ELEMENTARY TREATISE
oir
PLANE TRIGONOMETRY,
WITH NUMEROUS EXAMPLES AND ArrLICATIONS.
^cetgncb for the net ot ^iglt ^rhoolg mib CoUcQee.
BY
J.'^^ORMSON, M.D., M.A.,
rniNOlPAIi OF THB WAIiKKUTON HIGH BCHOOTi; BX-PBINCIPAL OP THIS NRWMAUKl.;
uioH scHoor,; late mbmbkb of the medical council ajtd examiner
ON TTTKOKKTICAL AND PRACTICAL CHBMISTBT IN THE CULLKGE
Oi' i*HySIOLANS AND BUaaHONS OW ONTAiUO.
TORONTO:
CANADA PUBLISHING CO. (LIMITED).
1880,
''.H., ^^-1
..*?»
Entered according to Act of Parliament of Canada, in the year one thousand eight
hundred and eighty, by The Canada Publishing Company (Limited), in the office of
the Minister of Agriculture.
C. B. ROBINSON,
PRINTER,
JORDAN STREET, TORONTO.
if-
* ■
PREFACE.
The following course of Trigonometry has been prepared with
the view of meeting the wants not only of High School but
also of Collegiate classes, and in pursuance of this end, the
more elementary portions — such as are required for the matricu-
lation examination in the University of Toronto — are printed
ift a larger type, and form a connected treatise independently
of the portions printed in the smaller type, which are intended
for candidates reading for University honors. ;f, ;
In discussing the trigonometrical functions I have adopted
the method of ratios or the Cambridge method ; but on the
recommendation of several eminent teachers in our colleges,
I have given a full account of the older method or that of the
"'.M.n.9 definitions," and have employed the latter method in
solving right-angled triangles and in tracing the variations in
sign and magnitude of the several functions in the various
quadrants. Of course, teachers who prefer to use the " ratio
definitions " only, can omit the portions treating of the " lino
definitions " without breaking the continuity of the ciourse.
The solution of right-angled triangles by the natural func-
tions, and the application of Trigonometry to the determina-
tion of heights and distances, have been introduced at a very
iv. PREFACE.
early stage, with the view of leading the sMident to take a
deeper interest in the study of the science at the outset.
Numerous examples and solutions, chiefly of a practical
character, are given, at frequent intervals throughout the
work, for the purpose of fixing in the mind of the student the
principles of the text, and of illustrating the practical bearing
and utility of Trigonometry in Surveying, Navigation and
Astronomy. ' '
The more advanced portions of the work have been dis-
cussed from an elementary point of view, and will be found
to include all that is re(juired on this subject from University
students generally. -
As my aim has been to produce a work of utility, I have
not hesitated to consult, in the preparation of this volume, the
various English and foreign works on the subject ; but those
to which I am most indebted for suggestions, are the works of
Hind, Todhunter, Snowball, Wiegand and Chauvenet. i •
The examples have been obtained from various sources ;
some are either entirely original or original in the form in
which they are presented ; some have been selected from other
trigonometries, and some from examination papers set during
the last twenty years in the Universities of Cambridge, London,
and Toronto, and in McGill College, Montreal, and Harvard
College, U.S. The answers have been given wherever neces-
sary, and great care has been taken to secure their accuracy;
but it is quite possible that some errors have crept in, and I
shall be very thankful for any corrections anyone may send to
myself or the publishers. I shall also feel very grateful to my
PREFACE. V.
professional confreres for any remarks regarding difficulties or
omissions in the text or examples. Great care has been taken
to render the text clear and explicit, and I am sure my readers
will unite with me in an expression of thanks to the publishers
for the almost faultless style in which they have executed the
typographical portion of the work.
A second part, on Spherical Trigonometry, with numerous
examples and applications to Nautical and Spherical Astronomy
and Geodesy, is in course of preparation, and will appear if
the present work meet with a favorable reception.
J. MORRISON.
High School, Walkbrton,
June lift, m^o.
•r J ...
CONTENTS
CHAPTER I.
PAOB
Definitions • • 1
Sexagesimal Division of Angles 3
Centesimal Division of Angles 4
Circular Measure 4
''^ CHAPTER II. •,,.'•' ■ ■:.,-■:..:;•
Definitions of the Trigonometrical Functions 8
General FormulsB 10
Line Definitions of the Trigonometrical Functions 12
Functions of 30", 60% 45% 18% 54°, &o 18
Funtions of half an Angle in terms of the whole Angle 21
Examples...... 24
...... .•"'7'X;i!, 'Ai:: ;■' ;.i; .■.:'.;•.,.■,, ;;,;„ii;,v,
CHAPTER III. -r
Solution of Right-angled Triangles 28
Heights and Distances 32
CHAPTER rV. V ...
Extension of the Definitions of the Trigonometrical Functions,
use of the signs + and — 37
Variations in Sign and Magnitude of the Trigonometrical
Functions 38
Line Definitions 43
Functions of 180° -A 45
Functions of 90° + ^ 46
Functions of Negative Angles 47
Groups of Angles having the same sine, dco. 48
Examples 50
viii. CONTENTS.
CHAPTKll V.
FAOB
Functions of th« sum and difference of two Angles 54
Functions of Coniiiound Angles CO
General Fonuulit; 61
Functions of the Multiple and Subniultiple of an Anglo 64
Functions of mA 66
Functionsof (vl+i?+C), ( -A+B+C), &c 67
General FormuLe 69
Forniuhe of Verification 71
Examples 72
CHAPTER VI. ' " .
Circulax Measure of an Angle > sine, but < tangent 80
- - sin ^ , tan
Limiting values of and ~ 81
^ 6
03
Smd>e-- ..,........: 82
4
Calculation of the Numerical Values of the Trigonometrical
:•: J Functions 82
Trigonometrical Tables 86
Increments of the Trigonometrical Functions corresponding to
a given Increment of the Angle 89
Examples 92
Dip of the Horizon 94
Examples 96
CHAPTER VII.
Properties and Usea of Logarithms . » c 99
Modulus of a System of Logarithms 103
Relation between the Bases of two Systems 104
Common Logarithms 105
Determination of the Charactoristic 1 06
Logarithmic Tables 108
Arithmetical operations 112
Arithmetical Complement of a Logarithm 114
Logarithms of the Trigonometrical Functions 115
Logarithmic Functions of Angles near the limits of the Quadrant 120
Examples 123
CONTKNTS. ix.
CHAPTER VIII.
PAGE
Formuloe for solving Right-angled Triangles 127
FormuliB for solving Oblique-angled Triangles 129
Area of a Triangle 139
Perpendicular of a Triangle 140
Exaniplea 140
CHAPTER IX.
Solution of Right-angled Triangles 151
Solution of Oblique-angled Triangles 154
Examples 109
CHAPTER X.
Application to Surveying, Navigation and Astronomy 176
Mariner's Compass 184
Length of a Degree of Longitude 185
Moon's Distance from the Earth 186
Parallax 187
Examples 190
'' CHAPTER XI. •■ _;•••':" -'.^ n!.f
i\ .. ''i ■
Radius of the Inscribed • "' cle of a Triangle 201
Radii of the Escribed Circxorf 203
Radius of the Circumscribed Circle 207
Distance between the Centres of the Inscribed and Circum-
scribed Cir'os 209
Distance between the Centres of the Escribed and Circum-
scribed Circles 210
Perimeter and Area of Inscribed and Circumscribed Polygons. 212
Circumference and Area of a Circle 213
Area and Angles of the Inscribed Quadrilateral 214
Examples 215
CHAPTER XII. -
Inverse Tiigonometrical Functions .*; . V 225
Examples 227
X, CONTENTS.
CHAPTER XIII.
PAOH
Division of Angles 231
Solution of Equations 236
Auxiliary Angles 238
Examples 241
Quadratic Equations 242
Cubic Equations 244
Eliminati jn of Trigonometrical Functions 246
Examples 248
CHAPTER XIV.
The Exponential Theorem . . 250
The Logarithmic Series 252
The Napierian Base 255
Calculation of Napierian Logarithms 257
Calculation of Common Logarithms 258
Jheory of Proportional Parts 258
Examples 262
CHAPTER XV.
De Moivre's Theorem , 263
Trigonometrical Expansions 266
The sine and cosine of an Angle expressed in terms of the
Circular Measure 268
Expansion of the Integral Powers of the cosine 269
Expansion of the Integral Powers of the sine 270
Expansion of cos tid in terms of Descending Powers of cos ft . . 272
Expansion of cos n6 in terms of Ascending Powers of cos 6 . . 274
Expansion of sin x and cos x in terms of the Circular Measure,
independently of De Moivre's Theorem 275
Sines and tangents of Small Angles 277
Examples 278
CHAPTER XVI.
Exponential Values of the sin, cos and tan 280
Circular Measure expressed in terms of the tangent — Gregory's
Series , . . 281
CONTENTS. Xi.
PAOB
Euler's Series for computing tt 282
Machin's Series for computing tt , . . 283
Solution of Equations by Series 284
Summation of Trigonometrical Series 290
Examples 295
CHAPTER XVII.
1
Resolution of x^n - 2 cos . ac« +1=0 into Simple and Quadratic
Factors 299
Resolution of cc" - 1=0, n being odd 302
Resolution of x" - 1=0, n being even 303
Resolution of x'^ + 1=0, 7i being odd . . . . 304
Resolution of x^ +1=0, n being even 304
Resolution of sin x and cos x into Quadratic Factors 307
Computation of Logarithmic sine and cosine 308
I Miscellaneous Examples qiq
[
[ Appendix ^ ; ; ;; ggg
Numbers often used in Calculations 332
^. a-
•^■■ryem;
THE GREEK ALPHABET.
The Greek Alphabet is here inserted to aid those unac-
({uainced with Greek in reading the parts of the text in which
its letters are used.
A a Alpha.
B /3 Beta,
r y Gamma.
A 8 Delta.
E € Epsilon.
Z ^ Zeta.
Rr) Eta.
e Theta.
I t loto.
K K Kappa
A A Lambda.
Mfji Mu.
NV Nil. :
H I Xi. ■
O o Omicron.
Hit Pi.
P p Rho.
S cr Sigma,
T T Tau.
Y V Upsilon.
$ <^ Phi.
X X Chi.
* i/^ Psi,
Q 0) Omega.
u.
PLANE TRIGONOMETRY.
CHAPTER I.
ON THE MEASUREMENT OF ANGLES. '
Article i. — Trigonometry (from rpLywvoV) a triangle, and
fieTpiio, I measure) is that branch of Mathematics which treats
of the methods of determining the unknown parts of a triangle
when a sufficient number of them is given ; but in its more
extended signification, it embi'aces the investigation of the
various relations of angles in general, such investigations being
carried on by means of certain quantities called trigonometrical
ratios or functions. It is divided into two branches, — Plane
Trigonometry and Spherical Trigonometry ^ — the former treat-
ing of angles and triangles drawn on a plane, and the latter of
those on the surface of a spliere*
2. — Definition of an Angle.
In Plane Geometry an angle is the inclination of two straight
lines to each other, and must therefore be always less than two
right angles; bu^- in Trigonometry an angle may be of any
magnitude whatever, and is conceived to be described by a
straight line revolving about a given
point from one position to another.
Thus, the lines AC, JSC, would, in
Geometry, bound only one angle
ACB; but in accordance with the
more extended definition of an angle, B^
they may also be regarded as containing an angle subtended by
the arc DE, and described by a line revolving about O^ In the
PLANE TllIGCNOMETRY.
direction of the arrow-head from the position AO to the posi-
tion CB. Again, in order to effect this, tlie revolving line may
be supposed to make any number of revolutions. Thus, the
minute hand of a watch, at half-past one o'clock, will, since
twelve o'clock, have described an angle whose magnitude is six
right angles.
3. Let AC and BD be drawn at right angles to each other,
then the whole angular space about is divided into four right
angles, AOB, BOO, COD, BOA,
which are respectively called the
first, second, third and fourth quad-
rants. Let OPi, OP^, OP3 and OP4
be different positions of the revolv-
ing line OP, by the revolution of
which about from its primitive
position AO, in the direction from
A to B, the various angles at are
described. The extremity of the revolving line will evidently
describe the circumference of a circle. When the revolving
line coincides with AO, it is said to make with it an angle zero ;
when it reaches the position OP^, it will have described an angle
AOP;, which is called an angle in the first quadrant ; when it
coincides with BO, it will have described the right angle AOB;
when it takes the position OPj, it will have described the angle
AOP2, greater than one right angle, but less than two, which is
called an angle in the second quadrant, ; when it coincides with
CO, it will have described the trigotioinetrical angle AOG, or
two tight angles. In like manner, when it takes the position
OP3, it will have described the trigonometrical angle AOP^, sub-
tended by the arc ABCP3, which is an angle in the third quad-
rant^ when it coincides with OP^, it will have described the
trigonometrical angle AOP^, subtended by the arc ABC DP i^
DEFINITIONS. S
which is an angle in ih^ fourth quadrant, and when it coincides
with AO, it will have described four right angles.
If the revolution be still continued until the position OP^ is
again reached, an angle greater +han four and less than five right
angles will be described, and 33 on. The angles described by
the progressive revolution of OP in the direction from A to B,
are regarded as +, or positive; while those described by OP
revolving in the opposite direction from A to D, are regarded
as - , or negative, the signs + and - here indicating merely
contrariety of direction. Positive and negative angles are
always understood in this sense, unless the contrary be stated.
4. — Sexagesimal Division of the Right Angle.
A degree is the ninetieth part of a right angle ; a minute is
the sixtieth part of a degree ; and a second the sixtieth part of
a minute, and so on according to a sexagesimal subdivision, but
in practice all subdivisions beyond seconds are expressed as
decimal parts of a second. The characters used to denote
degrees, minutes and seconds are °, ', " ; thus 17° 5' 24". 4, repre-
sent seventeen degrees, five minutes and twenty-four and four-
tenths seconds.
The comjylement of an angle containing A degrees is the
remainder obtained by subtracting it from 90°, and is written
90° --4°. llius the complement of 40° 10' is 49° 50'. If
the angle is greater than 90°, its complement will be negative ;
thus, the complement of 130° is 90° - 130° = - 40°.
In every right-angled triangle, each of the acute angles is the
complement of the other. ■ '
The supplement of an angle containing A degrees is the
remainder obtained by subtracting it from 180°, and is written
180°-^°; thus the supplement of 70° is 180° -70 = 110"; and
of 204°, - 24".
Since the sum of the angles of a triangle is equal to two
4 PLANE TRIGONOMETRY.
right angles, each of the angles is the supplement of the sum
of the other two.
5. — Centesimal Division of the Right Angle.
By some of the Frenci: mathematicians the right angle was
divided into 100 equal angles, called grades; each grade into 100
minutes, each minute into 100 seconds, and so on according to a
centesimal subdivision.
K D denote the number of degrees in an angle, and G the
number of grades in the same, then, since 90 degrees = 100 grades,
we have
P : Gf : : 90 : 100
; or, a = \^ D, and D = ^qG. (1)
The centesimal division is now abandoned oven in Franco, as its
general adoption would involve a change in our tables, and in the
graduation of astronomical and other instruments.
6. From Article 4 it will be seen that the unit of angular
measure there adopted, viz : 1°, cannot be directly compared
with any lineal unit, such as an inch, a foot, &c., since they
are magnitudes of different kinds. In the following articles it
will be shown how a lineal unit may be applied to determine
the magnitude of an angle.
7. — Circular Measure.
The circumference of a circle
varies directly as its radius.
,_. In the two concentric circles in-
scribe regular polygons ABC... abc...
having the same number of sides ;
then from the similarity of the tri-
angles ABO, ahO, we have
; ?: ah Oa :^
CIRCULAR MEASURE. 5
and since the polygons are equilateral,
perimeter of the polygon ABC ... _0A
perimeter of the polygon abc... Oa
If the number of the sides of the polygons bo increased, their
perimeters will approach more nearly to those of the circles in
which they are inscri})cd. Now, let the number of the sides be
inoieased indefinitely, then the perimeters of the polygons will
ultimately coincide with the perimeters of the circles, and, since
the above proportion will still hold true, we shall have
circumference of the circle ABG ... _ radius OA
circumference of the circle abc... radius Oa
Therefore the circumference of a circle varies directly as its
radius, and therefore, also, as the diameter; hence it follows
1.U i xi- i.' circumference , i • • • 1 1 i
that the ratio, ; , has a certain invariable value.
diameter
It will be shewn hereafter that this ratio is, to five places of
decimals, the number 3-14159 which is usually denoted by the
Greek letter tt.
If, then, r denote the radius of a circle, its circumference
will be 27rr.
8. In the circle BCD, take any
angle BAG = A°: let CB the arc which
subtends it = a, and the radius AB = r.
The semi-circumference BCD will be
Trt' which subtends the angle 180°;
p.n J since " the angles at the centre of
a circle are proportional to the arcs
which subtend them " {Euc. VI. 33), we have
180° : ^° : : arc ^(7i) : arc ^(7
9 PLANE TRIGONOMETRY.
Whence
A"
—
180° a
IT r
r-
1=
180'
3-14159 &c.
a
X —
r
= 57''-2957795... x.^. (2)
r
Therefore if a and r be given in lineal measure, the number
of degrees in the corresponding angle A can be obtained.
The invariable angle 57°*2957795... is usually denoted by
w", therefon (2) may be written
^" = a)°x-. . (3)
r
■ ' • ■
9. If a = r, then ^° = a)°, that is, the angle which in any
circle is subtended by an arc equal to the radius, contains a)°
or 57'''2957795... , which is called the unit of circular measure,
and the fraction is called the circular measure of the
radius
angle.
From (3) we have
0)
r = — a (4)
A' ^ '
A"
and a = —j- r. (5)
w
arc
10. If the circular measure be denoted by Oy we have
radius
ii° = 57*'-2957795... x^^ : - u,
^'=3437'-74677... x^V (6)
X" = 206264"-806... x^) •' -
If r denote the radius of a circle, its circumference is 27rr,
EXAMPLES. (
and iherefore the circular measure of four right angles is _.
or 277.
