J (g* - e-*) = a + J^ + ^ + «..
^^ ^"^
15. tanh-ia + tanh-ii = tanh-i-^-tA.
1 + cw
SPHERICAL TRIGONOMETRY.
-•>oj»:o«-
CHAPTER X.
SPHERICAL TRIAITOLE8.
I
93. Spherical trigonometry is concerned chiefly with the
solution of spherical triangles. Its applications are for the
most part in geodesy and astronomy.
The following definitions and theorems of geometry are
for convenience of reference stated here.
A great circle is a plane section of a sphere passing
through the centre. Other plane sections are %mall circles.
The shortest distance between two points on a sphere is
measured on the arc of a great circle, less than 180^, which
joins them.
A spherical triangle is any portion of the surface of a
sphere bounded by three arcs of great circles. We shall
consider only triangles whose sides are arcs not greater than
180^ in length.
The polar triangle of any spherical triangle is the triangle
whose sides are drawn with the vertices of the first triangle
as poles. If ABC is the polar of ^'5'(7', then A'B^O^ is the
polar of ABC,
In any spherical triangle,
The sum of two sides > the third side.
The greatest side is opposite the greatest angle, and conversely.
.Each angle < 180^ , the sum of the angles > 180°, and
< 540^
JEach side < 180° ; the sum of the sides < 360°.
112
SPHERICAL TRIANGLES. 113
The sides of a spherical triangle are the supplements of the
angles opposite in the polar triangle^ and conversely.
If two angles are equal the sides opposite are equals and
conversely.
The sides of a spherical triangle subtend angles at the
centre of the sphere which contain the same number of angle
degrees as the arc does of arc degrees ; i.e. an angle at the
centre and its arc have the same measure numerically.
The arc does not measure the angle for they have not the satne unit
of measurement, but we say they have the same numerical measure ; ue.
the arc contains the unit arc as many times as the angle contains the
unit angle.
*
The angles of a spherical triangle are said to be measured
by the plane angle included by tangents to the sides of the
angle at their intersection. They have therefore the same
numerical measure as the dihe-
dral angle between the planes
of the arcs.
In the figure the following
have the same numerical meas-
ure :
arc a and angle a ;
arc ( and angle fi ;
arc c and angle 7 ;
plane angle A'BC;
spherical angle B and dihedral angle A-BO-0;
spherical angle O and dihedral angle B^CO^A ;
spherical angle A and dihedral angle C-A O^B.
A^C'B and C'A'B have not the same measure as spherical angles
C and A, for BA^, A'C'f C'B are not perpendicular to OA or OC.
94. In plane trigonometry the trigonometric functions
were treated as fimctions of the angles. But since an angle
and its subtending arc vary together and have the same
114
SPHERICAL TRIGONOMETRY.
numerical measure, it is clear that the trigonometric ratios
are functions of the arcs, and may be so considered. ^11
the relations between the functions are the same whether we
consider them with reference to the angle
or the arc, so that all the identities of
plane trigonometry are true for the func-
tions of the arcs.
Thus in the figure we may write,
y . y
sin a ss - or sin a = - ;
r r
sin^a + cos*a = 1, or sin^a + cos^a = 1 ;
cos 2 a = 2 cos^ a — 1, or cos 2 a = 2 cos^ a — 1.
GENERAL FORMULiE FOR SPHERICAL TRIANGLES.
95. The solutions of spherical triangles may be effected
by formulae now to be developed:
First it will be shown that in any spherical triangle
cos a = cos 6 cos c + sin 6 sin c cos ^^
cos 6 = cos c COS a + sin c sin a cos By
COS c = cos a cos 6 + sin a sin 6 cos C*
The following cases must be considered :
I. Both 6 and <? < 90^ III. Both 6 and (? > 90^
II. h > 90% c < 90^ IV. Either lore = 90^.
V. J = <? = 90^
The figure applies to Case I.
Let ABO be a spherical tri-
angle, a, 5, c its sides, and
the centre of the sphere.
Draw AC^ and AS tangent
to the sides 6, c at A. (The
same result would be obtained
by drawing AB\ AO perpen-
dicular to OA at any point to
GENERAL FORMULiE. 115
meet OjB, 0(7.) Since these tangents lie in tbe planes of
the circles to which they are drawn, they will meet 00 and
OB in (7' and B\ and the angle C^AB^ will be the measure
of the angle A of the spherical triangle ABC. Since OAB\
OAC* are right angles, AOB\ AOO^ must- be acute, and
hence sides ^, h are each < 90^.
In the triangles 0* AS and COB,
(7'5'2 ^AC'^ + AB'^ --2AC^ ^ Aff cos C^AB\
and B' 0"^ = 0(7'a + p5'2 ^2 00' ^ OB' cos (7'05'.
Subtracting and noting that
cos CAB' = cos A and cos COB' — cos a,
we have
= pC'a - ^ C"2 + 05'2 - ^IjB'a
. +2AC 'AB> cos A-^OC -OB' co6a.