Hence it follows that the circular measure of two right angles
is TT, and of one right angle --.
Examples.
1. Find the number of degrees in an angle which ia sub-
tended by an arc of 5 feet, the radius being 2 yards. ..
. Ans. 47M4'47"'34 "
2. Find the circular measure of 42°.
Ans. -73303
3. How many degrees in an angle whose circular measure
^^^^' . ' Ans. 30° 0' 43''-45.
4. Find the radius when an arc of 1° measures 10 feet.
Ans. 572-957795 feet.
5. The arc of a railway curve is 250 yards, and subtends an
angle of 6J° at the centre. Find the radius.
Ans. 1-302 miles.
6. How many degrees are in an angle subtended by an arc
of one inch to a radius of one foot ?
Ans. 4" 46' 29".
7. What angle will an arc of one inch subtend to a radius
of one mile 1
Ans. 3"-254.
1
[8]
CHAPTER II.
TRIGONOMETRICAL RA.TIOS — FUNDAMENTAL PORMULiE.
II. Let QAP be any angle, and in AP take any points B,
Bif J?2, <fec., and draw BC,
£iCi, <fec., perpendicular to ~
AQ, then the right-angled tri-
angles ABC, AB,C,, AB,G„
are similar, and we have by
Geometry,
BC :AB :: B,C^ -.AB,:'. B^C, : AB^
BG B,C\ B,C, ^ \
or
and also
AB
BG
AG
AB
AB,
AG,
AB,
AB,
Ml
AG,'
AB,
AG AGi AG,
These ratios, then, depend entirely on the magnitude of the
angle, and not at all on the absolute lengths of the sides contain-
ing it. Both they and their reciprocals are, therefore, indices or
functions of the angle^ and have received special names as follows :
Let QAP be any angle
in the first quadrant; in
AP take any point B and
draw BG perpendicular to
AQ. Represent the angles
of the right-angled triangle
ABG by Ay B and (7, and the sides opposite them by the small
^ THE TRIGONOMETRICAL RATIOS. 9
letters a, h and c respectively : then whatever may be the absolute
lengths of the sides :
(1) The sine of the angle A is the quotient of the opposite
side by the hypothenuse.
Thus, sin i =— , and similarly sin j5 = —
c c
(2) The tangent of the angle A is the quotient of the oppo-
site side by the adjacent side.
6
Thus, tan A = -, and similarly tan ^ = _.
b a
(3) The secant of the angle A is the quotient of the hypothe-
nuse by the adjacent side.
Thus, sec A = —, and similarly sec ^ = — . '
h a
(4) The cosine of the angle A is the quotient of the adjacent
side by the hypothenuse.
h ft
Thus, cos -4 = — , and similarly cos B — —.
c ■ 0- ; ;
(5) The cotangent of the angle A is the quotient of the
adjacent side by the opposite side. .;
Thus, cot A = —, and similarly cot B= —.
a b
(6) The cosecant of the angle A is the quotient of the hypothe-
nuse by the opposite side. - . } v .; ' i-.
Thus, cosec A= ~, and similarly cosec B = —.
a b
12. Since the angle B is the complement of the angle A, Art.
4, we observe from the above six definitions of the Trigonome-
trical ratios, that the cosine, cotangent and cosecant of an angle
are respectively the sine, tangent and secant of its complement.
In fact it is from this circumstance that the terms cosine,
cotangent and cosecant derive their names, being respectively
10
PLANE TRIGONOMETRY.
abbreviations for " sine of the complement," " tangent of the
complement," and "secant of the complement."
Hence wo have from the last article :
sin A = cos /? = — .
tan A = cot B = —,
b
sec A = cosec B — —.
•coa A — sin /? = — .
cot A = tan />' = _.
a
cosec il = sec i5 = — .
a
<i)
From these equations it is seen that the sine and cosecant of
the same angle are reciprocals ; so also are the cosine and secant,
the tangent and cotangent, reciprocals. That is,
\
sin A=^
cos A =
cosec xi'
1
I
tan A =
sec A'
1
cot A'
cosec A
sin
A
sec A
1
cos
A
cot A
1
tan A
(8)
or thus.
sin A cosec A =cos A sec ^ = tan A cot ^ = 1.
13. From the figure of Art. 11, we have (Eiic. I. 47.)
(9)
or
or, by the definition of the sine and cosine (7) . ■ . \ .
sin^ ^ + cos2 ^ = 1. . (10)
where sin'^ A signifies " the square of the sine of A" &c.
FUNDAMENTAL FORMULAE. 11
Henco wo find sin A= J\ - coh'' A
J{i+coHA){i-cofiA). (11)
and also cos A ^^ Jl - Hin'^ A
= 7(l+sinyl)(l-8in.l). (12)
14. From {7', vvc have
. a
tan A — • 7-
a
sin A
(13)
COS A
' '1
And since the cotangent is the reciprocal of the tangent,
(1*)
or
cot A =
cos A
sin 2
15-
The
figure
of Art.
11
c2
gives
b'
' and therefore by the definition of the secant and tangent (7)
sec* ^ = 1 + tan^ A. (15)
- ft
16. Again, dividing by c^ gives
52
- = -l+l>
a^ a^
o» cosec- ^ = cot^ -4 + 1. (16)
5 r
12
PLANE TRIGONOMETRY.
Line Definitions.
17. The Trigonometrical ratios defined in Art. 11 may be
represented geometrically by means of a circle, as follows :
Let DHr be a circle whose radius = 1, and let DAB be any
angle in the first quadrant DAII.
From B, the extremity of the
radius AB^ draw BG perpen-
dicular to AD, and from D draw
DE perpendicular to AD, to meet
AB produced in E, Also draw
BG and HF perpendicular to
AH, the latter meeting AB pro-
duced in F. Representing the angle DAB by ^, we have
sin A
BG
AB
BG
= BG.
(1) That is, to a radius of unity, the sine of an angle is the
perpendicular drawn from one extremity of the arc subtending
it, to the diameter passing through the other extremity.
cos A ==^:=^=. AG = GB,
AB 1
(2) or, the cosine of an angle is the distance from the centre
to the foot of the sine, or the sine of the complement.
, ED ED „^
(3) or, the tangent of an angle is the line touching one ex-
tremity of the arc subtending it, and terminated by the diameter
produced through the other extremity.
THE LINE DEFINITIONS. 13
cot A = t2in BAH, (Art. 12)
~AH 1
(4) or, the cotangent of an angle is the tangent of the com-
plement.
(5) or, the secant of an angle is the produced radius drawn
through one extremity of the arc subtending it, and terminated
by the tangent drawn from the other extremity.
cosec ^ = sec j5il//, (Art. 12)
AF AF
. ■■■ -AH=-T==^^^ ' '- ■:-^:
(6) or, the cosecant of an angle is the secant of the com-
plement.
l8. From the similar triangles ABG, AFH, we have
AG:AH::AB::AF,
or, since AG^BC^sinA, and AF= cosec i,
sin ^ : 1 : : 1 : cosec A.
^■"■'- .■■■/■-■''■I' "■' . '■'.-" .
Hence :, .^ sm^^^^^^^.
From the similar triangles ABC, AED, we have
'^ ':\^f'''''"' AC.AD-.'.AB.AE, ,"' '
or, cos il : 1 : : 1 : sec -4
H«ncp cos ^
sec A
The angle AFH=i\iQ angle BAD, therefore the triangles
ADE, IHF, are similar, and we have
14 PLANE TRIGONOMETRY.
DE:AH::AD: IIF,
or, tan ^ : 1 : : 1 : cot A.
Henco tan A =
cot -4
The results of this article agree with (8).
19. From the similar triangles ABC, AED, we hav«
AC '.AD .'.BG '.DE,
or, cos A : \ : : sin -4 : tan A.
• A
Hence tan A = -, which agrees with (13).
cos^ ^ ^ ^
Again, from the similar triangles AGB^ AIIFj
cos A
cot A
sin A
From the right-angled triangles ABG^ AED, AHF, we
obtain, by Euc. I. 4:7,
sin^ -4 + cos^ -4 = 1
sec^ ^ = 1 + tan^ A '
cosec^ A = l + cot^ A.
20. Besides the ratios already defined, GD, the portion
of the diameter intercepted between the foot of the sine and
the extremity of the arc, is called the versed-sine, abbreviated
"versin;" and HG the versed-sine of the complement is the
coversed-sine (coversin).
The versin and coversin can have no existence according to
"ratio definitions" given in Art. 11, since the ratios there
defined are independent of the radius ; nevertheless the terms
yersin and coversin are used as convenient abbreviations for
1 - cosine and 1 - sine, respectively.
EXAMPLES. 15
21. By means of the formuloe (8) (16) any ratio of an angle
may be expressed in terms of any other ; or if any ratio is given,
all the others may be found.
^o;.— Express all the trigonometrical ratios in terms of the
tangent. , .
; 1 1
sec ^ = ± J{1 + tanM) cos A = ^^^ - ±-j^i+ tan^ a)'
tan A
sin A=±J{1- cos^ ^) = ± -jQ-^^A)
, 1 ^ ^(l+tan^^)
cosecJ. = -: — 7=—' — +r^"l '
sm A tan A
cot A
tan A
The significance of the double sign will be explained here-
after.
'...,- Examples. '' ;■■:^■ '.•■■• ^'^^ ,*■-;
1. Given tan A = -r^j find the sine, cosine and secant.
,5 ,12 .13
Am. sm A = -rj, cos A = -y^, sec A = -j^-
2
2. If tan A = ^> find the vcrsin and cosec.
5
^Tis. (1±— )and ij29.
3. Given sec ^ = 4, find sin ^ and cot 6.
•4* / 1 Fi" T
^ris. sin ^ = — -. , cot ^ = ±
4 ' ^15
4. Given cot 6 = 2, find sec and cosec $,
Ans. sec ^= ±~-, cosec 6=± ^.
16 PLANE TllIGONOMETRY.
5. Given versin = —, find tan 0. ■
o
Atis. tan^=±-^-.
6. If cos 6 = tan ^, find sin 6.
Ans. sin 6 = —
7. If tan ^ + 3 cot ^ = 4, find cos e.
Alls, cos y = ± -—• or ±
2 VlO
8. Prove cot^ B - cos^ ^ = cot^ Q cos^ 0.
cot - sec 1 ^ , . ^
9. If —;, = -r-r-> find sm B.
cot w 16
3
Ans. sin ^= ± — .
5
, - _, sec ^ cot ^ - cosec B tan B ^ -
10. Prove - - - . — ;, = sec Q cosec B.
cos ^ - sm
11. If tan ^ + cot ^ = 2, find the value of sin B + cos B.
12. Determine sin a from the equation
;. 9 sin^ a - 4 tan^ a = — . '
2
^ws. -iin a = ± — or ± ——.
o 2i
18. If sin 03 = m sin 2/> a-nd tan x = n tan y, shew that
m^- 1
cos' 03 »: -
?l2-l
1 i T» .1 . (cosec 03 + sec oj)'^ ,. ,„
14. Prove that — , ~ = (sm x + cos a;)'.
sec^ 03 + cosec'* x ^ '
EXAMPLES. 17
15. If 2 tan a = COS a, find ain a.
Ans. sin o= J^— 1.
16. If sin^-cose=-^r^, findtan^.
Ans. tan 6^= J^.
17. Given sec ^- tan 6=—, find sin Oi
A'ns. sin 6=—.
5
22. To radius unity, it will be seen that the resirlta of
Articles 18 and 19, deduced from the "line definitions," agree
exactly with those previously obtained from the " ratio defini-
tions." If, however, the radius is any other number than unity,
the sine, tangent, &c., of any angle have not a fixed value as
they have in the^ "ratio definitions" or in the "line defini-
tions" when the radius is unity, but vary with the radiuS
employed.
Thus, if AB in the figure of Art. 17 be represented by r,
then JBO, ED, AE, &c., become the sine, tangent, secant, <fec.,
respectively of the angle BAD to the radius *, and therefore
the formula \
sin^ A + CQS^ ^ = 1, becomes, to radius r,
" sin^ il + cos^ il = r^.
Also, tan^ = r, becomes, to i^adius r,
COS il ' '
. . r sin^
tan A = 7".
• ' COS A
' ■■ s '
and shiil:= :, becomes, "to radius r,
cosec A
sin il =
cosec u4'
and so on for all the other formulae of Articles 18 and 19,
8
18
PLANE TRIGONOMETRY.
This inconvenience, arising from the introduction of the
radius, has led to the almost general adoption of the "ratio
definitions," by which the trigonometrical ratios (or functions,
as they are also called), become abstract numerical quantities,
dependent only on the magnitude of the angle, and altogether
independent of the radius.
The " line definitions " to radius unity ^ however, give rise
to no inconvenience, and may be employed with advantage in
the solution of problems, and especially in illustrating geometri-
cally the changes in magnitude of the difierent ratios through
the four quadrants. ' '
23. To find the sine, cosine, <&c., of
30° and 60°.
Let ABCj be an equilateral triangle,
each of whose angles is therefore 60°.
Draw AB perpendicular to J5C, then
the angle BAB = 30°, and ^5 = 2 BB. g AsQ'
1>
/ sin 30°-
BB
Tb^Ybb^I'^'''^^^' ^"*^2.
v/5
cos 30° = VI - sin^ 30" - -^ = sin 60°.
tan 30° =
sin 30°
J.
2
cos 30° ^3 ^3
= =cot 60°,
cot 30°=.
sec 30° =
tan 30
1
^3 = tan 60°.
q^o = -7= = cosec 60°.
cos 30 ^3
1
cosec 30° = - "-57r, = 2 = sec 60°.
sm 30
TRIGONOMETRICAL FUNCTIONS OF 18°, 3G°, ETC. 19
24. To find the sine, cosiiie, <Cr., of 45°. -
Since 45° is the complement of 45°, we have ,
sin 45° = cos 45°;
and since by (10) sin^ A + cos^ ^ = 1,
we have
sin2 45° + cos^ 45° = 2 sin^ 45° = 2 cos^ 45° = 1
sin2 45° = cos2 45° = i, .
and
sin 4f° = cos 45'= J^ = -^,
,^- sin 45° ,
tan 45° = r^ = l.
cot 45' =
cos 45°
1
sec 45° = -
coseo 45°
tan 45'
1
cos 45
1
= 1.
/
sin 45
-.= J^'
25. To find the sine, cosine, <hc., of 18°, 36°, 54° cmd 72°.
I Let ABB be an isosceles triangle having each of the angles,
■ ABB, ABB, double of the angle BAB ;
hence, it follows that the angle BAB is
the fifth of two right angles, and therefore
contains 36°; therefore each of the angles
ABB, ABB, contains 72°. Draw AH per-
pendicular to BB, then the angle BAB
contains 18°. ■ -^ ,.,,,,,.. ...
Now, if ^5 be divided in C, such that B-
AB. BG = AG\ then by Euc. IV. 10, ^(7 =
BB = BC, and therefore if CF be drawn perpendicular to -42>,
the angle ^Ci^= 54°.
Since AB. BC = AC\ we have
■(••:■■ ;;^
20 PLANE TRTfJONOMETRY.
AB{AB-AG) = AG\
AC AC
°''' -A^^'Air^'
By solving this quadratic, wo oLtain
AC -1±./^
AC
Since -~ is a positive quantity, the upper sign must be
taken j
therefore,
AB 2
■is fi.
AB
AC ^ J6-1
AB 2 '
i
^ . ^., BE 2 BE AC
Now, sin 18=^ = 2-1:^=2-18
= -^^ = cos72°.
cos 18°= 7(l-sinM8°)
^^('0^^^)= sin 72'.
Again, cos 36 =
AC 2 AG 2 AC
^^ 1 J^ + l ^ V^+1
x/5-1 (n/5-1)(n/5 + 1) 4 •
= sin 54°.
sin 36°= 7(l-cos2 36°) < ' '
'; ' ;, , =J 7(10-2^5) = cos 54°. ;■•,;-
The other trigonometrical functions of these angles can now
be found. The student should verify the following :
1. cot 18°= x/5 J2 + Jq. =3-07768. = tan 72°.
2. sec 72°= ^5 + 1. =3-23607. =coseo 18°.
THE FUNCTIONS OF HALF AN ANGLE.
3. tan 54°
4. cosec 36'
5. tan 36°
6. tan 18°
v/(2 + ^ JW).
= 1-37638.
= 1-70130.
= -72654.
=^ -32492.
= cot 36°.
= sec 54°.
= cot 54°.
= cot 72°.
26. To express the functions of half an angle in terms of
those of the whole angle, and vice versa.
Let BAC be any angle in the first quadrant, represent it
by A, and in AG take any point
P, and with yl as a centre and AP
as radius, describe a circle ; pro-
duce BA to meet the circumfer-
ence in D, join DP and draw PE
perpendicular to AB,
By Euc. III. 20, the angle
BDP is half of the angle A ;
then
tan - = tan i?i>P
2i
PE
DE
PE
PE
AD + AE
AP + AE
PE
~AP
71
l^lp
sin A
1 + cos A
(17)
Or thus, by the "line definitions." Take JP as unity and
with Z) as a centre and ^i> as radius, describe the arc AH and
22 PLANE TRIGONOMETRY.
draw AF pcrpciiclicular to AB^ thon ^^=tan — -, FE — ^xn A,
and yl^=cos A. From the similar triangles BAF^ DEP^ we
have
DE'.AD-.'.EP: AF,
or 1 + cos -4 : 1 : : sin A : tan — ,
Z
whence tan — = :; r, as before.
2 1 + cos A
27. The formula just proved enables us to find the trigono-
metrical functions of IS'' and 75°, from having given those
of 30".
. Thus, let A = 30", then we have
tan 15" =
1_
sin 30° 2
1 + cos 30" Jo
= -— i-p = 2- ^3 = cot 75".
2+ n/3
Hence, by Art. 21, we easily find
cot 15° = 2 + J3 = tan 75°.
sec 15"= J2{ J^- 1) =, cosec 75°.
cosec 15" = J2{ 73 + 1) = sec 75°.
sin 15" = -^^5^^ = cos 75°.
2 V2
cos 15" = 4^ = sin 75°.
28. Squaring (17) we have
THE FUNCTIONS OF H^F AN ANGLE. 23
tan-
A
2
sin' A
~ (1 + cos A )'^
1 - cos" A
~{l + cosAy
1 - cos A
~ 1 + cos A'
Bee'^
A
2
1 - cos -4
1 + cos -4
2
1 + cos A
1 + cos ^
= 1 -
1 - cos A
2
From (18) we easily find
(18)
.Al + cosA ,- -X
therefore cos^ — = x •. (ly;
sin'^ -^ = 1 - cos' -T
(20)
l-tan4
cos^= " ■ (21)
1+tan-—
. » ' . ^^ VI' / ! I ■ • ■• ■ ; , ■ ' ' , •■
Multiplying (19) by (20), and extracting the square root,
we find J . . .. ,.
sin -4 = 2 sin -cos —. . ,; (22)
2'^- 2
Subtracting (20) from (19) we have
A A .
cos ,4 = cos' — - sin'—. (23)
2 2 ^ *
24) PLANE TRIGONOMETRY.'
If wo write 2 A for A, \^iich we are evidently at liberty to
do, then (22) and (23) become
siu 2A = 2 bin A cos A. (24)
and cob 2A = cos'^ A - sim'^ A
= 1 - 2 sin"'* ^
= 2cob^^l-l. (25)
Also, tan 2A ^ ^ ,-
cos 2 A
2 sin A cos A
y^
^
cos'^ A - sin'^ A
2 sin A
cos A
siii^ A
cos'^ A
2 tan A
1-tanM*
(26)
29. The formiilsB of the last three Articles will be found
extremely useful hereafter, and the student is requested to pay
particular attention to them. They will be again deduced by a
more general process iii a subsequent chapter.
Examples.
• 1. Find A from the equation tan'^ ^ - 4 tan A + 1 = 0.
Adding 3 to both members we have
tan" J: - 4 tan ^ + 4 = 3
hence tan A -2= ± J^
tan ^ = 2+^ or 2-^
= tan 75° or tun 15'^
therefore A - 75° or 15°.
EXAMPLES. 25
2. Given cosec ^ - sin d =^ ^""q » ^^^ ^^^ ^'
By (8) and (17), the given equation becomes
1 . ^ sin
- sin =
am ^ 1 + cos
sin'^ ^
1 + cos ^
1 - sin^
2 /}_
or
1 - cos- ^ ^
COS^ ^ = , ;,- = 1 - cos ^
1 + COS »
COS- ^ + cos ^ = 1
COS^ ^ + COS ^ + i = -r
* 4
cos^ + J=+-^
+ A-1
hence cos =
3. Given tan + cot = 2 sec 0, find ^.
By (8), (13) and (14), the given equation becomes
sin cos 2
-;; +
cos sin cos
sin' 6 + cos- ^ = 2 sin d
or 1 = 2 sin 0,
hence sin ^ = J = sin 30",
therefore ^ = 30'.
0'
4. Prove that cosec 6 = ^ (cot -- + tan — ).
1 cos2-+sin2--
cosec = -r—^ = 7^., ^y (10) and (22)
sm ^ ^ . ' -^ ^ ' ^ '
2 sin- cos— ,
* ■
s=J(cot-+tan--).
26 PLANE TrvIGONOMETRY.
^ _ , ^ 1 - cos 6 sec ^ - 1
0. Prove that tan -- = — : — —- = — ; -r—.
2 sin tan (^
9 6
6. Prove that cot 0=^ (cot — - tan — ).
2, A
7. Given sec x tan x = 2 ,^3, find x.
Ans. 03 = 60°.
8. Given 6 cof^ 03-4 cos^ a;= 1, find x.
Ans. X = 60°.
9. If sin A Hec B= /— , and cos A cosec 5 = ~-, find
'^ ^""^ ^' Ans. A = 60°, B = 45°.
10. If cos (2^+5)=-^, and sin {2>A-B)=]-, find
^^^^^' Ans. ^ = 12°, j5 = 6°.
11. If tan (3i + 2^) = 2 + J%, and tan (5^ - ^) = 2 - ^'3^
find A and 5.
Ans. A = %i^\ 5 = 25xV-
12. Find the tangent of 1\\
Ans. J6- J3+ j2-%
13. Find the tangent of 37^°.
Ans. JQ+ J'S- V2-2.
14. Given 2 cosec ^ - cot ^= ^, find 9.
15. Prove that sin A =
Ans. ^=60°.
1
cot — ■ - cot A
40
10. If tan (9= ^2"- 1, find cos 29.
Ans. cos 29
\/2
EXAMPLES.
27
17 If cos ^ = —^-^, shew that tan -= /^t tan -■
1,. itcos(; ^_^cos</>' 2 n/ a-b 2.
18. Prove that tan ^ - tan - = tan - sec e/.
19. Prove that 1 + tan ^ tan - = sec A.
20. If sec + cosec B^m, and sec 6 - cosec ^ = ti, find tan 6.
m + n
m-n
Atis. tan 6
21. Given sin ^ cos ^ = -—-=, find ^.
Am. ^ = 22|°.
22. In the fig. of Art. 26, join PK, and prove geometrically
that
A 1 - cos A
tan — = — r — J-, •• '^^
2 sm il
sin ^ = 2 sm — COS — ,
^ ^ ' ,, •
1 +COS ^ = 2 cos^— , - '■
A ' • ■■'■■
1 -cos ^ = 2 sin'^ ^.
23. Prove that -■■• ;. . :■ -■;^''i':'n.:--',- , ^,
sin ^ (1 + tan A) + cos ^ (1 + cot A) = cosec -4 + sec A.
24. Prove that
tan ^ + sec ^ - 1
= tan 6 + sec 0,
tan t; - sec (; + i
+.n.Ti fl - sin B
and
tan ^ - sec ^ + 1
tan ^ - sin ^ sec
sin^ ^ 1+cos^'
28 PLANE TltlGONOMETKY.
CHAPTER III.
APPLICATION OF THE PRECEDING PORMULiE.
30. The formula) of Article 12 are directly available for the
solution of right-angled triangles. Thus, if ABC he a. right-
angled triangle, G being the right angle, and a, 6, c the sides
opposite the angles A, B, C respec- ^B
tively, we have according to the
definitions oLthe sine, &c.,
' A ^ A ^ 4. A ^
sm A = —: cos A—-; tan A = —.
c c b A ~- ^
Each of these equations expresses a relation between three
parts — an angle and two sides — hence, when any two of these
parts are given, the other can be found.
In order, however, to solve a triangle trigonometrically, it
will be necessary to know the values of the sine, cosine, &c., of
any given angle. The trigonometrical tables supply these values
for every minute^ and sometimes for every ten seconds, from 0°
to 90°. At present, however, we will use only those trigono-
metrical ratios which we have already computed in the preced-
ing chapter. The construction and use of the tables will be
explained in a subsequent part of this work. ,.
31. The solution of right-angled triangles presents the fol-
lowing cases :
Case I. — Given the hypotlienuso and one ant/le, as c and A.
To find a, we have
sin A = ~'f whence a = c sin A. . (27)
c
SOLUTION OF IIIGHT-ANGLED TEIANGLES. 29
To find J, wo liave . .
if h
I cos^=— , whence 6 = c cos -4. (28)
c
To find B, we have B = 90°- A.
Examjde. — Given c = 24, and the angle A = 60°, to find the
other parts.
By (27) we have
a = 24xsin 60°
= 24x-^ = 12V5.
By (28) we have •
6 = 24cos60° = 24x^ = 12. - ' "
and 5 = 90° -60° = 30°.
If the angle B had been given instead of ^, we might first
find A, thus, A = 90° - B, and then proceed as above.
Case II. — Given the hypothenuse and one aide; c and a.
To find A, we have
sin A=—. (29)
To find B, we have B = 90'' -A.
To find b, we have
cos A = — , whence 6 = c cos A, (30)
or b= Jc'^ -a?= J{c + a) (c- a). (31 ^
Example. — Given c = 10 and a = 5 J% find other parts.
By (29) sin^=^=JL = sin45°,
• -^y V2
hence ^ = 45° and i? = 45°.
By(30) 6 = 10cos45° = 10x-V-5 ^2.
. . \/2
30 PLANE TRIGONOMETRY.
Case III. — Given an angle and its adjacent side ; A and h.
To find a, we have
tan -4 = -r-j whence a = h tan A. (32)
To find c, we have
h h
COS A = —. whence c = = h sec A. (33)
c cos A
Example. — Given -4 = 75° and 6 = 15, find a and c.
By (32) o= 15 tan 75°
= 15(2+ V3) = 55-98.
By (33) * c = 15sec75'
n/3
= 15.— ^-=15(76+ ^) = 57-648.
*./ 51 — 1
Caie IV. — Given.an angle and its opposite side ; A and a.
We may first find B^ which reduces this to Case III., or
^« may proceed thus :
To find 6, we have
tan -4 = — , whence h = : = a cot A. (34)
6 tan A ^ '
To find c, we have
sin ^ = — , whence c = -: — - = a cosec A, (35)
c sin A -^ '
Or, using h just found, we have
, cos A = — , whence c = h sec A. (3G)
C
Case Y. — Given the two sides ; a and b.
To find A and ^, we have
tan ^ = cot 5 = ^. (37)
6 ^
SOLUTION OF RIGHT-ANGLED TRIANGLES.
To find c, we have
31
sin A =i -, whence c = -: — r
= acosec.4. (38)
32. The formula of the last Article may be illustrated
geometrically by the " line definitions," as follows :
With A as a, centre and a
radius unity, describe an arc
DG, draw DU and GF perpen-
dicular to AG ; then, Art. 17,
i>.^ = sin A, AF = cos A, GF^
tan A, and AF = ^qg A. From
the similar right-angled trian- ^-
gles, AED, AC 23, we have
EG
that is,
whence
AD :AB :: DE : BG
1 : c : : sin J[ : a,
a = c sin A, which is (27),
and
a
sin A = —t
c
which is (29).
Again
that is,
whence
and
AD:AB::AE:AG
1 : c : : cos J. : 6, >,
6 = c cos ^, which is (28),
coSil = — , which is (36).
Again, from the similar right-angled triangles, AGF, AGB,
we have
AG:AC :: GF : GB
that is, 1:6 :: tan^ : a, . • .
whence ' ^^ "I a = 6 tan A, which is (32),
and
tan A =
a
which is (37>
82 PLANE TRIGONOJIETRY.
Examples.
1. The radius of a circle is 10 feet ; find the side of the
inscribed regular pentagon.
Ans. 5( J5 - 1) ft.
2. The base of an isosceles triangle is 20 yards, and the
vertical angle is 108° ; find the sides.
Atis. 10(^5-1) yds.
3. The side of a regular octagon is 14 feet ; find the radius
of the inscribed circle.
Ans. 7(^+1) ft.
4. If the earth be a sphere whose diameter is 7912 miles,
find the radius of the 45th parallel of latitude.
Ans. 1978 ^2 miles.
5. The angular elevation of a tower at a place A due south
of it is 45°, and at a place B due west of A, and at a distance
a yards from it, the elevation is 30° ; find the height of the
tower and the distance from £ to the foot of the tower.
Ans. Height =^-^ ^/^ yds. ; distance from B = ^j6 yds.
6. A person observes the angle of elevation of a mountain
to be 27°, and approaching one mile nearer, the elevation is 54°;
find its height and distance from the first station.
Ans. Height = J( ^ + 1) miles.
Distance from first station = J(4 + ^/lO - 2 ^5^) miles.
7. A tower 60 feet high casts a shadow 20 ^^feet in length ;
find the altitude of the sun above the horizon.
Ans. 60°.
5 • ' , EXAMPLES. ' * 38
^' 8. In a triangle ABC, sin A =* —5--, sin B = -— and the
isi
sides opposite to these angles are ^X5 ^^^ ^i > ^^^ the other
side.
9. A church spire subtends an angle of 60° at a certain dis-
tance from its base, and 80 feet farther off it subtends an angle
of 45°; find its height, allowing 5 feet for the observer's height.
Ans. 125 + 40 J^ feet.
10. The angular elevation of a steeple standing on a hori-
zontal plane is observed, and at a point 10 yards nearer to it
the angular elevation is found to be the complement of the
former. On advancing 3 yards nearer, the elevation is double
of the first. Find the height of the steeple, the distance of the
first point from its foot, and the tangent of the angle of eleva-
tion at the third point. '
Atis. Height = 12 yds. ; distance = 18 yds. ; tan of
the angle of elevation at third point = 2 '4.
11. Wishing to know the distance from a given point A,
to an inaccessible point B, on the opposite bank of a river,
a base lijie AG, 84 rods, is
measured along the bank of
the river at right-angles to
ABj and from its extrem-
ity C, equal distances CD, '
CE, 6 rods, are measured ^^
on the lines GB, GA. Finally,
DE is measured and found to be 4 rods. Find the distance
between A and B, and the tangent of the angle AGB.
3
'34,
PLANE TRIGONOMETRY.
Bisect DE in F, join OF; then, in the right-angled tri-
angle CDF,
sin DCF^
DF_2^_\_.
CD~ Q~V
henco
2 /o~
COB BCF^ Jl-i^ 1±»
tan J)CF=
sin DCF _ 1 3 V?
cosi>Ci^~3'2^2"*^ *
The angle DGB is double of the angle DCF, therefore,
by (26),
2 tan DCF 2 4 ^5
taniC5 =
l-tan2 2>(7i^
-^
Hence, by (32), ii5 = ^(7 tan AGB
= 84xi^=48^rods.
12. If, in the figure of the la^t problem, CB = 25 yards,
and CD = 0F=5 yards, and DE=i yards, find AB and the
tangent of i5.
— 17 —
Atis. AB = 4: 721 yards ; tan 5 = -- ^21.
13. From the summit of a tower whose height is 108 feet,
the angles of depression of the top and bottom of another tower
standing on the same horizontal plane, are observed to be 30°
and 60° respectively ; find the height of the second tower, and
the distance between their summits.
Ans. Height = 72 ft. ; distance between summits = 72 ft.
14. One side of a right-angled triangle is 60 rods, and the
EXAMPLES. * V ^^ 35
cosine of the adjacent angle equals the cotangent of the opposite
angle ; find the hypothenuse and the other side. . .
Ans. Hypothenuse = 30 J2{ Jl+^ I); side = 30 j'^ij^^i).
15. The base of a right-angled triangle is 60 yards, and the
tangent of the opposite angle equals three times the cosine of
the adjacent angle j find the perpendicular.
Am. 15 ,^2 yds.
16. In a right-angled triangle whose right angle is C,
shew by (17) that tan — = ^— — = ; '
•^ "^ ^ 2 b +c a
A Ic-h " ' 'v
and by (20) that sin — = / -^-« , , ,
17. In any right-angled triangle, given A and c, shew that
the perpendicular from the right-angle on the hypothenuse is
--sm 2A, .,. ., „:;, _ . ,: .-, ,, , ,, , j
18. The sides of a right-angled triangle are 3, 4 and 5, find
the length of the perpendicular from the right angle on the
hypothenuse.
'"' ■' '■'• ;■-'■-,' '::'- ''-"■'■ ' ' ' Ans. 2.4.
19. A person in a line svith two towers, and at distances of
100 and 150 yards from them, observes that their apparent
altitudes are the same. He then walks towards them a distance
d of j^O yards, and finds that the angle of elevation of the nearer
is just double of that of the former. Find the heights of the
towers. ^*:.vJ>^.-;.;
20 _
Ana. The nearer, -— ^7 yds.; the other, 10 7^ yds.
• 20. A tower of unknown height stands on a horizontal
plane, and at a distance ot 80 yards from the foot of the tower
36 . PLANE TRIGONOMETRY.
•
a mark which is known to be 50 feet high, has an angular
elevation just half of that of the tower. Find the height of
tower, the observer's eye being on a level with the plane.
Ans. 104 '537 feet.
21. The sides of a right-angled triangle are in arithmetical
progression, and the area is 108 ; find the sides.
Ans. 9, 12 and 15.
22. Standing straight in front of a house, opposite one
corner, I find that its length subtends an angle whose sine is
2 3
— J'5, while its height subtends an angle whose tangent is -^i
the height of the house is 51 feet ; find its length.
Am. 170 ft.
23. When the altitude of the sun is 54°, the shadow cast
by a church spire is 150 feet. Find the height of the spire.
Ans. 68.819 yds.
24. The base of a right-angled triangle is one-third of the
hypothenuse, and the perpendicular from the right angle on the
hypothenuse is 6 ^2^; find the sides.
Am. 9, 18 ^2" and 27.
25. A May-pole being broken off by the wind, its top struck
the ground at an angle of 36°, and at a distance of 30 feet from
the foot of the pole ; find its whole height.
Am. 58-878 ft.
EXTENSION OF DEFINITIONS.
37
CHAPTER IV. ^
EXTENSION OP THE DEFINITIONS OF THE TRIGONOMETRICAL
FUNCTIONS — CHANGES IN THEIR SIGN AND MAGNITUDE.
33. In chapter II., the sine, tangent, &c., have been defined
for angles in the first quadrant only, or for those less than
90°. We will now extend these definitions so as to include
all angles.
Draw the lines PQ, MN,
intersecting at right angles
in A ; let AB \i<& different
positions of the revolving
line in the first, second, third
and fourth quadrants, and
from any point oi AB as B
draw BG perpendicular to
PQ. In the first and second
quadrants BC lies above, and
lyr
K
B
B
C C
-P
V
in the third and fourth quadrants, below PQ ; while in the first
and fourth quadrants, AG lies to the right, and in second and
third quadrants, to the left of Jfif. These opposite directions are
indicated by the signs + and - ; thus, lines which lie above PQ
are regarded as positive, consequently those which lie below
it must be regarded as negative. Again, lines which lie to the
right of MJ^ are regarded as positive, consequently those which
lie to the left must be regarded as negative. Hence BG is posi-
tive in the first and second qiiadrants, and negative in the third
38 PLANE TRIOONOMKTRY.
and fourth, and AC in ponltive in tho first and fourth quadrants,
and negative in the second and tliird.