But OfT'a - ^ (T's = OA" and 0B>^ - AB'^ = OAK
Hence, = O^" ^. ^(7/ . ^^f cos A-OC • OB' cosa;
cosa = — .— + _._co8A
••• cos a =s cos h cos <? + sin h sin c cos -4..
Similarly,
cos h = cos a cos (? + sin a sin e? cos J?,
and cos (? = cos a cos 6 + sin a sin 6 cos C.
These formulae are important, and should be carefully
memorized.
II. 6 > 90^; (?<90°.
In the triangle ABO, let h >90*
and c < 90°. Complete the lune
BACA'. Then in the triangle
AVB the sides a and J.'(7are both less than 90®, and by (I)
cos A'B S3 cos A'O cos a + sin J.'(7 sin a cos A' OB.
116
SPHERICAL TRIGONOMETRY.
But ^'^ = 180^-c, ^'(7= 180^- J, and ^'C!B=180*'-.a
.-. cos (180^ - c)= cos (180^ - 6) cos a
+ sin (180^ - J) sin a cos (180*^ - C) ;
or — cos (? =a (— COS h) COS a + sin 5 sin a (— cos (7),
and COS c = cos a cos i + sin a sin I cos (7.
A similar proof will apply in case e > 90°, I < 90*^.
III. Both J and (? > 90^
In the triangle AB(7, let both
I and (?>90®. Complete the
lune ABA'C. Then since A^O
and ^'5 are both < 90'',
cos a = cos A^O cos ^'5 + sin A^C sin ^'jB cos J.'.
But A' = ^, ^'(7= 180° -ft, ^'5=180°-c.
.-. cos a = cos (180° - 6) cos (180° - c)
+ sin (180° - 6) sin (180° - <?) cos A ;
or cos a ^ cos 5 cos c? + sin I sin ^ cos A,
Cases IV and Y are left to the student as exercises.
96. Since the angles of the polar triangle are the supple-
ments of the sides opposite in the
first triangle, we have -4'
a' = 180°-^, 6' = 180° -5,
c'=180°-(7, -4' = 180° -a.
Substituting in
cos a' = cos V cos <?'
+ sin V sin c' cos A\
we have
cos (180° - ^)= cos (180° - jB) cos (180° - (7)
+ sin (180° - 5) sin (180° - (7) cos (180° - a);
or — cos-4.=s( — cos£)(— cos(7)+8in£ sin (7(— cos a).
GENERAL FORMULA 117
Changing signs,
eos ^ = - eo8 B cos C + sin B sin C cos a.
Similarly, CMB = -co8^co8C + 8ln^8iiiCGOs&9
and eos C = - cos^ cosB + sin ^ sin B cos c#
tvn T u • 1 4. • 1 i. sin A sin B sin C
97. In any spherical triangle to prove — : =« — : — r= — r
Since cos A =
sin a sin 6 &mc
cos a — cos I cos c
sin I sin c
,*. sin^-4.
_ ^ __ /cos a — cos 6 cos gV
\ sin 6 sin c J
__ sin^ 6 sin^ c — (cos a — cos b cos g)^
sin^ b sin^ c
_ (1 — cos^ 6) (1 — cos^ c) — (cos g -^ cos b cos g)^
sin^ b sin^ g
1— cos^a— cos^J — cos^g+2cosacos6 cose?
sin^ 6 sin^ c
Hence,
^ Vl — cos^a— cos^J— cos^g— 2cosa cosJ cose
sin -A = : — ;— :
sin 6 sm c
J sin^ __ VI — cos^g — cos^ft—cos^g— 2cosa cos 6 cose
sin a sin a sin 6 sin c
By a similar process, 51 — _ and 51 — will be found equal
sm sin c
to the same expression.
• sin ^ _, sin B _, sin C^
" sina sin 6 sine
118 SPHERICAL TRIGONOMETRY.
9B. Expressions for sine, cosine, and tangent of half an
angle in terms of functions of the sides.
We have 2 sin^— = 1 — cos A
_ 1 cosg — cost cosg
sin b sin c
_ cos h cos € + sinb sin c — cos a
sin b sin c
_ cos (6 — g) — cos a
Bin b sin c
Then 2 sin'^ = ^Binl(^a+b-c)Bin^(a-b+e) ^^^ ^^^
2 sin b sm g
^
_ 2 sin (g — t) sin (g — e)
sin ( sin g
when 2s = a + b + c.
2 "-^ sin^sinc
SimUarly, «in f = Jii*Hii^^EZ,
'' 8 ^ gin a slue
and gin C^J^^iiLz ^Tan^s - a) ,
2 ^ Bin a sin 6 ;
Also from the relation
2cos2^=l + cos^
-I , cos a — cos b cos ^
= IH : — i— r- — '
sm sin g
we have em 4= >fa*8i» (*-«) .
Also, cogf=A^pi^i^^,
2 ^ sine sin a
and cwig=J»M£MiL=^,
2 ^ sin a sin 6
GENEKAL FORMULA. 119
From the above, . a
8in -:r-
1^;^^;; 2 ^ J8iii(« - 6) sin (« - c)
2 ^g:4 ^ 8in«8lii(«-a)
2
Also, tan f = v/gtoIJbi- «) "l" (* - «) ,
2 ' 8ill«8lll(« — 6}
and t^gJglnl^^ a) gin (*-&) .
2 ^ 8iii«8iii(tf-c)
Compare the formulaB thus far derived with the torresponding for-
mulae for solving plane triangles. The similarity in forms will assist
in memorizing the formulae for solving spherical triangles.
99. From the formulae of Art. 96, the student can easily
prove the following relations :
sin ^ - J - C08 S cos (;S - A)
2 ^ sin £ sin C
wherjB 2Sf = J. + J5+C*
i Bln| = l,
rin|=f.
2 ^ sinBBinC
C08| = f,
eos| = t.
2
tan — -^ / - cos H cos (S — ^)
2 ^cos(S-B)co8(S-C)
tonf^l.
tan| = ?.
120 ' SPHERICAL TRIGONOMETRY.
100. Napier's Analogies.
Since
2 _ ^ sin g sin (^ — a)
tan:? J sin(g- g)8in(g~ a)
2 ^ sin « sin (« — 6)
•
_ /sin^(8 — 6) _ sin (g — S) ^
^sin^ (« — a) sin (« — a) *
by composition and division,
A B
tan — + tan— . , ,
2 2 __ sm (« — 6) + sin (« — a)
tan^-tan^"®^^(*-*)-«^^C^-^)'
2 2
sm — • sm -rr-
COS — COS —
2 2 ^ sin j^(2 g - g - 6)cos ^(g - V)
sin- sin-"^^®^<^^*'"^'"*)®^^i<^^~*)'
2 2 (Art. 61)
A B
cos- cos-
sin ^jA + ^) _ tan ^(2 g -- g - 6)
sin ^(^ - J5) "" tan J(g - 6)
tan-
tan ^ (g — i)
, since 2« — g — 6 = (?.
.-. tan I (a - 6) = — ^ tan|.
To find an expression for tan J (J. — jB) we have only to
consider the polar triangle, and by substituting 180° — A for
g, etc., 180° — a for A, etc., we have the following relations :
i(g - J)= K180° - ^- 180° + 5) = - JC^ - 5);
GENERAL FORMULA 121
also, J(^-5) = -i(a-6);
J(-4 + 5) = i(180''-a + 180°-i)=180»-J(a + J);
and £. = 90°-^.
2 2
The formula then becomes, applying Art. 29,
ton^(^-B) = — ? coti.
8liil(a+6) *
Formulae for tan J (a + 6), tan J (A + B) are derived as
follows :
Since
tan^ ^ tan^ ^ J 8in(g~6)sin(i?-c) ^ / 8in(i?.^g)8in(j^~a)^
2 2 ^ sin « • sin (« — a) ^ sin « • sin (« — J)
sm — sin —
2 2 ^ 8in(^-c)
cos4cos^ si^«
By composition and division,
A B^ . A . B
cos— cos— + sm— sm—
2 2 2 2 _ sing + sin(g-g)
^^A^^B '^A'^B sin « — sin r« — {?")
cos— cos— — sm— sm-— oxx^^o ^/^
2 2 2 2
whence ^"^f^^--g) = ^^^K^ + ^), (Art. 61)
cosi(A + 5) ^^^_c
since 2« — <?=a + J»
or, • taii|(a + &) = — ? tan^.
C08|(^ + B) *
122 SPHERICAL TRIGONOMETRY.
The value of tan ^- QA + 5) is derived by substituting in
terms of the corresponding elements of the polar triangle.
cosK^-^) ^ -tan^(^ + ^)
-ooH^ia + b) cot-
2
.-. tsakhA + B)= — f-^ i^cot^.
cwjCa + ft) *
Similar relations among the other elements of the triangle
may be derived, or they may be written from the above by
proper changes of J., B^ (7, a, 6, c in the formulae. The stu-
dent should write them out as exercises.
101. Delambre's Analogies.
Since sin J (-4 + -B) = sin -5- cos — + cos — sin •-,
then
«a^UA+B\ = 8in (»-&)+ sin Co -g) ^ /sin « ■ sin (« -_£2.
sin c ^ sin a • sin 6
(Art. 98)
Hence s^p^(^ + -S) ^ sin (« - i) + sin (« - a)
cos^' "«'•'
2
2 sin ^ cos J (« — 5)
2» c c
sin-cos-
cog I (a - 6) f,
and aaUA + B)= — ?— - — co8|j
C08| 2
In like manner derive
8ini *
2
^ (Art. 61)
BIGHT SPHERICAL TRIANGLES. 123
C08| 2
C08|(-4-B)=— S— Sillf.