Tho revolving line AB, in always regarded as positive, being
measured olF in tho direction from A towards B; if, however,
AB be produced backwards, tho produced part must be con-
sidered iieyativc. The student will not fail to perceive tliat tho
signs + and - , as thus employed, indicate, as in Art. 3, merely
contrariety of direction. So it is in geography and astronomy —
if north latitudes and east longitudes be considered positive,
south latitudes and west longitudes must be considered nega-
tive. The student must be careful, however, to observe that
no absolutely fixed direction is here meant. Any line what-
ever may be regarded as the initial one, ■ "'e object of the conven-
tion being merely to indicate that lines measured in opposite
directions are affected with ojjposite signs. Thus, in the
obtuse -angled triangle
ABC, the obtuse angle
C is measured from BC
as the initial line, and
is an angle in the second
quadrant ; while the angle B is measured from AB as the initial
line, and being less than a right angle, is an angle in the first
quadrant.
34. To trace the Variations in Sign and Magni-
tude of the sine, cosine and tangent of an Angle,
as it varies from 0° to 360°.
First Quadrant. — Let a, 6, c represent the sides of the
right-angled triangle ABC, opposite to the angles A, B, C
respectively.
VARIATIONS OF TRIOONOMETRICAL FUNCTIONS. 39
gin A = — , which is therefore positive,
c
When ^ = 0^a = 0,
hence sin 0° = 0.
As the angle increases, a in-
creases, and therefore sin A in-
creases in magnitude.
When A = dO%a^c,
hence sin 90° =1.
cos A = , which is therefore positive.
c
cos 0° = 1.
When
hence
As A increases, 6 decreases, and therefore cos A decreases in
magnitude.
When ^ = 90°, 6 = 0,
hence cos 90° = 0. %
+ a
tan A= — 7, which is therefore positive,
+ h
When
hence
^ = 0°, rt = and 6 = c,
tan 0° = 0.
As A increases, a increases and h decreases, and therefore
tan A increases in maijnitude.
When
hence
A = 90°, a = c and & = 0.
tan 90° = X .
Second Quadrant.
sin A =
+ a
which is therefore jjositiv^.
40
PLANE TRIGONOMETHY.
As A increases, a decreases,
and therefore sin A decreases
in magnitude.
When A = 180\a^0,
hence sin 1 80^ = 0.
COS A = , which is thore-
B
•*«:
\
~c^-T
Xi
foro negative.
As A increases, h increases, and therefore cos A increases in
magnitude.
When ^ = 180°, 6 = c,
hence
aos 180° = = -1.
c
tan A
+ a
J, which is therefore negative.
As A increases, a decreases and b increases, therefore tan A
decreases in magnitude
When A = l80%a = and b = c,
hence tan 180° = 0.
Third Quadrant.
sin A , which is tbere-
c
fore negative.
As A increases, a increases,
and therefore sin A increases in
magnitude.
rss
/
V
When
hence
A = 270° (3 right angles), a = c,
- c
sin270^
1.
VARIATIONS OF TRIGONOMETRICAL FUNCTIONS. 41
COS A = , which is therefore negative.
As A increases, b decreases, and therefore cos A decreases in
magnitude.
When
hence
il = 270^6 =0,
cos 270° = 0.
tan A = — - =+-=-, which is therefore positive.
-00 -*
As A increases, a increases and b decreases, and therefore
tan A increases in magnitude
il = 270°, a = cand6 = 0,
tan270':z:oc.
When
hence
Fourth Quadrant.
bin ^ = , which is therefore negative,
c
As A increases, a decreases,
and therefore sin A decreases in
magnitude.
When ^ = 360", « = 0,
hence sin 360" = 0.
a
"^+B (
n
N
\
■ ■
c\
-a
\
B
cos A , which is there-
c
fore positive.
As A increases, b increases, and therefore cos A increases in
magnitude.
When . ;. < ^ = 360°, 6 = c,
hence cos 360° = 1.
42
PLANE TRIGONOMETRY.
-a
tan A = —7-, which is therefore negative.
As A increases, a decreases and h increases, therefore tan 4
decreases in magnitude.
When A = 3G0°, a = and 6 = c,
hence tan 360° = 0.
35. As the cosecant, secant and cotangent are the reciprocals
of the sine, cosine and tangent respectively, they require no
special examination.
The variations of all the trigonometrical functions, both in
sign and magnitude, are given in the following tabular form :
1st Quadrant.
2nd Quadrant.
3rd Quadrant.
4th Quadrant.
sine
0° to 90°
90° to 180°
180° to 270°
270° to 360°
positive.
to 1
positive.
1 to
negative.
to -1
negative.
-1 to
cosine
positive.
1 to
negative.
to -1
negative.
-1 to
positive.
to 1
tangent
positive.
to oc
negative,
oc to
positi\^e.
to cc
negative,
oc to
cosecant . . .
positive,
oc to 1
positive.
1 to oc
negative,
oc to -• 1
negative.
-1 to oc
secant
positive.
1 to oc
negative.
oc to -1
negative.
-1 to oc
positive,
oc to 1
cotangent. .
positive,
oc to
negative.
to oc
positive.
X to
negative.
to oc
36. From the above table it is seen that the values of the
sine and cosine always lie between and ± 1, and those of the
secant and cosecant between + 1 and + oc , and between - 1
and - oc ; while those of the tangent and cotangent lie between
VARIATIONS OF TRIGONOMETRICAL FUNCTIONS. 43
and ± oc , and therefore may be of any magnitude whatever,
positive or negative. It is also seen that each of the functions
changes its sign when its value passes through zero or infinity ;
hence we may write cos 90° = + or - 0, and tan 90° = + a
or - oc , &c.
37. The preceding results of this chapter are easily obtained
by means of the line definitions. In a circle of radius unity,
draw two diameters, PQ and MN, at right angles to each other,
and let PAB be an angle taken in each of the four quadrants.
Draw the various lines according to the definitions of Art. 1 7,
and let, as usual, the positions of the lines in the first quadrant
be taken as the positive positions.
Fic. 3.
p
M
, : ■
N^-J
\
o(-
^
"^
Tk
' V
! A
\y
P
Fie. 2.
\^
y
^
c
' \f.
N
„/^<;
^ ^
Fic.4.\
V^
B7>
N
p
The sine BG is positive in the first and second quadrants,
and negative in the third and fourth.
The cosine ^C is positive in the first and fourth quadrants,
and negative in the second and third. •
44 ' PLANE TRIGONOMETRY.
The tangent PD is 'positive in the first and third quadrants,
.and negative in the second and fourth.
The cotangent MB is positive in the first and third quad-
rants, and negative in the second and fourth.
The secant AD, in the first and fourth quadrants, is obtained
by producing the revolving radius AB forwards ; while in the
second and third, it is found by producing AB backwards. It
is therefore positive in the first and fourth quadrants, and Tiega-
tive in the second and third.
The cosecant AE, in the first and second quadrants, is
obtained by producing AB forwards ; while in the third and
fourth it is formed by producing AB backwards. It is there-
fore positive in the first and second quadrants, and negative in
the third and fourth.
38. With regard to magnitude, it is seen from the figures
of the last Article, that as the angle PAB increases from 0° to
90°, the sine BC increases from to AM or 1 ; the cosine AC
decreases from 1 to ; the tangent PD increases continually
from 0, and when the secant AD coincides nearly with AM
both the tangent and secant become exceedingly great, and
infinitely great when AD actually coincides with AM, or when
the angle PAB = 90°, AD being then parallel to PD. Hence
the tangent varies from to « , and the secant from 1 to oc .
Again, when the angle PAB begins at 0°, AJE is parallel
to MB, or the cotangent and cosecant begin by being infinitely
great, and then decrease continually as the angle increases,
until at 90° the cotangent MB vanishes, and the cosecant
becomes AM or I.
In the same way we may proceed to find the magnitude of
the various iunctions in the other quadrants ; but this we leave
to the student. The results will be found to agree exactly with
those obtained in Art. 35, ^ .. ; / i , - > . ■ »
FUNCTIONS OF 180^— A.
45
From the figures of the last Article it is seen that the versed-
sine GP, which equals 1 - ^C, or 1 - cosine, is always positive,
and varies from to PQ or 2, in the first two quadrants, and
from 2 to in the other two quidrants.
The coversed-sine, which equals 1 - sine, is also always posi-
tive, and varies from 1 at 0° to at 90°, then from at 90° to
1 at 180°. In the third and fourth quadrants it varies^ from
1 at 180° to 2 at 270°, and then from 2 to 1 at 360" or 0°.
39. To find the Trigonometrical Functions of
i8o°-A. •
Let PAB be any angle A ; produce PA to Q, and make the
angle B'AQ = to the angle
P^5; then Pil-B' is the sup- b' • B ,,
plement of PAB or PAF
.-180°-^. Take AB' = ^a
AB, and draw B'C and BG
perpendicular to PQ ', then
B'G' = BG,&.ndiAG'^AG.
C - 6 A + * C
sin P^i?'
+ a
-h
= ain PAB.
cos PAB' = — = - cos PAB.
tan PAB' = ^ = - tan PAB.
or, we may write these results thus :
sin (180°-^)= Bin A.
cos (180°-^)= - cosyl.
tan (180° -i)= -tan^.
The cosecant, secant and cotangent being the reciprocals of
the sine, cosine and tangent respectively, we have
VsfSl/- ■).■
PLANE TRIGONOMETRY.
cosec (180°-^) =
• sec (180°- J) =
cot (180"-^) =
1
sin (180" -^1) sin^
1 1
= cosec A.
cos (180° -^1) -cos A
1 1
= - sec A,
— - cot A,
tan (180° - yl) -tan^
The same results are readily obtained from Figs. 1 and 2
of Art. 37, where it is seen that BG is the sine of PAB, and
also of BAQ, which are supplementary angles.
Also, cos PAB = AC = cos BAQ, but with the contrary sign.
Hence, sin 70° = sin 110°; cos 160°= -cos 2(f j tan 170° =
- tan 10°, &c.
40. To find the Trigonometrical Functions of
90° + A.
In the figure, let the angle
MAB = A, then the angle PAB =
90° + A. .
Now, the radius being unity,
sin
cos
tan
cosec
sec
cot
B
D
V
/
\
\
/
\
^
V
! A
H
7
1
dO° + A)=+BO
= +^i> = cos^.
= -BD= -sin A.
90 +A) = )-— — -;= , — = -cot^.
cos (90 +A) - sin A
90° + y1) =
90° + J) =
90° + ^) =
1
sin (90° + J) cos^
1 1
cos(90° + yl) -sin^l
1 1
tan (90° + A) - cot A
= sec A.
- - cosec A.
= - tan A.
FUNCTIONS OF NEGATIVE ANGLES. 47
These results, as well as those of the preceding Article, will
be deduced by a more general process in the next chapter.
41. To find the Trigonometrical Functions of a
Negative Angle.
In the last figure, let the negative angle PAK be denoted
by -A. Draw KH perpendicular to PQ ; then, radius being
unity, we have
sin ( - ^) = HK, which is negative (Art. 33.)
= - sin A.
m
cos {- A) = AH^ which is positive
= cos A.
. . sin (-^) -sin ^
tan {- A) = ) — -f = -— = - tan A.
cos ( - ^) cos A
COSeC (- A)^ — ; jr- = ; — - = - coscc A,
sin ( - ^) - sm j1
sec(-^) = — — - = j = sec^.
COS ( - ^) cos A
11
cot(-^)=-- — - — rr = — I — T-=-eot^.
tan {-A) - tan A
-1- .
In the same way, let the student find the various functions
of negative angles in the other quadrants. „ ;.
42. It is evident that any given angle has only one set of trigono-
metrical functions ; but from the preceding Articles it is seen that,
for any given function, there are more than one corresponding angle.
We will now proceed to determine the groups of angles correspond-
ing to any assigned value of each of the trigonometrical functions.
To find a general expression for all Angles which
have the same sine.
In a circle of radius unity, let PAB be an angle whose sine BC
is given. Make the angle jB'^g=to the angle PAB, and draw
48
PLANE TRIGONOMETRY.
DC perpendicular to PQ ; then B'C'-BC, and therefore the
angles PAB, PAB' have the same sine.
Moreover, if to each of these angles wo
add 3G0°, 720°, &c. , the positions of the
revolving lines AB, AB' will be the same.
Hence, if we represent the circular
measure of the angle PAB by a, and that
of two right angles or 180°, by tt, the
positive angles which have the same sine
are, a, tt - a, and these increased by 2M7r where w is any integer posi-
tive or negative, including zero.
For negative angles, the revolving line, in passing from the posi-
tion AP to AB', describes an angle whose magnitude is - (Tr + a), the
negative sign being prefixed to shew the (direction of revolution.
When it attains the position AB, the angle described will be
~{2T-a). Hence the negative angles which have the same sine
are, -(tt + o), -(27r-a), and these increased by 2n7r.
Therefore the positive angles are
2mr + a and 2mr + tt - a, or, 2mr + a and (2n + l)7r - a.
The negative angles are
-2n7r-(7r+a) and - 2n7r - (27r - a), or -(2n + l)ir-a and
-2(n + l)Tr+a.
Now, observe that when a is positive the multiple of tt is even,
and either positive or negative ; and when a is negative, the mul-
tiple of IT is odd, and either positive or negative. Hence all these
angles are included in the general expression,
mr + {-lf'a,
since. ( -1)"^=+! or -1, according as n is even or odd.
Therefore sin a = sin (mtt +(-!)** a). (39)
43. To find a general expression for all Angles
which have the same cosine. v
Let PAB be the angle whose cosine AC h equal to the given
one, and let a be its circular measure as before. Make the angle PAB^
ANGLES IIAVlxXG THE SAME TANGENT.
49
equal to PAB; then, since the cosine is positive in the first and
fourth quadrants, AC, the given cosine, is
the cosine of all angles corresponding to
the positions AB, AB' of the revolving
line.
Therefore the positive angles which have
the same cosine are, a, 27r~a, and these
increased by 2mr.
The negative angles which have the same cosine are, -a,
- (27r - a), and these increased by 2nir.
The posit *;^'C angles are therefore •
27i7r + a and 2/t7r + 27r-a, or 2mr + a and 2{n + l)ir-a.
The negative angles are
-2mf-a and -2mr-{2Tr-a), or -2mr-a and -2(;i + l)7r + a,
which are all included in the general expression,
2mr zfc a.
Therefore cos a=co3 (2w7r± a). (40)
44. To find a general expression for all Angles
which have the same tangent.
Let PAB be the angle whose tangent PD is equal to the given
tangent. The positive angles which have the same tangent are
evidently PAB and PAB' , or a and ir + a,
and these increased by 2mr.
The negative angles are manifestly
-{tt-o) and - (27r - a), and these increased
by 2mr.
Therefore we have for positive angles,
2mr + a and 2n7r + Tr + a, or 2mr+a and {2n + l)7r + a, .
And for negative angles,
- 2/i7r - (tt - a) and -2)^- (2Tr- a), or -(2/i + l)7r + a and
i . -2{n + l)rr + a,
50 i'lanp: tuigonometuy.
which are all inclutled in the gonorul expression,
mr + a.
Therefore tan a=tan (nTr + a). (41)
45. From the last three results we evidently have
cosec a = cosec (mr + (-!/' a). (42)
sec a= sec (2w7r±a). (43)
cot a= cot (n-TT+a.) (44)
In deducing these expressions, the least positive angle which has
the given value forthe trigonometrical functions, has been employed ;
but they are equally true for any angle which has the given value
for its sine, cosine, etc. , as the case may be.
Examples.
1. Shew that sin A, when determined from cotj Ay has two
values, equal in magnitude but opposite in sign.
From (10) we have
sin ^ = ± s/l - cos'^ A.
This is also seen from the figure of Art. 43, where, ii AC
be the given cosine, BC = + sin it, and B'C= - sin A.
2. Given tan 0= J3 , shew that sin $ will have two values.
From Art. 21 we have
sin 6 =
tan 6 J^
± J{l+tan'&) - 2
= sin 60° or sin 240°,
since the tangent is positive in the Jirst and third quadrants.
3
3. The cosine of an angle =» - -^, and the tangent 1 J, in
what quadrant is the angle ]
'• -4 ns. The third.
EXAMPLES. 51
4. Given sec 0= - 2, find the general value of 0.
2
Here ^= 120° = .ttt ; hence we have from (43)
o
$=(2n7r±^-) or (Gn±2)~.
5. Given tan 56= V^ , fiiid the general value of 0.
Here we find 5^ = 60° or - , and the general expression for
o
all angles which have the same tangent is mr + a, where a is
the circular measure. •
1 TT IT
Hence, the general value = -- (nrr + -) = (3w + l)^-^-
6. Find the sines and cosines of 300° and 162°. •
/J
A71S. sin 300" = - ----, cos 300° = J,
sin 162°== ^ , cos 162°= -Jn/10 + 2^5-.
7. Find the trigonometrical functions of 1098°. ',c^'
Ans. The same as those of 18°.
• A-
8. The sine and tangent of an angle are both negative, and
the tangent = 2 sine; find a general expression for all angles
having this property. '■' tt
Ans. {6n - 1)—.
o
9. Express in terms of 6 all the angles whose sine is - sin 0.
Ans. nTT + { - ly+'^e.
10. Find the values of which satisfy each of the following
equations: - . - -.- .
(1) cos 6= - 1. . , . J .,j..^,, Am. {2n+ l)-n: :
(2) ta.nO== -1. Ana. mr- —
4-
52 PLANE TlUaONOMKTltY.
(3) H[n'e = l. A7i3. (Gn.±])Z-.
(4) soc ^ = 2 tan (9. Ans. (Qn + ( - lf)''r.
(5) 2 cosec - cot ^=- J3. Ans. (Gn ± 1)-.