8ill| *
These formulae are often called Gauss's Formulse, but they were first
discovered by Delambre in 1807. Afterwards Gauss, independently, dis*
covered them, and published them in his Theoria Motus,
102. Formulae for solving right spherical triangles are
derived from the foregoing by putting = 90°, whence
sin (7=1, cos (7=0.
cos e = cos a cos 6 + sin a sin b cos (Art. 96)
becomes cos c = cos a cos 6. (1)
Substituting the value of cos a from (1), and simplifying,
.._ . cos a -cos 6 cos <? (Art. 96)
becomes cos ^ = ^^. (2)
in the right triangle is
8in
I
sine?
COS^
tanc*
sin J.
sin a
sin (7
sine?
\ is
gin /I
_glna
sine*
Again, Siii^ = ^^^ (Art. 97)
(3)
Dividing (3) by (2),
sin a cos h sin a cos a cos h sin a
tan J. =
cos c sin h cos c cos a sin h cos a sin h
since cos a cos h = cos <?.
.*. tan^ gin 6 = tana. (4)
124 SPHERICAL TRIGONOMETRY.
From (4) tan a = tan J. sin 5,
also, tan b s= tan B sin a.
Multipljring, tan a tan b =: tan A tan B sin a sin b^
or, oot^cotB = oasaoas& = oosc. (5)
From (2) and (3), by division,
tan 5
cos A tan c cos c
— : — - = -: — r = 7 = cos a.
sm ^ sm cos 6
sin c
.*. cos^ = cosa sinB. (6)
Let the student write formulae (2), (3), (4), (6) for B.
It will be noticed that (1) and (5) give values for c only,
while (2), (3), (4), (6) apply only to A and B.
103. Formulae (l)-(6) are sufficient for the solution of
right spherical triangles if any two parts besides the right
angle are given. They are easily remembered by comparison
with corresponding formulae in plane trigonometry. Two
rules, invented by Napier, and called Napier^s Mules of Cir-
cular Parts, include all the formulae of Art. 102.
Omitting (7, and taking the comple-
^*^-S ments of A, <?, and B, the parts of thie
triangle taken in order are a, 5, 90*^—^1,
90^ - (?, 90^ - B. These are called
the circular parts of the triangle.
Any one of the five parts may be
selected as the middle part, the two
parts next to it are called the adjacent
parts, and the remaining two the opposite parts. Thus, if
a be taken as the middle part, 90® — B and b are the
adjacent parts, and 90® — <?, 90 — A the opposite parts.
NAPIER'S RULES. 125
Napier's Two Rules are as follows :
The sine of the middle part equals the product of the tangents
of the adjacent parts.
The sine of the middle part equals the product qf the. cosines
of the opposite parts.
It will aid the memory somewhat to notice that i occurs in
sine and middle, a in tangent and adjacent, and o in cosine
and opposite, these words being associated in the rules.
The value of the above rules is frequently questioned,
most computers preferring to associate the formulsd with
the corresponding formulae of plane trigonometry.
These rules may be proved by taking each of the parts as
the middle part, and showing that the formulsB derived from
the rules reduce to one of the six formulae of Art. 102.
Then, if 5 is the middle part, by the rules,
sinJ = tanatan(90*' — A)=tanacotA, or tanJ.=-r^,
sin
sin I = cos (90^ - <?) cos (90^ - B) =sin c sin B,
• i> sinJ •
or sm -B = -: — ,
sin<?
results which agree with (4) and (8), Art. 102. If any
other part be taken as the middle part, the rules will be
found to hold.
104. Area of the spherical triangle.
If r = radius of the sphere,
• U^ spherical excess of the triangle = A+-B+ (7— 180%
A = area of spherical triangle, then by geometry
A = Er^ X -?-.
ISO
If the three angles are not known, U may be computed by
one of the following methods, and A found as above.
126 SPHERICAL TRIGONOMETRY.
Cagnoli's Method.
sinf = sin J(A + B^C- 180^)
Si
= sin J (^ + 5) sin^- COS ^ (^ + ^) COS §
.
sin— cos—
= [cob J (a - J) - cos J (« + ^)] (^Ajij. 101)
cos^
2sin-sm- .
__ 2 2 ^ V sin 8 sin (« — a) sin (« — J) sin (« — c)
cos - sin a sin 6 *
2 (Arts. 51, 98)
^■E_. Vglng gin (8 - a)gln(8 - 6)rin(8 — e)
^ 2GOg|cog|GOg|
8 8 2
Lhttllier's Method.
4 008^(^ + 5+ (7- 180")
Now, multiply each term of the fraction by
2coa{CA + B- + 180°),
and by Art.. 51, (1) and (3), the equation becomes
fi
^ sinJ(^ + 5)-co8-
tan- = \ ,
* cosJ(^ + -B)+8in-^
[cos J (« - *) - cos|1cos -
-•- ^-^ ^ (Art. 101)
cos J(a + J)4- co8| sin-
sin 1^(8 — })sin \(^8 —
cos -cos J (
in J(« — a) / sin<sin(g — g)
,(s-e) ^sin(» - a)sin(« - 6)'
(Art. 51)
AREA OF SPHERICAL TRIANGLES. 127
By Art. 52, introducing the coe£Bcient under the radical,
tan^= Vtan ftaii^ (« - a)taiik8 - &)taii^ (« - e).