O
(C) tan ^ + cot ^= 2 soc ^. yfvis. (Gn + ( - 1)")'^-.
11. Trace the variations in tlie sign and magnitude of the
tangent and secant, as the angle varies from 90° to 270°, by
means of the line definitions.
12. Prove that (sec - + tan %) (cosec J + t^n '^ ) = 5.
o o
13. Find the general value of in sin + cosec ^=2.
Ans. ^=:(4?i + 1)--.
14. Shew that tan 52|° = ( ^3" + >J~i) ( v/2 - 1).
15. Shew that sec 2^ - tan 2^ = -?-"**"^
1 + tan 0'
9
16. Shew that tan Q ■■=^ -
cot ,- - tan - -
2 2
17. Shmv that tan ^. = ''" ^^^ ""^ ^
18. Shew that sin ^ =
2 I + cos 2yl 1 + cos A
2 1
, e~
tan ^ + cot — tan -- +cot 6
19. Shew that cos ^ = sin ^ cot 1.
20. Find all the values of $ which satisfy 1 - cos ^ = 2 sin^ $.
Am. e=^{{2n+\)±\}~.
o
EXAMPLES. 53
21. Trace the variations in sign and magiiitudo of the secant,
coflocant and cotangent, in the third and fourth quaih'ants, by
the lino definitions.
22. Shew that cos 1 1 J^ = J ^2+ V2T75'
23. Shew that sin — and cos -v-, when determined in terms
of COS yl, will each have two values, but in terms of sin A, four
values.
1- v/T
24. Shew that cos 105" = ^~-.
2^/2
25. Find tan 1G5° and sin 1G5°.
26. If tan 6= J^ + 1, find cos 26, and the general value
of e.
Ana. cos 26— r=:, 6 = mr± .
J2 8
27. Find the general values of the limits between which 6
lies when sin^ 6 is greater than cos'' 6.
IT 3
Am. Between 2mr + -- and 2mr + -' tt :
4 4
5 7
and between 2mr + —ir and Inir + — tt.
4 4
28. Prove that sec* 6 + tan* ^ = 1 + 2 sec" 6 tan^ 6.
54
PLANE TKIGONOMETRY,
CHAPTER Y.
TRIGONOMETRICAL FUNCTIONS OP THE SUM AND DIFFERENCE OF
TWO ANGLES, AND OF MULTIPLES AND SUBMULTIPLES OP
ANGLES.
45. To find sin (A + B) in terms of the sines and
cosines of A and B.
Let the angle BAG = A and the angle QAD = B^ then the
s^rvglQ BAD =^ A + B.
In AD^ one of the bound-
ing lines of the compound
angle {A + B), take any point
.P, and draw FM, PQ perpen-
dicular to AB and AC respec-
tively, and from Q draw QK
perpendicular to PM; then
the angle QPK = the angle
KQA = A.
8in(yl + ^) = —
MK+PK
M V
AP
^QN PK
"AP^AP
jig' AP ^ P(i'AP
= sin A cos J5 + cos -4 sin B.
(45)
FUNCTIONS OF {A-\-B). 55
46. To find cos (A + B) in terms of the sines and
cosines of A and B.
Employing the figure of the last Article, we have
AM
cos {A ■{■ B) =
AP
_ AN-NM *
AP
AN QK
"AP'AP
ANAQ QK QP
~ AQ'TP'qP'AP
— cos A cos J5 - sin -4 sin B. (46)
47. To find tan (A + B) in terms of the tangents
of A and B.
sin {A + B) * •
tan (A-\-B)^ -)--. — ^f
^ ^ cos {A + B)
sin A cos B + cos -4 sin ^
cos A cos B - sin A sin B'
Dividing the numerator and denominator of this fractiqji
by cos A cos By we have
sin A sin B
+
, . „. cos A cos B
tan {A + ^) = -
sin A sin B
COS ^ COS B
tan ^ + tan B
1 - tan ^ tan B'
(47)
56
PLANE TRIGONOMETRY.
48. To find sin (A - B) in terms of the sines and
cosines of A and B.
Let the angle DAC = A and CAD = B, then BAD = A -J3.
In AB, one of the bounding lines of tho compound an!^'l(>
(A - B), take any point P,
and draw PM perpendicular
to AB, PQ perpendicular to
AC, (^iV perpendicular to AB,
and PK perpendicular to QN;
then the angle PQK being the
complement of AQN^, is equal
to BAG or A.
Bm{A-B) = ~
QN-QK
AP
^AQAP Pq AP
= sin ^ cos i? - cos A sin J5. (48)
49. To find cos (A - B) in terms of the sines and
cosines of A and B.
From the last figure we have
cos (^-.5) = ^'^
AP
_ AN^ NM
~AP
AN PK
^AP'^AP
^AQAP '^ PQ'AP
= cos A cos B + sin A sin B.
(49)
FUNCTIONS OF {A-{-B) AND (A—B). 57
50. To find tan (A-B) in terms of the tangents
of A and B.
tan {A-B) = )- — j/
cos (^l - JJ)
sin A cos B - sin B cos A
cos A cos li + sin A sin B'
Dividing numerator and denominator by cos A cos B, we
have
4. /A n\ tan ^ - tan B
tan IA-B) = - — -— j. (50;
^ "^ 1 + tan A tan B ^ '
51. Formul.'B (46) (48) and (49) can be easily deduced from
(45), as follows :
In (45) write - B for B and we have
sin {A - B') = sin A cos ( - J5) + cos ^ sin ( - B)
= sin A cos 5 -cos A sin ^5. (Art. 41.)
The complement of {A + B) is 90" - (^ + i5), and since the
cosine of an angle is the sine of the complement (Arts. 1 2 and
19), we have
cos (^ + i?) = sin {90" - (^ + ^)}
= sin {(90"'-^)-i?}
- sin (90° - ^ ) cos ^ - cos (90° - ^) sin i?,
■■ cos A cos B — sin A sin B,
If we write - B for B in the last formula, we have
cos {A-B)^ cos A cos ( - ^) - sin ^ sin {- B)
= cos ^ cos B + sin vl sin B.
In a similar manner, from any one of the four formulas (45),
(46), (48) and (49), the others may be deduced.
52 . The preceding f ormulte of this chapter have been established
only when A-\-B is less than 90^. We will now proceed to shew that
58 PLANE TRIGONOMETRY.
they are true for all angles, whether positive or negative ; and from
the results of the last Article it is evident that it will be sufficient to
shew this in the case of (45) alone.
I. Let A-\-B be > 90°, and let ^=90°- a; and JB=90°-2/,
and hence a;=90''-^ and 2/=90°-B.
Then ^+J5=180°-(x+2/), and therefore (ac+y) < 90'-.
Hence the formula (45) is true for (x-\-y).
Now, sin {A + B)=am { 180° - (x+y) }
=sin (x+y) (Art. 39.)
=sin X cos y-\-coB x sin y
=sin (90° - A) cos (90° -B]+ cos (90° - A) Bin(90° - B)
=co3 A sin B+sin A cos B.
Hence the formulae are true when {A-\~B) > 90°.
n. Agam, let a;=90°+^, and therefore ^= - (90° -cc).
Then x+B=90°-^A+B
and 8in(a;+-B)=8in {90°+(^+i?)}
=co8 {A-{-B) (Art. 40.)
=co8 A cos JB— sin -4 sin B
=cos { - (90°- a) } cos A'- sin { - (90° - «) } sin B
=cos (90°-a;) cos B+sin (90° -x) sin B
sin X cos jB+cos x sin B.
Hence the formulae are true for 90°+ -^4 and B
In the same manner it may be shewn that they are true for A
and 90°4--B ; hence they are true if 90°, or any multiple of 90°, be
added to either or to both, that is, they are true for all angles, since
by adding multiples of 90° we may increase the angles A and B to
any magnitude whatever.
In a similar manner we miy show that the formuliu are true for
all negative angles. *
53. The preceding formulae furnish us at once with the
trigonometrical functions of all compound angles.
* For a gooiaetrical proof iu auy aasiguod case, see Appendij.
FUNCTIONS OF (90°H B) AND {180<^-B). 59
In (45) and (46) let A = 90°, then
sin (90° + B) = sin 90° cos B + cos 90° sin B
= 1 X cos j5 + X sin i?
= cos B.
cos (90" + B) = cos 90° cos B - sin 90° sin B
= X cos i? - 1 X sin ^
= - sin B.
/^^o T.^ sin (90° + B) cos J5 ^ _
ton (90° + B) = y^-jr, — jf. = ^-p = - cot B.
^ ' cos (90 + B) - sm B
cot(90° + i5) = - — ,,.^o , „. = — „= -tani?.
^ ' tan (JO + ^) - cot i?
1 1
sec (90° + 5) = — .^r^r^ — pT = — ; — jj = - cosec ij.
^ ' cos (90 ■\-B) - sm B
1 1
cosec (90° + 5) = . ,„, - — 5r = ^ = sec B.
^ ' sm (90 + B) cos ^
These results agree with those of Art. 40.
54. In (48) and (49) let A = 180°, then
sin (180° -B) = sin 180° cos B - cos 180° sin B
= X cos 5 - ( - 1) X sin B
= sin 5.
cos (180° -B) = cos 180° cos 5 + sin 180° sin B
= - 1 X cos J5 + X sin B
= -COS B.
Hence tan (1 80° - 5) = ^ tan B, cot ( 1 80° - 5) = - cot B,
. sec (180' -B)= - sec B, cosec (180° -B) = cosec B.
These results agree with those of Art. 39.
Here we may observe that if B bo less than 90', its supple-
ment 180° - ^ is greater than 90° or obtuse, and tliat the sine
and cosecant of an obtuse angle are positive, while the cosine,
tangent, cotangent and secant are negative.
GO PLANE TRIGONOMETRY.
55. In (45) and (4G) let A = 180°, then
ft
sin (180° + /?)=- sin B, cos (180° + i?) = ~ cos B,
tan (180° + B) = tan B, cot (180° + /?) - cot B,
nee (180° + B) = - sec B, cosec (180° + //) = - cosoc /;.
56. In (48) and (49) let A = 270°, then
sin (270° -B)= - cos 7?, cos (270° -/?)-- sin B,
tan (270° -B) = cot 7?, cot (270° - B) = tan B,
sec (270° -B)= - cosec i?, cosec (270° -B)= - sec /?.
57. In (45) and (46) let A = 270°, then
sin (270° + B)= - cos B, cos (270° + i?) = sin B,
tan (270° + B)= - cot i?, cot (270° + i?) = - tan B,
sec (270° + 7^) = cosec B, cosec (270° + 7?) = - sec 7?.
58. In (48) and (49) let A = 360°, then
sin (360° - 7?) = - sin B, " cos (360° - 7?) = cos B,
tan (360° - B)= - tan B, cot (360° -B)= - cot 5,
sec (360° - 7?) = sec B, cosec (360° - 7?) = - cosec 7?,
or, the functions of 360° - B are the same as those of - B.
59. In (45) and (46) let A = 360°, then
sin (360° + B) = sin B, cos (360° + B) = cos B,
or, the functions of an angle greater than 360° are the same as
those of the excess above 360°.
60. General Formulae involving the Functions
of two Angles.
From (45), (46), (48) and (49) we have
Hm{A + B) + 8m(A-B)= 2 sin ^ cos 5. (51)
sin (A + B)- sin {A-B)= 2 cos A sin B. (52)
cos (A + B) + cos (A- B)= 2 cos A cos B. (53)
cos (A + B)-coa(A-^^ - 2 sin ^4 sin B. (54)
GENERAL FORMULAE. Gl
If Ave put A+Ji:= A' and A~B = B',
we have 2A=A'^B' and 2B = A'-B',
whence A = l{A' + B') and 5-i(^'-J5').
Making these substitutions in the above group, and omitting
the accents, since A' and B' may be any angles whatever, we
have
sin ^ + sin ^ = 2 sin }, (A + B) cos I {A- B). (55)
sin tI - sin 5 = 2 cos | (yl + B) sin | (^1 - B). (56)
cos ^ + cos i? = 2 cos I (^ + J5) cos ^ (yl - i?). (57)
cos ^ - cos i? = - 2 sin \{A^B) sin ^ (vl - ^). (58)
These formulse are of very great importance, and the student
sliould make himself quite familiar with them under this form.
They are especially valuable in trigonometrical computations in
transforming a sum or difference into a product, and vice versa.
Each of them may be expressed as a theorem ; thus, " The sum
of the sines of any two angles is equal to twice the sine of half
the sum of the angles multiplied hy the cosine of half their dif-
ference,^^ and similarly for the others.
6i. Divide (55) by (56), then we have by (13) and (14)
^n ^ + sin 5 2 sin }^ {A + B) cos \ (A-B)
sin ^ - sin ^ ~ 2 cos | (-4 + B) sin ^ {A-B)
= tan \{A + B) cot \ {A - B)
• _ tan|(^ + ^)
"tan^^-^)' ^ ^
In the same manner let the student verify the following :
cos ^ - cos 5 , , , ^, , , , T,. • ,»r.s
-. 5- = - tan i (^ + B) tan -h (A - B). (60)
cos ^ + cos ^ ^ ^ ' -^ / \ /
sin ^ + sin ^ , ., „, • .^,,
;; ^ = tanH^ + ^)- (61)
COS -4 + cos B ^ ^ ' ^ '
n
C2 PLANE TRIGONOMETIIV.
sin -4 - sin J9 , , ^
-, ,-- = - cot ^ (.1 + B). (G2)
cos ^ - cos i> '' ^ ' ^ '
sin ^ - sin i?
cos ^ + cos i?
sin vl + sin 5
= tan^(^-Z?). (63)
„ = -cot H^ + i?). (64)
cos -4 - cos ^ ^ ^ ' ^ '
62. Divide (45), (46), (48) and (49) by cos A cos B, then
wc have by (13)
sin {A + B) sin ^ cos -B + cos A sin B
cos ^ cos ^ ~ cos ^ cos i^
= tan ^ + tan B. (65)
cos (A + B\
1—77 = 1 - tan A tan B. (66)
cos A cos B ^ '
sin {A - B)
cos -4 cosi?
= tan ^ - tan B, (67)
f?liilL^_ = 1 + tan ^ tan 5. (68)
cos -4 COS i}
Tn a similar manner we find
sin (A + B) „ , ,^^,
-■ ■ \ . ',- = cot i? + cot A. (69)
Bin il sin /j ^ '
COS (^ + i?) , „ , ,^^,
■ . ^, . ,; = cot il cot j5 ^ 1. (70)
sinil HUiB ^ '
sin {A - B)
. , . „=coti?-cotil. (71)
sin il sin ^ ^ '
cos (A - B) , . T, ,
. \ . ; =cot il cot i? + 1.
sin^ sm-B
(72)
. sin (^ + -B) , , _ ' . ,^_.
-r— ^- ^ = 1 + cot ^ tan B, (73)
sin A cos B
f?li^ = cot^-ta„a (74)
sin -4 cos JO
GENEEAL FOUMULiE. 63
^(±'^?1 = 1 - cot i tan B. ♦ (75)
sin A cos i^
_^_^y_:U?I = cotJ + tan5. (76)
sin A cos j6'
63. Divide (45) by (48) and (4G) by (49), then we Lave
sin (A + B) sin A cos B + cos A sin i?
sin {A - i^^ ~ sin ^ cos 7^ - cos A sin i?
tan A + tan ^
tan A - tan i!^'
cot B + cot vl
cot A cot ^ - 1
~ cot A cot i^ + r
64. In (47) and (50) let A = 45°, then
„ ^, tan 45" ± tan i?
tan (45° ± iJ) = , — r—Tf^ «
^ 1 + t^ii "^5 tan B
1 ± tan i?
1 + tan B'
_cot7?±2
~ corZTqpl*
In (50) let i? = 45°, then
tan ^ + 1*
. 1 - cot A
i + cot J.'
65. From (47) and (50) we have by (8)
(77)
(78)
cot B - cot A'
cos (A + -B) _ 1 - ta n A tan B .^^.
~coii{A-B)~l+isinAtiiiiB' ;: ;
(80)
(81)
(82)
, ^-. tan ^ — 1 ' /oo\
(84)
/ . T-.N cot B cot A=f-1 yQp,v
cot {A±B)= ^^^ ^ . m
G4 PLANE TRIGONOMETRY.
The secant and cosecant of (A^ Ti) or (A — /?) are easily cleriveil
from those of the cosine and sine respectively, by means of (8);
thus we have
sec (^4-J5)= . . , „v
^ ' cos {A-\-IJ)
1
cos A cos ii — sin A sin B
sec A sec B
1-tan A tan B'
(80)
by multiplying numerator and denominator by sec A sec B. In a
similar manner the others may bo derived.
66. The product of (45) and (48), and of (4G) and (49),
are
sin (A + B) sin {A - B) = sin" A cos- B - cos'' A sin' /?,
cos (A + B) cos {A — B) - cos^ A cos'' B - sin'^ A sin^ B.
But cos' ^ = 1 - sin' J, and cos" 7i = 1 - sin''' i>,
therefore sin (^ + J5) sin (yl - B) = sin' yl - sin' B. (87)
= cos' 7? -cos' J. (88)
and cos (i + B) cos {A'B) = cos' yl - sin' JS. (89)
= cos'^-sin'^. (90)
67. General Formulae involving the Functions of
the Multiples and Submultiples of an Angle.
In (45), (40) and (47), let jB = J, then
sin 2 A = sin A cos A + cos A sin A
= 2 sin ^ cos .4. (91)
cos 2 A = cos A cos A - sin A sin A
= cos'^-8in'^. (92)
tan A + tan A
tan 2 A =
1 - tan A tan J^
2 tan .4
1 - tan' A'
(93)
GENERAL FORMULAE.