If two sides and the included angle are given, U may be
determined as follows :
cos:f = cos iCA + B + 0- 180^)
= cos ^(A + £) 8in-^+ sin ^(A + B) cos-^
= COS K« + *) sin^-^ 4- cos J (^ - *) ^os^-^ (Art. 101)
2 2
cos - COS ;- 4- sm - sm - cos C
2 2 2 2
cos-
2
«
sm - sin - • 2 sm-- cos—
But sin— = . (Cagnoli's Method)
cos-
2
Dividing this equation by the above,
mn - «w - 8in U
, U 2 2
2 « 6 , . a . 6 /v
cos - co« - 4- ««w :r 8in-co8 U
2 2 2 2
.a,
This formula is not suitable for logarithmic computations.
Usually it is better to compute the angles by Napier's Analo-
gies, and solve by A = JEr^ x
180
128 SPHERICAL TRIGONOMETRY.
EXAMPLES.
1. Show that cos a = cos ( cos c + sin b sin c cos A becomes
sec ^ = 1 + sec a, when a = h = c.
2. Ifa + ft + c = T, prove
B C
(a) cos a = tan -- tan ■—.
/1.x ^^„2 A cos a
{p) cos^ •— = -:-
2 sin 5 sin c
(c) sin^ — = cot 6 cot c.
(d) cos -4 + cos 5 + cos C = 1.
(e) 8in« I + sina | + sin^ |^ = 1.
8in:=^cos i (^ — ^ ) sin^sini (« — a)
3. Prove ^ \ =- = (Art. 104)
Sin -— cos -
2 2
4. Show that cos a sin & :± sin a cos 6 cos C + sin c cos A.
CHAPTER XL
SOLUTION OF SPHERICAL TRIANGLES.
105. According to the principles of Bfpherical geometry
any three parts are sufficient to determine a spherical tri-
angle ; the other parts are computed^ if any three are given,
by the formulae of trigonometry. The known parts may be :
I. Three sides, or three angles.
II. Two sides and the included angle, or two angles and
the included side.
III. Two sides and an angle opposite one, or two angles
and a side opposite one.
It will appear that, as in plane geometry. III may be
ambiguous.
The signs of the functions in the formulae are important
since the cosines and tangents of arcs and angles greater
than 90° are negative; whether the part sought is greater
or less than 90*^ is therefore determined by the sign of the
function in terms of which it is found unless this function
be sine. In this case the result is ambiguous, since sin a
and sin (180® — a) have the same sign and value. Thus if
the solution gives log sin a = 9.56604, we may have either
a = 21'* 33', or 168^*27'. The conditions of the problem
must determine which values apply to the triangle in
question.
The negative signs, when they occur, will be indicated
*^^® • log cos 115® 20' = 9. 63185-,
indicating, not that the logarithm is negative, but that
in the final result account must be made of the fact that
cos 115® 20' is negative.
129
180 SPHERICAL TRIGONOMETR¥w
106. FormuloB for the solution of triangles.
I. sin ^ _ gin B _ gin C
sin a sinfr ginc*
II. tiiii^=\^^2^ZE35lZES.
2 ^ sin 8 gin (« — a)
III. tan^-V"^^^^^^"^^,
2 ^cog(iSf-B)cos(iSf-C)
glnJC^-B)
IV. tani(a-6) = -^ tanf.
* gln|(^ + B) 2
c©gl(^-B)
* cogJ(-4 + B) 2
ginl(a-6) ^
VI. tan|(^-B) = — f; --cot|.
VII. tan|(^ + B) = — I- -cot|.
* co9|(a+6) *
VIIL A = Er^
180'
where H is determined by
tan^=A/*«n5tani(«-a)tain|(«-6)tan|(«-c).
4 « 8 8 8 8
Right triangles may be solved as special cases of oblique
triangles, or by the following :
(1) cogc =co8acos&. (4) tan ^ sin & = tan a.
(2) C08^=^. (5) cotAcotB = c©9C.
(3) 8*n-^=f^* (6) cog^ = co8a8inB.
The formula to be used in any case may be determined by
applying Napier's Rule of Circular Parts.
107. In solving a triangle the student should select formulae
MODEL SOLUTIONS. 181
in which all parts save one are known, and solve for that
one (see page 77). Referring to Arts. 105 and 106, it will
appear that solutions are effected as follows :
Case I by formulaB II, or III, check by I.
Case II by formulsB VI, VII, I, or IV, V, I, check by IV
orVL
Case III by formulaB I, IV, or I, VI, check by VI or IV.
MODEL SOLUTIONS.
lOa 1. Given a = 46<> 24', b = 67^ 14', c = 81^ 12'. Solve.
tan^=\^HiLE^}ililIiZ£i, tan ^ -J^^ (' " ^) "^ (' ~ ^,
2 ^ sin 5 sin (« — a) 2 ^ sin « sin (s — b)
^^ C Jsm(s^a)sin(8--b)^ ^^^^^ . iBina ^ jm6 .