65
If, in the last three equations, we write A for 2 A, and there-
fore U for A, which we are at liberty to do since A is any angle
whatever, we have ■:■■■'■
Bin -4 = 2 sin ^ cos -. (9*)
cosil = cos2 .^ ..sin^-. (95)
A
2 tan -
tan^ = ^. W
1 - ton' ^-
From (10) and (92) we have .• : ,
cos- A + mii^ ^ = 1 -•
cos- -4 - sin'^ ^ = cos 2il,
the sum and difference of which are ••
2 cos^ yl = 1 + cos 2.4 .->. (97)
2 sin*'' ^ = 1 - cos 2i. (98)
A ^'
Writing A for 2 A and — for ^, these become
^cor'-^I+cobA, (99)
2 8in'''4 = l-cos^, (100)
the quotient of which is ^ fe
tan'i = ^^:^. -""^ (101)
2 1 + cos il
Multiplying the numerator and denominator of the second
member of (101) by 1 + cos A, we get
^A (1 -cos' .4)
'^"^ 2"(l+cos^)»
sin" A
(1 + cos -4 )
,2>
06
PLAXE TRIGONOMETUV.
therefore
tan.
sin A
(102)
2 1 +008.1'
Multiplying tlio numerator and denominator of the second
member of (102) by 1 - cos A, we get
( 1 - cos A ) sin A
1 - COS'* .4
(1 - cos A) sin A
sin'^ A
1 - cos A
sin A
(102 bis)
The formulae of this Article have already been deduced for
angles less than 90°, in Art. 28, by a less general process, and
the student will now see that they are true for angles of any
magnitude.
68. To find expressions for sin mA and cos mA.
From (51) and (53) we have
sin mA-\-ain (m-2)A=2 sin (m— 1)A cos A
and cos inA-{-cos {m—2)A=2 cos (m — 1)/1 cos A,
hence sin mA=2 sin (?n. -1)A cos A - sin (m — 2)A
and cos mA=2 cos {m — l)A cos A - cos (w-
.-2)A I
,-2) A. I
(103)
If we make m successively 1, 2, 3, &c. , these give
sin il=8in A
sin 2A=2 sin A cos A
sin 34=4 sin A cos^ 4 — sin A,
&c., &c.
and cos A=cos A
cos 2il=co8'* A -sin^ A
COB 3i4=4 cos^ A-3 cos A,
&c., &c. : , i :
u^J Other general formulae for sin mA and cos A will be given in a
subsequent chapter. j , » '
(104)
GENERAL FORMULA.
67
69. General Formulae involving the Functions
of three Angles.
Lot A, B,C bo any three anglos, then by (45) and (46)
Hin(/l+5+C)=8in {(J-^B)-\-C\
=Hin {A ]B) cos C'+co8 {A^ /?) sin
=8in A cos B cos C+cos A sin B cos C
+C08 A cos J5 sin C-sin ^ sin B sin C. (105)
cos{A+B+C)=coti {{Ai-B)+C\
=cos (^ + -6) cos (7 -sin (^ + B) sin (7
=co8 A cos B cos C7— sin A sin fi cos C
- sin ^ cos B sin C - cos A sin -B sin C. (106)
In the same way we may develop the sine and cosine of
(A—B-^C), &c.
If we divide (105) by (106) we get, after dividing the numerator
and denominator by cos A cos B cos C,
tan ^ +tan B^t an C - tan ^ tan ^tang^
tan U^-B+C)=^^^^ ^ ^.^^ B-tB.nAt&n (J- tan JS tan C ^ '^
If Ai-B+G=mr, then tan (A+B-^G)=0, and therefore from
(107) wo obtain , . . .
tan ^+tan B+tan C=tan A tan B tan C. (108)
If A-\-B-\-C=(2nirl)-- , that is an odd multiple of --, then
tan (^ f B+0)=oc , and therefore the denominator of (107) must
be zero,
hence tan A tan B+tan A tan C+tan B tan C=l. (109)
If (109) be divided by tan A tan B tan C, we get
cot A + cot JB+cot C=cot A cot B cot C. , (110)
70. Again, let A+B~\-0=mr=2n-, an wen multiple of -,
then . "
sin ( A-{-B+C)=Bm htt = ^
Bin (-^+B+C)=8in (mr- 2A)= - ( - 1)« sin 2A. _ ,^^^^
sin( ^-B+C)=sin (n7r-2B)=-(-l)" sin25.
; * sin ( ^+B - C)=8in (wtt - 20)= - ( - 1)" sin 20.
68
PLANE TRIGONOMETRY.
The sum of this group is, by (55) and (58),
4 sin A sin B sin C=— (— 1)« (sin 2^+sm 2B+8in 20). (112)
Again, we have
cos( A-\-B-{-C)=coa nv =(-1)". \
cos {- A+B-\-C)=co& (nir-2^)=(-l)"co8 2J[. I
cos ( A—B-{- C) =cos (nTT -2B}={- 1)" cos 2B. 1
co8( ^4-B-C)=co8 (nTT -2a)=(-l)" cos 20./
The sum of this group is, by (57),
4 cos A cos B cos C={ - 1)»* (cos 2^+co8 25+008 2C+1). (114)
(113)
TT
ir
sin( A-{-B+C)=Bm (2n+l)-
' (1116w)
71. If ^+J5+C=(2»+1) 2", an odd multiple of ^ , we have
8in(-^+iJ+C?)=8in{(2n+l)^-2^}=(-l)»»cos2J[.
sin ( A-B+O^Hin {{2n-\--')'^-2B} ={-!)*' cos 2J8.
sin ( Ai-B-G)=ain{{2n-\-l)~-2C] =(-1)" cos 2 J. ^
Subtracting the difference of the first two of this group from the
sum of the last two, we find
4 sin ^ sin B sin = ( - 1)" (cos 2A-{-cos 25 +co8 20 - 1). (112 bis)
Again,
co8( .4+B+0)=cos (2n4-l)~ = 0.
co8(-^+5+C)=cos ((2»+l)~-2^}=(-l)»» sin 2A.
cos( ^-J5+0)=co8{(2/i + l)-|-2jB}=(-l)" sin 2B.
co8( J[ + B-0)=co8{(2n+l)|-20}=(-ir sin 20. ^
The sum of which is
4 cos A cos B cos 0=( - 1)»» (sin 2^+ sin 2J9+sin 20). (114 bis)
■ (113 6t»)
OSNERAL FORMUUE. ¥»
Bv combining the eauations of group. (lU) and (113), m»ny
„te irp"rt"rre.uUs a.0 obUined. Thus, in (113), takmg ha
difflnce between the sum of the first two and the sum of the
second two, we find
4 cos A sin B sin C=( - !)« ( " cos 2^+cos 25+cob 2(7-1). (115)
In a similar manner, from (lU) we obtain
4 sin A cos B cos C= - ( - 1)« ( " b^ 2^+sin 2B+sin 2C). (116)
Again, from groups (111 Us) and (113 his) the following are
easily obtained, which will serve as exercises for the student.
4sin^cosBcosC=(-l)M-co8 2^+cos2B+cos2C+l).^
4co8^sin5cosC = (-l)«( cos 2^ -cos 25+ cos 2C+1)
4co8Acos5sinC=(-l)M cos2^+cos2B-cos2C-fl). ^ ^^^^^
4cos^sin5sinC = (-l)M-8in2^+sin2B+sin2a)
4 sin ^ cos 5 sin C= ( - 1)« ( Bin 2^ - sin 254-8in 2C).
4 sin ^ sin 5 cos C= ( - 1)'» ( sin 2^+Bin 2B - sm 2(7).
72 The following six formxilffi are of great utility in com-
puting the values of the trigonometrical functions, and in trans-
forming a sum or a difference nto a product.
cos A t'n A
cot^+tan^= ^v^ + ^^^
CDs'* ^ + s in'' A
~ sin A cOii ^
2 \_
sin 2^1
2 sin A COS -4
= 2 cosec 2 -4.
(118)
COS A sin il
cot ^ - tan A = -: — r 7
sin A cos ^
2 ( cos' A - sin'' ^)
~ 2 sin A cos -4
2 cos 1A
~ sin '2 A
« 2 cot 2il.
(119)
70 PLANE TK1G0N0METK\
'-,f
.,i'»
, . , 1 cos A •■'■'!■ ' ''
COSec ^ + cot ^ = -: + -: . . , , .
sm A am A - , .im
1 +C08 A ' ,
sin A
2 cos^* |-
, by (99) and (94)
I'H.' .,. .v.- )., 2 Sin — COS —
= cot-. i . , (120)
. , . 1 cos A
cosec ^ - cot A = -7
sin A sin ^
;. ,, ,.■ ,■ ,,;,. ■ _ 1 -COS A
■ -::■ . sin ^
A
>W!V"
• 2 sin2 -
= — ^1 J, by (100) and (94)
2 sin — COS
2 2
From (81) we have
= tan|-.
(121)
, /.Ro .V l+tan^
tan (45" + A) =
^ ' 1 - tan ^
tan (45 -A)=- ,
^ ^ 1+tani'
hence -
tan (45- + ^) . taii (45" -A) = l±^^ ^ Ll^^
1 - tan A 1 + tan A
_ 2 (1+tan^^)
1 - tan^ A
2(cos^^+sin'J)
cos^il-sinM ' ^y(^^^
2
cos 2A
= 2 sec 2^. ^ (122)
F011MULJ5 OF VERIFICATION. 7l
4 tan A
tan (45* + A)- tan (45° -A) = Yltan^A
cot ^ - tan ^
i__, by (119)
.., "2 cot 2^' ^^ '
- =2tan2il. G^S)
Formulae of Verification.
The following four formula are useful for testing the accu-
racy of the trigonometrical tables :
sin(36V^)-sin(36»-^) = 2cos36°sin^ = |(^4-l)sinX
sin(72°-fl)-sin(72°-^) = 2cos72'sinl = KN/5-l)«i^^-
The difference of which is
sin (36» ^ A) - sin (72° + A) + sin (72° - A) - sin (3G° - A)
= sin^. (Euler's formula.) \^^^)
Substituting (90° - A) for A in (124) we obtain
sin (54° + ^) - sin (18° 4- ^) + sin (54° - ^) - sin (18° - i)
= cos^. (Legendre's formula.) \}^^)
By (55) we find
sin (30°-^) + sin (30° + ^) = cos A (126)
Similarly, by (58) we find
cos (30° -A)- cos (30° + ^) = sin A. (1 27)
' • Examples.
1 + sin A , ,, . A.
1. Prove that y^^^^ = K^ + *^" 2 ^ '
72 PLANE TRIGONOMETRY.
From (91) and (99) we liave
. ^A A A A
... sin^-- + 2 sm— cos— + cos'^ —
1 + sin ^ 2 2 2 • 2
1 + cos ^ ^ ^A
2 cos^
sin-+cos-
=M ^1 —
cos-
= Hl+tan|y.
2, Prove that tan — =
2 1 + sec 61*
From (102) we have
sin
sin cos 6 tan ^
tan
2 1 + cos (9 1 , 1 + sec ^'
cos 6'
6. Trove that tan — - tan — - tan — = tan- tan —tan -.
•s o fa 2 3 6
'an~-tan- •tan- = ^ ^ ^ , by (13) and (G5)
cos— cosmos —
. e
sin — ^ ^ ^
cos —cos --cos —
J o fa
.
sin—
^ 5 ^ { Kcos g - cos-) } , by (63)
cos-cos -cos
^ 0\
tan ^ tan- tan-, by (58)
EXAMPLES. •{ *^^
, Tf ..« /} - ££i^±£^, prove that tan ^ = tan |tan ^.
From (101) we have ^ >" **-^-^' ^^ ^
^ ' cos g + cos P _
^ 1 - cos (9 _ l+cosaco3^
1 + cos a cos /3
/■*' 1 + cos g cos j8 - cos g - co s g
- ^ 1 + cos g cos /3 + cos g + cos ^
' " 1 - cos g 1 -cos^
, ' ~ 1 + cos g 1 + cos ^
= tan^|tan^|, by (101),
therefore , tan - = tan- tan-^-
*,.>
e ^ ^
5. Prove that cos (30° - ^ ) " ^os (30" + -^) = sin -.
6. Prove that cot (30' + |-) - tan(30- + | ) = 2 tan (30" - 0)
o ^\ 2 cos ^ - 1
7. Prove that tan (30° + j) tan (30° - ^) = gcos^+l
1 - sin 20
8. Prove that tan'^ (45° -6)^ fTsin '2d'
, sin 20 + sin d 3d -^ '■
9. Prove tliat ^^^jf:^-^J = ^^^ T' ,
tan a + sec g , g . „„ / i eo . ^\
10. Prove that ^ = tan - tan (45 +-^).
XV. J.XV/ cotg + cosecg ^ -^
■ sin d + sin 3d + sin 5d ^^
11. Prove that ^^^^^^^^-^-^^,^ = ta- 3d.
. ■ ; sin' g - sin^ /?
12. Prove that tan (g + (S) - -^-^^ ^ ^ sin )8 cos )3*
' = cos d- cos 2d ^ d 3d
13. Prove that ^^^^"^^^ = *- 2 """ T'
t^l
74 PLANE TRIGONOMETRY.
, . ^ , sin 30 + sin 4^ .
14. Prove that ^— 2^ = t.ana '
cos ^0 + coa |6'
15. Express by means of a sum or difference
cos 26 cos B ; sin 2d cos 30 ; sin 50 sin 9.
Ana. |(cos 30 + cos 0) ; |(sin 50 - sin 0) ; ^(cos 40 - cos 60).
16. Reduce sin^ 30 -sin'' 20 and
6 50
cos'' (45° + -^) - sin'' (45° — -) to products.
Ans. sin sin 50 ; sin 20 cos 30.
17. Prove that (1+cot + co8ec 0) (1 + cot 0-cosec 0)
= cot — — tan — .
18. Express by a product each of the following :
sin + sin (0 - 2^) ; sin" - sin'' (0 - 2</)) ; sin + cos ;
sin (0 + <^) + cos (0 - ^) ; cos" (0 - <^) - sin" (0 + 0).
Ana. 2 sin (0 - <;^) cos ^ ; sin 2(0 - ^) sin 2^ ;
^2" cos (45° - 0); 2 cos (45° - 0) cos (45° - ,^) ;
cos 20 cos 2<^.
19. Prove that cos" A - cos" ZA = sin 4-4 sin 2^.
20. If tan" = tan (a - 0) tan (a + 0), shew that
sin 20 = ^2^ sin a.
21. Shew that tan (30° + 0) tan (30° - 0) = tan cot 30.
22. If sec (^ + a) + sec ((^ - a) = 2 sec ^, shew that
.-— a
cos 9=^2 cos — .
23. If tan (^ + a) + tan (^ + y8) = 2 tan 0, shew that
tan <^ = |(cot a + cot /3).
24. Shew that cos 25° - cos 1 1° + cos 47° - cos 61° = sin 7°.
25. If a, /?, y be in arithmetical progression, prove that
(sin a - sin y) sin /8 = (cos y - cos a) cos /S.
: ' i EXAMPLES. 75
2G. Prove that
tan'^ - iaii^ <^ - sin (^ + </>) sin (^ - </>) sec" sec^ </>.
27. Sliew that cos- (^ + </>) - sin* ^ = cos <)!) cos {26 + </>).
28. Shew that sin (6* + <^) cos d - cos (^ + <^) sin = sin </>.
29. Prove that sin a + sin /3 + sin y - sin (a + /3 + y)
= 4 sin |(a + ^) sin ^(a + y) sin ^{(3 + y).
30. Prove that cos a + cos /? + cos y + cos (a + ^ + y)
= 4 cos ^(a + /3) cos |(a + y) COS |(^ + y).
31. Prove that tan a + tan ^ + tan y - tan a tan ji tan y
sin (a + y8 + y)
cos a cos /3 cos y
32. Prove that cot a + cot )8 + cot y - cot a cot j8 cot y
cos (a + )8 + y)
sin a sin ft sin y'
33. Prove that cos (a + /3 + y) + cos(a + ^3- y) + cos (a - ^ + y)
+ cos ( - a + |8 + y) = 4 cos a cos yS cos y.
34. If a + /? + y = 1 80°, prove that
sin (a + /8) sin (/? + y) = sin a sin y ;
sin 2a + sin 2/3 + sin 2y = 4 sin a sin /3 sin y.
35. If a + ^ + y = 90°, prove that
. 1 + tan —
cos a + sm y - sin /» 2
cos )8 + sin y - sin a /8'
1 + tan -^
36. If a, j8, y be in arithmetical progression, prove that
sin a - sin y = 2 sin (a - /?) cos /? ;
cos a - cos 7 sin a - sin y
tan (a -3) = : — ^ = ^.
V sin a + sm y cos a + cos y
37. If sin </) = m sin (2^ + ^), prove that
tan (0 + <{>) = tan 6.
76 PLANE TRIGONOMETRY.
38. If tan J(a + /?) tan ^{a - )9) = tan'' ^, prove that
cos a = cos )S cos y.
or> Ti. ^ 1 - tan''rf> -
oy. It tan - = , n— ., prove that
2 1 + tan-* (f) '^
2 cot 2</) = ^sec 6 + tan - t'^sec ^ - tan $,
40. If cos 6= Jl, ami cos ^ = — ^ — — — , shew that
cos (^ + <^) = ^.
/J
41. If tan = ^ and tan <f> = \, prove that sin {2$ + ^) = -~.
42. If cot 29= - tan <^, prove that tan (6 - <fi) = cot 6.
. _ -r « 2 sin sin i/r , . ,
4:6. It tan d> = — ; — -^, then tan 0, tan and tan il/ are
in harmonical progression.
44. If cos (6 - \j/) cos (f> = cos (^ - <^ + \p), then tan 6, tan </>
and tan ^ are in harmonical progression.
45. If tan a tan 6 + sec a sec 6 = sec ^, shew that
sin /? :; sin a
tan ^ = ± —~ —.
cos a cos p
46. Prove that sin" 30° = sin 18" sin 54°.
47. Prove that tan 50° + cot 50° = 2 sec 10°.
48. Prove that tan 52|° = ( V 3 -+ J 2) {s/2 -\).
49. liA+B + C= 180°, prove that
sin A + {^in B -sin C A B
= tan — tan — :
sin .4 + sin -S H- sin C 2 2
B C A
sin (il + — ) + sin (5 + — ) + sin (C + — ) + 1
2i ^ ^
= 4 cos \{A - B) cos \{B - C) cos \{C - A).
60. If sin ^ = sin a sin (/? + 6), prove that
tan (^ + ^) = tan | tan" (45° + ^).
ii ^ ^
EXAMPLES. 77
51. If sin (2C + y) cos « = 2 cos (y - 2) sin x, prove that
cot £c - cot y = 2 tan 2.
52. Prove that 2 cosec 4« + 2 cot ix = cot x - tan aj.
53. Prove that
2 sin ^(.4 + B - 90°) cos ^(^ - ^ + 90°) = sin A - cos B.
54. Prove that (cot'^ _ - tan^ — ) tan ^ = 4 cosec 0.
55. Prove that sin (A - B) + Bin (B - C) + sin (C-A)
= 4 sin }^{li - A) sin ^(C - i5) sin J(^ - C).
56. Prove that
sin A CO8OC (^ - B) cosec (i - C) + sin J5 cosec (B - (7) cosec (B-A)
+ sinCcofiec(O-A)cosec(G-B) = 0.
57. If cos (A + B) sin (C + i>) = cos (^-5) dn (C - J)),
shew that tan i) = tan A tan 5 tan C.
58. If sin (a -7) cos /3 + sin (/8-y) cos a=0, shew that
tan a + tan (3—2 tan y.
59. If cot a + cos 13= JJ cot e sin /i^, ' ••-
and cot y + cos ^= ^2^ cot (3 sin ^, prove that
sin = sin a sin /3 cosec y. ' ' '
cos A cot ^ cot J - cos A
60. Prove that -—r = -. — TT'-
cos A + cot A cos A cot ^
61. Prove that cosec 4:6 + cot 4:6 = ^(cot - tan 6). . t
62. If cot 6=2 tan <^, then cos (0 -</>) = 3 cos (^ + <}>), and
sec (0 + </)) = 2 sec 6 sec <^. ' ■' ■■ ■ ' ' - '
63. If ^+5 + C= 180°, shew that sin^ A + sin" B + Bixi'
= 2(cos il sin J5 sin C + cos B sin A sin C + cos C sin ^ sin JJ),
. ^A . ,B . ^G ^ . A . B . C
and sin'* — + sm" — + sm'* -- + 2 sm - sm — sm ^^ = 1-
64. Prove that
sin {x + y) sin (y + ») = sin a; sin » + sin y sin (x + y + s).
78 PT-ANK TRIOONOMETRY.
05. U A ¥ n + (1 180°, prove tlmt
• 1 - cos A + cos B + cos (J : A G
— — -•- — --; = tan -- cot --,
1 + cos A + cos Ji - cos C « 2 2
and sin' A + sin' B - sin' C = 2 sin A sin B cos C.
66. Prove that cos {x + y) sin y - cos (a: + z) sin s
= sin {x + y) cos y - sin {x + s) cos z.
67. Determine ^ from the equation
sin a + sin (^ - a) + sin (2.6 + a) = sin (^ + a) + sin (20 - a),
Ans. = (lOn±l)~ or {5(2n+ 1)± 1 }^.
68. Prove that sin' (a? + y) + cos' (x - y) = 1 + ain 2x* sin 2y
69. If ^ + 5 + C = 1 80°, prove that
cot A + cot B cot ^ + cot C cot ^ + cot C
tan .4 + tan B tan 7i + tan C tan il + tan C '
and
cot A + cot i5 + cot C — cot ^1 cot B cot C + coaec A cosec 5 cosec C
70. If Jj^J^^I sec 2e, she^v that = i(n7r + ( - If ^ ) '
1 — tan 6' 2 b
71. liA+B-^C = 180°, sheAv that
cos' A + cos' i? + cos' (7 + 2 cos ^ cos B cos C =» 1.
„„ ^, , 1 - cos 2a; cos 2y tan' .r + tan' y
72. Shew that , -^ = -/-.
1 + cos 2a; cos 2y 1 + tan-'a; tan- y
73. Shew that 1+cos (a-/8)+cos (/3-y)+cos (y-a)
= 4 cos i(a - /8) cos i(/3 - y) cos ^(y - a),
and sin (a - P) sin y - sin (a - y) sin /8 = sin (y - /3) sin a.
■ 74. if 2 tan + tan (a - 0) = tan (y3 + 6), shew that
tan = ^ sin (a - ;8) cosec '\ cosec )8.
A B C
75. If tan — , tan — and tan — be in arithmetical pro-
gression, so also are
cos A^ cos B and cos (7, where A + B + C= 180°.
76. Given cos 20 - cos iO = sin 6, find 6.
Ans. = n7r or ^(w7r + (- 1)"— ),
EXAMPLES.
70
77. Find the general value of $ in sin (a - ^) ^ cos (a + 6).
ir
Am. ^ = (4/t-l) ..
4
7S. If a + ^ + y = 180°, shew that
tan a tan B tan v tun a tan y tan B
^ — a ^ L 4. 4. 1 + — -I- 2
tan (3 tan y tan a tan y tan p tan a
= sec a sec ft sec y.
79. Prove that cos (w + n + r)a + cos {vi + n - ?')a
+ cos (m - n h r)a + cos (in - n - r)a = 4 cos 7na cos na cos ra.
80. If tan (a + /?) = 3 tan a, shew that
sin 2(a + ^9) = 2 sin 2/3 - sin 2a.
81. If il + i? + C = 180°, shew that
A ABC B B A C
tan -- + cos — sec - sec — = tan — + cos — s(^c — s(;c .-
^ C C A B
— tan - + cos — sec -- sec .-,
2 2 2 2
and (sin ii + sin ^ + sin C) (tan - + tan . + tan --)
2i 2i Z
I i-x ' A , B . ij
= 4(1 + sm — sin - sin - ).
2 2 J •
- tan
{^-^-1-{--^}-
82. Prove that
cos B - sin <;^
cos + sin
83. If a, /?, y be in arithmetical progression, shew that
\ J^ \ ' 1_
tan a + tan y 2 tan ft cot a + cot y 2 cot y8*
84. \iAvB^G = 90", shew that -^
tan A + tan B + tan C = tan ^ tan B tan C + sec A sec ^ sec C.
85. Given cos ^ + cos 7(9 = cos 4^, find ^.
^?w. ^=(2w+l)Jor4(2n7r±^).
8t
PLANE TRIGONOMETRY.
CHAPTER VI.
ON THE CALCULATION OP THE NUMERICAL VALUES OF THE
TRIGONOMETRICAL FUNCTIONS.
73. The Circular Measure of an Angle is greater
than the sine, but less than the tangent of the
Angle.
. Let PBQ be an arc of a circle whose centre is A ; bisect the
angle PAQ by AB and draw the
chord PQ and the tangent DBE.
Now it is evident that the arc
PBQ is greater than PQ and less
than DE, therefore PB is greater
than PC and less than DB^ or
PB PC
—j^ is greater than -jv, j and less
A.Jl Air
than
DB
AB'
PB
But -j-p is the circular measure of the angle BAP, (Art. 9.)
PC '■■-•■ '■' - "'' "- ' ■ ■
-j^ is the sine of the angle -5 J/* : .?-; . . '■
DB . , , ^
and -T-^ IS the tangent of the angle BAP,
A B . i\
■•■y i i Af
v, +'
or, if the radius be unity, we have PC the sine, BD the tangent
and PB the circular measure of the angle BAP. Hence, repre-
senting the circular measure by 0, we have
$ greater than sin $, and less than tan d. ' ;*; -
SIN B TAN B _-
LIMITING VALUES OF -— — AND — r — . 81
u o
^ ^ - sin B , - tan .
74. To find the limit of -r— and of — z" » when
is indefinitely diminished.
In the last Article Ave have shewn that
sin B, B and tan &
B
are in ascending order of magnitude. Hence, 1, — — ~ and
1 B
are also in ascending order of magnitude, that is, —
cos B «"i ^
lies between 1 and -. whatever the value of B may be,
cos B
But as B is diminished, cos B tends to 1 and ultimately
becomes 1 when ^ = j hence we have r :^; :: ;m:
e • ■'■•:. '■!: ■■i.y
-: — ;:-= 1, when ^ = 0,
sin ^
and therefore also — 7,— = lj when ^ = 0. ' . '■
Since
B
ban B sin B 1
B ' B cosB'
wc have — v;- = ^> when ^ = 0.
B
From these results it follows that when B is very small,
sin B = B = tan ^, very nearly.
75. To shew that sin B>B-~, where B is the
4
Circular Measure of any Angle betv/een 0° and 90".
By Art. 73 we have ,, . - . ,
.,. , ■- . r :n. :■
B B ''""2 B
:■„ '.■ «OS 2 ^-,^ ' •■ • >' ^. ■>••■•
hence 2 sin — >^ cos — •
2 2 . .-- ^
82 . PLANE TRIGONOMETRY.
Multiplying both members of this inequality by cos — , we
have by (94)
sin 0>6 cos'* — '
>0{l-~), since -->
sin
76. To find the Numerical Values of the sines
and cosines of all Angles from 0° to 90° at inter-
vals of 10". „ r
Let 6 be the circular measure of 10", then since the circular
measure of 180° is tt, or 3.141592653589793 ... as will be shewn
in a subsequent chapter, we hav«
180° : 10" ..TTiO,
IOtt
or 6 =
180 X 60 X 60
3.141592653589793
~ 64800 •
= .000048481368. J ■
But sin e<^ and >^ - -r-, (Arts. 72 and 75.)
4
hence sin ^>.000048481368 - J(. 000048481368)' '• ; ' ^
>.000048481368-i(.00005)3 ■ ^:. '
>.000048481368 - .00000000000003,
from which it appears that the quantity to be subtracted from
does not affect the first twelve places of decimals. Hence, the
sine of 10" coincides with the circular measure of 10" to twelve
places of decimals, and therefore -
COMPUTATION OF TRIGONOI^rETRICAL FUNCTIONS. 83
sin 10" = circular measure of 10" = . 000048481368 ;
also sin 1" = circular measure of 1" = . 0000048481368.
If a be the circular measure of any angle -4°, then it is
evident that
A"=- " — - =-T-^ = ax 206264".806... (128)
cir. meaa. 1 sin 1
which agrees with (6). .
The cosine of 10" is obtained from the formula
cos 10"= v/l-sinU0"= v/(l-sinlO") (1^+ sin 10")
= .999999998824. •
77. From (51) and (53), we have
sin (^ + ^) = 2 sin A cos B - sin (^ - .5)
cos (^ 4- .^) = 2 cos ^ cos ^ - cos ( J[ - ^,
in which let B be constantly equal to 10"; ' * -
thus, sin {A + 10") - 2 cos 10" sin ^ - sin (i - 1 0")
cos {A + 10") = 2 cos 10" cos ^ - cos (^ - 1 0").
Now, if ^ =»= 10", 20", 30", <fce., in succession, we find for the
sines
sin 20" = 2 cos 10" sin 10" - sin 0",
sin 30" = 2 cos 10" sin 20" - sin 10",
sin 40" = 2 cos 10" sin 30" - sin 20", - - /
«fec., <fec., -''■>
and for the cosines
* ' '•; ; cos 20" = 2 cos 10" cos 10" -cos 0",
■ <\ •-■: cos 30" = 2 cos 10" cos 20" - cos 10",
-f •>•' cos 40" = 2 cos 10" cos 30" - cos 20",
.84 PLANE TRIGONOMETRY.
These results may be further simplified as follows :
2 cos 10"= 1.999999997648,
= 2 - .000000002352,
= 2 - m, suppose.
Making this substitution, we have
sin 20" = 2 sin 10" - sin 0" - m sin 10" = .0000969627,
sin 30" = 2 sin 20" - sin 10" - m sin 20" = .0001454441,
sin 40" = 2 sin 30" - sin 20" - m sin 30" = .0001939254,
cos 20" = 2 cos 10" - cos 0" - m cos 10" = .9999999953,
cos 30" = 2 cos 20" - cos 10" - m cos 20" = .9999999894,
cos 40" = 2 cos 30" - cos 20" - m cos 30" = .9999999812,
&c., . &c.
78. Having computed the sines and cosines by this process
up to 30°, those for angles greater than 30° may be easily found
by (126) and (127), thus, •
sin (30" + ^) = cos ^ - sin (30' - A)
cos (30° + A) = cos (30° -A)- sin A,
in which let A = 10", 20", 30", &c., in succession, and we have
sin 30° 0' 10" = cos 10" - sin 29° 59' 50",
sin 30° 0' 20" = cos 20" - sin 29° 59' 40",
&c., <fec.
cos 30' 0' 10" = cos 29° 59' 50" - sin 10",
cos 30° 0' 20" = cos 20° 59' 40" - sin 20",
&c., ^
It is not necessary to continue this process beyond 45°, since
the sines and cosines of angles above 45° are respectively the
cosines and sines of their complements ; thus, sin 46° = cos 44° ;
sin 50° = cos 40° ; cos 46° = sin 44° ; cos 50° = sin 40°, <fec.
COMPUTAXrON OF TRIGONOMETRICAL FUNCTIONS. 85
79. The tangent is found from the formula tan 6 = ~\
but when the tangents have been thus computed up to 45*^, the
rest may be found by the addition of those ah*eady computed ;
thus, from (123) we have
tan (45° + J) = tan( 45° - ^) + 2 tan 2^ *
in which put A = 10", 20", &c., in succession, and we haye
tan 45° 0' 10" = tan 44° 59' 50" -;- 2 tan 20"
&c., (fee.
The cotangent is already known from the tangent, thus,
cot r = tan 89°, cot 10° = tan 80°, &c.
80. The secant is easily computed from the tangent by
(122), thus,
sec 2^ = ^{tan(45° + ^) + tan(45°-^)} -^ .'.
in which put A = 10", 20", &c., in succession, and we have ,
,. , sec 20" = j{tan 45° 0' 10" + tan 44° 59' 50"} ..',...,:
&c., &&- ■ '■ . (..
The cosecant may be found from (118), (120) or (121).
From (118) we have
" , • ■ < ' "" : -,V'..,'
cosec 2^ =^(tan :24 +cot ^),
^ierefore cosec 20" = |(tan 10" + cot 10"), ' •'
<kc., , , , 4eo» ' ./..';-
' 81. The accuracy of the work should be verified from time
to time by separate and independent computations. If, for
example, the functions of 15°, 18°, 30°, 36° and 54°, agree
with those found by the process of Chapter II., the work if cor-
rect. Formulae (124) and (125) may also be used to verify the
results. Thus, if we write 6° for ^ in (125), we have
sin 60° - sin 24° + sin 48° - sin 12° = cos 6° = sin 84°,
a relation involving the sines of five angles, which will be satis-
fied if the computed values are correct.
S6 PLANE TRIGONOMETRY.
62. The tabulated numerical values of the sines, cosines,
ikc, for each given angle constitute what is called " A table of
natural sines, &c.," to distinguish it from "A table of logarith-
mic sines, tfec," to be described in the next chapter. The latter
table is by far the more useful, as nearly all computations in
trigonometry are carried on by logarithms.
The arrangement of both tables is most easily understood
from a simple inspection of them. As ample explanations are
always prefixed to every set of trigonometrical tables with
regard to their arrangement and to the manner of using them,
we shall here refrain from giving the usual mechanical direc-
tions for their use. We shall, he wever, give a few illustrations
shewing how to use them in accordance with the principles on
which they are computed.
It will be necessary, too, for the student to make himself
perfectly familiar with the arrangement of the tables he uses, in
order to understand any peculiarities which belong to them, and
to use them intelligently and not mechanically.
83. Since all the trigonometrical functions pass through all
their possible numerical values, as the angle varies from 0° to
90°, the tables are not extended beyond the first quadrant.
The functions of angles greater than 90° are found by the prin-
ciples of Chapter V., Arts. 53-59, thus,
sin 153° 10'= sin 26° 50' = .4513967,
cos 124° 30' = - cos 55° 30' = - .5664072,
tan 170° 42'= - tan 9° 18'= -.1637563,
(fee, (fee.
For angles greater than 180° and less than 270° we deduct
180° and take the same function of the remainder, being care-
ful to prefix the signs that belong to that quadrant. Thus, by
Art. 55, we have . _„ ,. .,.. ,. ,.
■■■•~1
USE OF TRIGONOMETRICAL TABLES. 87
sin 225" 15'= - sin 45° 15'= -.7101854
cos 250° 2'= -cos 70° 2'= - .3414734,
tan 192" 50'= + tan 12° 50'= +.22780G3,
Again, by Art. 57, we have
sin 300° 10'= -cos 30° 10',
cos 350° 0' = + sin 80° 0',
tan 354° 40'= - cot 84° 40',
<fec., (fee.
84. The tables are computed, for intervals of 10" or 1', and
to five, six or seven places of decimals. The sine, cosine, <fec.,
of any angle not given in the tables, can be found with suffi-
cient accuracy for most practical purposes by means of simple
proportion, except near the limits of the quadrant where the
disparity between the variation of the angle and that of the
function, changes too rapidly to admit of the application of this
principle ; and conversely, an angle whose sine, cosine, (fee, is
not found exactly in the tables, can be obtained in a similar
manner.
Ex. 1. Required the sine and cosine of 42° 17' 24". -.^
From the tables, we have \.. '..
sin 42° 17' = . 6727973
sin 42° 18' = .6730125
Difference for r or 60" = .0002152 ■■■
Hence the proportional part for 24" is |* x .0002152 = .0000861,
which must be added to the sine of 42° 17', since the sine in-
creases as the angle increases, (Art. 34) - ' -
therefore sin 42° 17' 24" = . 6728834. , i .Ki:-:: ^
88 PLANE TRIG(JNOMiOJ'UV.
Again, from the tables, we have
cos 42° 17'- 7398268
008 42° 18' = .7396311
Difference for 1' or 60" = .0001957
Hence the proportional part for 24" is J J x .0001957 = .0000783,
which must be subtracted from the cosine of 42° 17', since the
cosine decreases as the angle increases, (Art, 34)
therefore cos 42° 17' 42" = .7397485.