2 ^ 8in5sin(5— c) sin il sin^
Arrange and solve as in Example 1, page 80.
Ans. A = 46** 13'.5, B= , C =
Solve: (1) ^=96° 45', 5 = 108° 30', C = 116*»15'.
(Use formal» III in the same manner as in Example 1.)
(2) a = 108^14', 6 = 75° 29', c = 56°37'.
(3) ^ = 57*50', 5 = 98^20', C = 63°40'.
2. Given b = 113° 3', c = 82° 39', A = 138° 50'. Solve,
"^ cosl(6 + c) 2 ^ ^ sinj(6 + c) 2
i(B + Cr)±i(B--C) = B,OTC, 8ina=515LiiJ^.
^ sin J5
Check: tan2 = tanKft~c)8in}(^ + C)
2 sin J (5 - C)
6=113° 3' log cos 1 (6-c) =9.98453 logsm J (&-c) =9.41861
c = 82° 39' colog cos J (6 + c) = 0.86461" oolog sin i (6 + c) = 0.00409
J (&+<?)= 97° 51' log cot- =9.57466 log cot -=9.57466
J(ft-.c)= 15° 12' ^ 2 ^ 2
J -4 = 69° 25' log tan J (5 + C) = 0.42380" log tan J (5 - C) = 8.99736
} (5+ 0=110° 39' i (5-C)=5°40'.6
i(B-C) = 5°40'.6
.-. 5 = 116° 19'.6
and C=104°68'.4
132 SPHERICAL TRIGONOMETRY.
Check:
log Bin A = 9.81839 log tan i (5 - c) = 9.43408
log sin h = 9.96387 log sin J (5 + C) = 9.97116
cologs in 5 = 0.04756 colog sin J (5 - C) = 1.00474
log sin a = 9.82982 log tan - = 0.40998
a = 137^ 29' 2
a = 137^29'
Notice that tan I (A + C) is -. Hence, HB-h C)\b greater than 90^, i.e. IIQP 39^
Solve: (1) ^ = 68° 40', B= 56° 20', c= 84° 30'.
(Use formuliB IV, V, I. Compare Example 2.)
(2) a = 102° 22', ,b= 78° 17', C = 125° 28'.
(3) ^=130° 5', B= 32° 26', c= 51° 6'.
109. Ambiguotts cases. By the principles of geometry the
spherical triangle is not necessarily determined by two sides
and an angle opposite, nor by two angles and a side opposite.
The triangle may be ambiguous. By geometrical principles
it is shown that the marks of the ambiguous spherical tri-
angle are:
1. The parts given are two angles and the side opposite
one, or two sides and the angle opposite one.
2. The side, or angle, opposite differs from 90® more than ,
the other given side, or angle.
3. Both sides, or angles, given are either greater than
90®, or less than 90®.
In the right triangle ABO^^
sin a = sin A sin c, (formula (3))
Therefore there will be no solution, one
solution, or two solutions, according as
sin a = sin J. sin c, i.e. according as a =
the perpendicular j3. (See Art. 65.)
But the most expeditious means of determining the am-
biguity is found in the solution of the triangle. The use
of formula I gives the solution in terms of sine, so that it is
to be expected that two values of the part sought may be
possible ; and whether the triangle be ambiguous or not,
there must be some means of determining which of the two
\
AMBIGUOUS SPHERICAL TRIANGLES. 133
angles, a and 180° — a, that have the same sine is to be used.
If there are two solutions, both values are used.
This is determined in the further solution of the triangle
by formula V, which may be written
, J _ cos ^ (A -h 0) tan ^ (a + c)
^2 cos J (A -C)
Now -<90% whence tan - is 4-. Then if for both values
of (7, found by the sine formula, the second member is + ,
there are two solutions; if the second member is — for
either value of (7, there is but one solution ; while if both
values of make the second member — , there is no solution.
The various cases will be illustrated by problems.
3. Given a = 62° 15'.4, h = 103o 13* g^ ^ = 63° 42'.6. Solve.
gj^^^smJsin^ ^^^^c_ co8iM+^)tanH« + ft),
sin a 2 qo8\{A — B)
sin C = S2L^-51Ld. Check: cot^ = ^^^ K^ + ^) 8^^ J (« + ^).
sin a 2 sin ^ (a — 6)
Solving the first formula gives
log sin B = 9.94766,
whence B^ = 62° 24'.4,
5j = 117° 36'.6.
For each of the values B^ and B^
cos \(A -{-B) tan \(a + h)
cos \{A -^B)
is + and therefore equal to tan ^* Hence there are two solutions. Find
c = 163° 9'.6, or 70° 26'.4
and C = 155° 43'.2, or 69° 6'.2
4. Given a = 46° 46'.5, ^=73°ir.3, 5 = 61°18'.2. Solve.
gin ^- sing sin jB ^ ^q^ C _ tan j (^4 - B) cos ^(a + h)
sin ^ 2 cos i (a — &)
8inc=?lBJ«8inC^ ^^^^^. ^^^t^i^u\{a ^h)^m\{A + Bl
sin il 2 sin \iA-B)
134 SPHERICAL TRIGONOMETRY.
Solving for h gives log sin 6 = 9.82446,
whence 6i = 4P62'.6,
and 6j = 138° 7'.5.
For the value h^ the fraction
tan H-^ + B) cos j (o + 5)
cos i(a — b)
is +, but for 62 cos i (a + 6) is — , making the fraction — , and hence it
can not equal cot — , which is +• There is then but one solution. Find
C = 60* 42'.7, c = 41° 35M.
5. Given a = 162* 30', A = 49* 50', B = 57* 52'. Solve.
Solving gives log sin 6 = 9.52274,
whence b^= 19*27'.9,
ft, = 160* 32'.1.
For both values, h^ and h^, cos ) (a + ft) is — . Therefore,
tan ^ (^ + jB) cos^ (a + 6)
cos 1 (a — ft)
is — and not equal to cot — • Hence the triangle is impossible. «
Solve, testing for the number of solutions :
(1) ft = 106*24'.5, c= 40*20', C= 38*45'.6.
(2) a = 80* 50 , ^ = 131* 40', B = 65* 25'.
(3) a= 60*31'.4, ft = 147*32'.l, 5 = 143*50'.
(4) a = 55* 30', c = 139* 5', A = 43* 25'.
RIGHT TRIANGLES.
110. Right triangles are a special case of oblique triangles,
but are usually solved by formulae (1) to (6), Art. 106.
Students should have no difficulty in applying these.
Computers generally question the utility of Napier's Rules
of Circular Parts. For those who prefer the rules a problem
will be solved by their use.
SPECIES. 135
6. Given c = 86*» 51', B = 18° 3'.5, C = 90«.
The parts sought are a,h^ A, and it is immaterial which is computed
first, a and A are adjacent to c and B^ while h is the middle part of c
and £. Then by Napier's first rule
sin (90^ - 5) = tan (90* - c) tan a ; ^""^
or tan a = ?2?j5 = cos B tan c,
cote
which is formula (2).
By the same rule go—A
sin (90* - c) = tan (90* - ^) tan (90° - B),
or cot A = £2i£ = cos c tan 5, formula (5).
cot 5 . ^ ^
Finally by the second rule
sin ft = cos (90* - c) cos (90* - 5) = sin c sin 5, formula (3).
The solutions give a = 86* 41'.2, 6 = 18* 1'.8, A = §8* 68'.4. Verify.
111. Species. Two angles or sides of a spherical triangle
are said to be of the same species if they are both less, or
both greater, than 90°. They are of opposite species when one
is greater and the other less than 90°. Since the sides and
angles of a spherical triangle may, any or all, be less or
greater than 90°, it is necessary in solutions to determine
whether each part is more or less than 90°. The directions
already given are sufficient in oblique triangles. In right
triangles the sign of the function will determine if the solu-
tion gives the result in terms of cosine or tangent, but not
if the result is found in terms of sine. Thus in Example 6,
above, we have log sin 5 = 9.49068, whence 5 = 18° 1'.8, or
161° 58'. 2. By formula (4) sin 5 ==^5:54. Now sin b is
•^ ^ "^ tan ^
always +? therefore, tan a and tan A must be of the same
sign, whence in any right spherical triangle an ollique angle
and its opposite side must he of the same species.
Again by formula (1) cos c = cos a cos J. Now cos (? is +
or — according as c is less or greater than 90°. If then
c<90°, cos a and cos h are of the same sign, but if (?>90°,
cos a and cosi are of opposite sign. Therefore, if the
Bin a
8inc =
186 SPHERICAL TRIGONOMETRY.
hypotenu9e of a right spherical triangle is less than 90^, the
other sides, and hence the angles opposite, are of the same
species; Ivt if the hypotenuse be greater than 90*^, the other
sides, and the angles opposite, are of opposite species.
112. Ambiguous right triangles.
When the parts given are a side adjacent to the right
angle, and the angle opposite this side, the triangle is
ambiguous, for solving for the hypot-
4si, enuse by formula (8) gives
jj\^^^ ^y/ Sin -4.
Fio. 54. from which there result two values of c.
By the last rule of species it follows that
to the values of c, one <90°, the other >90®, there will cor-
respond two values for i, one of the same species as a, the
other of opposite species.
Clearly sin c ^ 1, according as sin a = sin A, and hence
there will be no solution, one solution, or two solutions,
according as sin a ^ sin A,
Solve the spherical triangles, right angled at (7, given :
(1) 5 = 73° 21'.4, <?= 84° 48'.7.
(2) (? = 54°28', 5 = 128°12'.6.
(3) I = 45° 42', B = 185° 42'.
(4) a = 108° 22'.8, I = 120° 14'.5.
(5) a = 70° 50', A = 170° 40'.
(6) h = 32° 8'.4, B = 46° 2'.8.
(7) I = 34° 28', c = 62° 50'.
(8) c = 102° 35', B = 17° 45'.
(9) a = 92° 16', c = 57° 35'.
■N
X.
EXAMPLES 137
EXAMPLES.
Solve, given :
a be ABC
1. 97° 35' 27° 8'.4 119° 8'.4
2. 67°33'.4 94° 5' 99°57'.6
3. 40° 20' 70° 40' 40°
4. 82°39'.5 116° 20' . 70° 7
5. 155°47'.l 110°46'.4 90°
6. 49°44'.3 121°10'.4 26° 6'.3
7. 144° 10' 41°44'.2 130°
8. 127° 30' 132° 16' 139° 44'
9. 155° 5'.3 110° 10' 70°20'.8
10. 62° 42' 50° 12' 58° 8'
11. 120° 30' 70°20'.3 69° 35'
12. 50° 15' 75° 30' 90°
13. 116° 20' 104°59'.l 138°50'.2
14. 84°14'.5 32°26'.l 36°45'.4
15. 100° 50° 60°
16. 87° 12' 88° 12' 90°
17. 63° 50' 80° 19' 50° 30'
18. 34° 15' 42°15'.2 121°36'.2
19. 50° 63° 15' 90°
20. 159° 50' 159° 43' 123° 40'
21. 124°12'.5 54° 18' 97°12'.5
22. 48°31'.3 62°55'.7 125°18'.9
23. 76° 36' 40° 20' 42°15'.2
24. 28°45'.l 44°22'.2 122°25'.l
25. 44° 53' 53° 52' 90°
26. 98°21'.7 109°50'.4 115°13'.5
27. 99°40'.8 64°23'.2 95° 38'.]
y"
188
SPHERICAL TRIGONOMETRY.
For
APPLICATIONS TO GEODESY AND ASTRONOMY.
113. Geodesy is concerned in measuring portions of the
earth's surface, considering the eartli as a sphere.
To find the distance on the earth's surface between two
points whose latitudes and longitudes
are known.
If A and B are two places on the
earth, P the north pole, ECDW the
equator, and PUP' the principal merid-
ian, e.(j. the meridian of Greenwich,
and if the latitude and longitude of A
and B are known, then AB can be
computed.
AP = 90° - latitude A,
BP = 90° - latitude B,
angle APB = longitude A — longitude B.
.". two sides and the included angle of the triangle APB are
known, and AB can be computed.
Ex. 1. Find the distance between Ann Arbor, 42^ 19' N., 83°43'.8 W.,
and San Juan, 18° 29' N., 66° 7' W.
2. How far is Manila, 14° 36' N., 120° 58' E., from Honolulu,
21° 18' N., 157^55' W.? Honolulu from San Francisco, 37°47'.9 N.,
122° 24'.5 W. ? San Francisco from Manila ?
•
114. The celestial sphere. The heavenly bodies appear to
be situated on a sphere of indefinitely great radius with the
centre at the point of observation. This is called the celes-
tial sphere.
A tangent plane to the earth at the point of observation
cuts the celestial sphere in a great circle called the horizon.
The points of the horizon directly south, west, north, east
are called the south, west, north, east points.
A vertical line through the point of observation cuts the
celestial sphere above in the zenith, and below in the nadir,
the zenith and nadir being poles of the horizon.
APPLICATIONS. 139
The earth's axis produced is the axis of the celestial sphere,
cutting it in the north and south poles of the equator.
The altitude of a star is its distance from the horizon
measured on an are of a great circle drawn through the star
and the zenith.
The azimuth, or hearing, of a star, is the are o£ the horizon
measured from some fixed point to the foot of the great
circle through the star and the zenith. The fixed point is
usually the south point.
The declination of a star is its distance from the celestial
equator. The circle d rawn through the pole and the star is
the hour circle, and the angle at the pole between the prime
meridian and the hour circle is the hour angle oi the star.
Let an observer be at on the surface
of the earth, and let P be the position
of a star.
Then Z is the zenith, Z' the nadir,
EQE' the celestial equator, iV its north
pole, S its south pole, BRH' the horizon,
NFS the meridian, or hour circle, of P,
and ZNP the hour angle. The deelina- ' '"" ""'
tion of the star is PQ, its altitude PR, and its azimuth, or
bearing, NZP. The astronomical triangle NZP can be
solved if any three of its parts are known.
EXAMPLES.
1. What will be the altitude of the hqii ai, 9 a.m. in Betroit,
lat. 42= 20' N., its declination being 17° 30'.5?
2. At what time will the sun rise at San Francisco, lat. 37° 47'.9, if
its declination is 12°46'.2?
3. Find the azimuth and altitude of a star to an observer in
lat. 42° SO" N., when the hour angle of the star ia 3 h. 42.3 m. E., and
the declination is 42° 31' N.
4. The latitude of Sayre Observatory is 40°3e'.4 N.; the sun's alti-
tude is 47''15'.3, its azimuth 80°23'.l. Find its declination and hour
angle.
5. At Ann Arbor, March 13, 18fll, the altitude of Regulus ia 32° 10'.3,
and the azimuth is 283° o'.l. Find the declination and hour angle.
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