Ex. 2 Given cos A = .7216446, to find A.
From the tables, we have
cos 43° 48' = .7217602
cos 43° 49' = .7215589
Difference for 1' or 60" = .0002013
cos A =.7216446, the given cosine,
cos 43° 49' = .7215589, the next less in the table.
Difference = .0000857
then .0002013 :. 0000857 :: 60": 25".5,
which must be subtracted from 43° 49',
therefore A = 43° 48' 34".5.
Ex. 3. Given tan ^=1.1726470, to find 0.
From the tables, we find ■ ,
tan 49° 32' = 1.1 722298 ?v ;
tan 49° 33' = 1.1729207 ^^
Difference for r or 60"= .0006909
INCREMENTS OF TRIGONOMETRICAL FUNCTIONS. 80
tan ^^1.1726470, the given tangent,
tan 49° 32'= 1.1722298, the next less in the table.
Difference = .0004172
then .0006909 : .0004172 :: 60" : 36".2,
which must be added to 49° 32',
therefore d = 49° 32' 36". 2.
The student must not fail to observe that the proportional
part must be added for the sine, tangent and secant, since these
functions increase as the angle increases; and subtracted for
the cosine, cotangent and cosecant, because they decrease as the
angle increases.
85. We will now deduce general formulae for the increments of
the trigonometrical functions of an angle corresponding to a given
increment of the angle.
Let any angle 6 be increased by a very small increment A^, ex-
pressed in circular measure, and let the corresponding increment
of the sine, &c., be denoted by A sin 6, A tan ^, &c. ; then, we
have
A sin ^ = sin {6 + M) - sin ^
= cos 6 sin A^ - sin B {\ - cos A0)
1 - cos L9 . ' ' •
= cos ^ sin A^ (1 - tan Q — ; — — — ) .
= cos 6 sin A^ (1 - tan 6 tan -— ), by (102 his)
sin A^
A6
A/3
= cos ^ A^ ( 1 - -^ tan $). Art. 74. (129)
Now, if tan is not very large, that is, if 6 is not near 90°,
\a ■
— — tan may be neglected, and we have approximately
' Aflin^ = co8^Aft .,v, , : ; ^ (130)
90 PLANE TUIGONOMETilY.
A cos $ = cos (0 f Ai9) - cos
= - sin $ sin A0 - cos 6(1- cos A^)
. /, . r -. /, 1 - cos A^,
•~ - sin ^ sm Af> (1 + cot 6 )
^ sin A6'
Ad
= - sin d sin A^ (1 + cot ^ tan — -)
Ad
= - sin d Ad ( 1 + --- cot 6). (131)
Ad
When d is near 90°, -— - cot d may be neglected, and we have ap-
proximately
Acos d= -sin dAd. * ' a32)
A tan d = tan (d + Ad) - tan d
_sin (d + Ad) sin d
cos (d + Ad) cos d
sin Ad
cos'^ d cos Ad (1 - tan d tan Ad)
sec^ d tan Ad
1 - tan d tan Ad ' >
sec'^^ Ad ,
~1-Adtand" , ^^^^^
When d is not nearly equal to 90°, Ad fan, d may be neglected,
so that we have approximately
A tan d = sec2 dAd. (134)
A cot.d = cot (d + Ad) - cot d
cos (d + Ad) cos d
"sin (d + A"d) ~ siiTd
- sin Ad
sin2 d cos Ad (1 + cot d tan Ad)
- cosec=* d tan Ad
1 + cot d tan Ad
-cosec'^d Ad
1 + Adcot d* v :;- vyui
\I36)
INCltKMENTS Ob' TRIGONUM ETHICAL FUNCTIONS. 91
When is not nearly c(iuul to 0", A^ cot $ may be neglected,
and we have approximately
A cot 0= - cosec2 Aa. (130)
A see ^ = sec (0 + A^) - sec
1 1
cos (^ + A^) cos
cos ^- cos (6 + A6)
cos $ cos (6 + A^)
sec 6 (tan + tan ) tan AO
1 — tan B tan A6/
A0
sec B (tan ^ + ) A^
1 - A6^ tan ^
(137)
When B is neither very small nor very nearly equal to 90°,
A^"
— — sec B and A^ tan ^ may be neglected, and we have approxi-
mately
A sec ^==sin B sec^ '^ A^. ■ (138)
A cosec B = cosec (^ + A^) - cosec Q
'_\ 1 ■'.>,-[■ ^■"
-i {B + A^)~8in B '■'. .;'"'•.'
sin ^ - &in (B + ^B)
amBam(B + M)
A6'
cosec B (tan — _ — cot B) tan Ad
1 - cot ^ tan Ad
Ad
cosec B (—r — cot B) Ad
(139)
1 -Adept d
When d in neither very small nor very nearly equal to 90°,
92 PLANE TllIGONOMETRY.
cosec 6 and A^ cot 6 may be neglected, and we have ap-
Ji
proximately
A cosec 6 = - COB $ cosec^ AO. (140
From these forinulse we observe, (1) that a small increment of
the angle causes a small increment of the sine, tangent and secant,
and a small decrement of the cosine, cotangent and cosecant ; (2)
that for all the trigonometrical functions, a small increment of the
angle produces a small proportional increment or decrement of the
function, with the exception of the particular cases noticed above.
Hence, if an angle is near 90°, it cannot" be found from the ordi-
nary tables with great accuracy from its sine, tangent or secant ;
nor if near 0° or 180°, from its cosine, cotangent or cosecant.
In the next chapter, however, we will explain the construction
of a special table by which the trigonometrical functions of an angle
near the limits of the quadrant, can be found with great accuracy
86. The results of this chapter enable us to solve many
interesting and useful problems in surveying and astronomy,
without using the trigonometrical tables.
Examples.
1. An object standing on a horizontal plane subtends an
angle of 3' 20" at the distance of two miles, find its height.
Let BC be the object whose •
height is required. The angle
^ = 3' 20"= e, and AC == 2 ^
miles = a.
Then BG = AC tAiiO, by (32)
= aO, by Art. 74,
— ax circular measure of $
ti"
.°'""206264".8' ^y^^^®>
2 X 200
206264.8
10| feet, very nearly.
EXAMPLES.
93
2. The equatorial radius of the earth is 3962.8 miles, which
is found by astronomical observations and calculations to sub-
tend at the sun's centre an angle of 8". 95. Find the sun's
distance from the earth. , .
Let S and E represent the
centres of the sun and earth.
The angle ASE=: 8".95 = 6,
AE = 3962.8 miles = r, and
SE = X.
Then
and
X sin = r
x =
sin 0'
r
cir. meas. '
206264".8
8". 95'"
= 91328000 miles.
— rx
by (35)
by Art. 74
t
. by (128)
r.»/,
3. The apparent semi-diameter of the sun is 16' 1".82 at
the earth's mean distance, as given in the last problem. Find
the sun's radius. ., ,
) In the figure of the last problem, let S be the earth's
centre and E the sun's. The angle ASE = 16' 1".82 = ^,
SE= 91328000 miles = d and AE the sun's radius -= r.
Then
r = d sin 0, by (27) "■' '
= dx cir. meas. of 0, by Art. 74
= dx
= d X
t
' .r.
206264".8 '
961".82
by (128)
206264".8
425860 miles.
f)4
PLANE TRIGONOMETRY.
E.
r'' Hence the sun's radius = 107.4 times the earth's radius and
as the volumes of spheres are as the cubes of their radii, we
have ■'''' I- "' ■■ ■'•. •' -- ■*■■ ■ -'■
volume of sun : volume of earth :: (107.4)' : 1
:: 1238833:1, ■
so that the sun is more than a million times as large as the
earth.
Dip of the Horizon.
87 . The dip of tlie horizon is the angle of depression of the
visible horizon below the true horizon, arising from the eleva-
tion of the observer's eye above the level of the sea. Let BBG
be a section of the eartli re-
garded as a sphere, and A the
position of an observer above
the surface. Join A C and pro-
duce it to 6r, also draw AE at
right angles to the vertical line
AG, and A F touching the sur-
face at J), then neglecting the
eflfect of atmospheric refrac-
tion, D is the most distant point visible from A.
ceive AE and AF to revolve about AG as an axis, AE will
describe the plane of the true or celestial horizon and AF will
describe the surface of a cone toucliing the earth in the small
circle called the apparent or visible horizon, and the angle FAD
is called the dip of the horizon.
Let 7t = ^-B, the height of the observer's eye,
r = DC, the radius of the earth,
/) = the dip of the horizon.
The angle FAD = ihe angle ACD = D.
PIP OF THE HORIZON. 95
AD JAG xAB
DG~ D
J{2r + h)h
Then ,.^ tani) = — = ^~-^^ , {Euc. III., 36)
[271 /h^^
As h is always very small compared with r, the square of
tlie fraction — is quite inappreciable, and therefore may be
r
neglected, so that we have . ^ •,'-.■ •■ •• : .
tan 2) = ^—. ' (141)
Again, as the angle D is always very small for all accessible
heights above the earth's surface, we may write the circular
measure of the dip for its tangent (Art. 74), and expressing it
in seconds by (128) we obtain
i?" = 206264".8.
The distance AD oi the visible horizon = the arc BD very
nearly ; let -4 i> = i)5 = c?, then .:,.;,
, . ^ . . d = DC X the circular measure of D
•';' ■ , =r^^=j2^,. . ' •■ (143)
The mean vp' e of r is 20888628 feet. Substituting this
in (142) and reducing the constant coefficient of vA we have
D" = 63".S2^fh, • (142 his)
where h is expressed in feet.
Owing to the effect of refraction, this coefficient is dimin-
ished by about ^ part. Deducting ^^ of 63.82 we have finally
i)" = 58".02>/A. (144)
06'", PLANE TRIGONOMETRY.
By this formula, the dip is computed and tabulated for
heights varying from 1 to 100 feet, in works on navigation and
iihitronomy.
Expressing d and r in miles and h in feet we have, by com-
puting the constant coefficient in (143),
d=l.2Ujh. (145)
which gives the distance, unaftected by refraction. The amount
of refraction however varies very much with the temperature
of the atmosphere, in ordinary states of which, this distance
is increased by about -^^ part. Hence, increasing the coefficient
1.224 by -jJj of itself, we have
c^= 1.317 s/A
= M V^j very nearly. (146)
In (145) let d= 1 mile, then we have
A = .6675 feet
= 8 inches, very nearly.
Hence, if the tangent DA in the figure, be one mile, AB va^
inches, which is expressed by saying that on the surface of still
water the depression. of the horizon is 8 inches for a horizontal
distance oi one mile ; therefore in levelling for canals, aqua-
ducts, <fec., a reduction of 8 inches per mile must be made for
the curvaoure of the earth's surface. v .'; .
Since k varies as the square of the distance, the depression
for 2 miles is 32 inches; for 3 miles, 6 feet, &c. '• '.•-(.'.; • ' ">
. ... Examples.
1. Given cos A = .6400566, to find A,
From the tables, we hare ' ' '- v , . ? .
EXAMPLES. 97
COS 50" 13' = .6398862
cos 50° 12' = .6401097
Difference for 1' or 60" = .0002235
cos -4 = .6400566, the given cosine,
cos 50° 13' = .6398862, next less in the table.
Difference = .0001 704
Hence we have
.0002235 : .0001704 :: 60" : 45".7,
Therefore ^ = 50° 12' 14". 3.
2. Find in a similar manner the angle whose cotangent is
2.1453675.
Ans. 24° 59' 28".2.
3. Given tan 45° 1' = 1.0005819, find tan 45° 0' 40".
Am. 1.0003879.
4. Find the angle whose sine is .7126666.
Ans. 45" 27' 8".2.
5. Find the angle whose tangent is .8952524.
, Am. 41° 50' 11".6.
6. Given sin 45° = -'^^--, show that tan 22° 30' -.41421 36,
and sec 22° 30' = 1.0823922.
7 The circular measure of an angle is .1047683; find the
number of degrees in it.
Ans. 6°0' 10".
8. Given cosec A + cor A = J^ , find A.
Ans. ^ = 60°.
9. Find the dip of the horizon for a height of 25 feet.
, ' Am. 4' 50".
98 PLANE TRIGONOMETRY.
10. How far can the top of a lighthouse 200 feet high be
seen at sea 1 '
Arts 18| miles.
11. From what height will the horizon be 44 miles distant 1
Ans. 1111^ feet.
12. From the top of a ship's mast whose height is 81 feet,
the top of a lighthouse 100 feet above the level of the sea, was
just visible ; required their distance, taking the effect of refrac-
tion into account.
Ans. 25^ miles.
13. What is the difference between the true and the appar-
ent level, at the distance of 1000 feet?
Ans. .2873 inch.
14. A ship whose mast is 90 feet high, is sailing directly
towards an observer at the rate of 10 miles per hour, and the
time from its first appearance till its arrival at the observer, is
1 hour 9§ minutes ; find approximately the earth's radius, sup-
posing the observer's eye to be on a level with the surface of
the sea and no allowance made for refraction.
Ans. 3958 miles.
15. When the moon's distance from the earth is 238824
miles, her radius subtends at the earth an angle of 15' 33".5 ;
find her diameter.
Ans. 2161.6 miles.
16. Find the distance and the dip of the horizon for a height
of 6 feet, taking the effect of refraction into account.
Am. 3.226 miles; 2' 22".
LOGARITHMS AND LOGARITHMIC TABLES. 99
CHAPTER VII.
ON LOQARITHMS AND LOGARITHMIC TABLES OF THE TBIGONO-
METRIOAL FUNCTIONS.
88. The logarithm of a number is the index of the power
to which a fixed number, called the base, must be raised, in
order to produce the given number. Thus, if
then X is the logarithm of the number R, to the base a, and is
expressed thus,
a; = logaiV;
or, if there be no occasion for mentioning the base, simply by
aj = log N.
Again, since 10" = 1 ; 10^ = 10 ; 10^ = 100, &c.,
logio 1=0; logio 10 = 1; logi„ 100 = 2, &c.
It is evident that any number except unity may be taken
as the base; and if for any base as 10, the logarithms of all
*
numbers be computed and registered in a table, the table thus
formed constitutes what is called "A table of a system of
logarithms to base 10." There may therefore be an infinite
number of systems of logarithms. ..
89. Although in theory any number except unity may be
used as the base, yet in actual practice it has been found most
convenient to use only two systems, viz. :
I. — Logarithms to base 2.71 8 J8 . . . (denoted by e) which are
called Napierian logarithms, from the name of the inventor,
100 PLANP] TRIGONOMETRY.
Lord Napier, Baron of Merchislon in Scotland, who published
the first table of logarithms in the year 1614, and*
II. — Logarithms to base 10, the radix of the ordinary scale
of arithmetical notation, which are called common logarithms.
These possess some peculiar advantages, and all the tables in
common use are calculated to this base. Hence when we speak
of logarithms, we mean logarithms to base 10, unless the con-
trary is stated.
go. Since a" = l, and a} = a^ we have
loga 1 = 0, and log^ a= 1. (147)
whatever a may bew
If a be greater than unity, then
a-a = ; a°= 1 and «+• = oc ,
therefore loga = - x ; log^ 1=0; loga oc = oc . (148)
Hence, the logarithm of any number will be between and
— oc , or between and + oc , that is, will be negative or posi-
tive according as the number is less or greater than unity.
Again, if a be less than unity, it may be shewn in a similar
manner that the logarithm of any number will be positive or
negative according as the number is less or greater than one.
Since a* and a—* are always positive quantities, negative
numbers have no real logarithms.
gi. In any System of Logarithms, the Logarithm
of the product of two numbers is equal to the sum
of the Logarithms of the numbers.
Let a be the base, m and n any two numbers, and x and y
their logarithms respectively. Then by the definition of a
logarithm, we have
V
PROPERTIES OF LOGARITHMS. 101
,2W. = a* or jc = loga wi
and n = ay or y = loga ^ •
and therefore mn ^a"^ .aV
but by the definition, x + y is the logarithm of mn to base a,
therefore log^i mn = x + y
= loga m + loga n.
92. In any System, the Logarithm of the quo-
tient of two numbers is equal to the difference of
their Logarithms.
Using the same notation as in the last Article, we have
ma* .
— = — = a«-l/:
n av '
but x-y is the logarithm of — to base a,
n
lliorefore loga— =*-y
= loga W- loga W.
Hence it follows that ' ^ •
loga— = loga 1- loga m • ' •
• =-logaW. by Art. 90.
93. In any System, the Logarithm of the zith
power of any number is tz times the Logarithm
of the number.
Let m = a* as before, that is, £C = logrt w». -> • • ^ .
Then -' = . ' m» = (««)♦* ^; ^^ vj . ^a^^-
' 102 PLANE TRIGONOMETRY.
)mt nx iH the logarithm of 7/1" to base a,
therefore loga 7/t" = nx
-= n loga m,
which is true whether n be an integer or a fraction.
94. We thus see from the last three Articles that, by means
of logarithms, the processes of multiplication and division are
reduced to those of addition and subtraction ; and those of
involution and evolution to multiplication and division. The
arithmetical operations of addition and subtraction cannot be
performed by logarithms.
•The following examples shew the application of the pre-
ceding principles :
. 189 , 3^x7 , -3 ^i
. log— = log— --= log 3^x7*
= 3 log 3 + J log 7.
log — 7- — = n log a^ — log c — p log h.
logQ^=^logi=i{log3-logl7}.
log sld'-h' = \\og{a'-h'')
= |log{(a + 6)(a-6)}'
. = i log (a + 6) + J log (a - h).
95. Having given the logarithm of any number to a given
base as e (where e denotes a certain number 2.71828... of which
more hereafter), to find the logarithm of the same number to.
any other base as a. ;, r
Let X and y denote the logarithms of any number N to the
bases a and e respectively, that is, ^.
PROPERTIES OF LOOARITHMS. 103
let a5 = loga y, and y = loge iV,
hence a* = iV^ and eV = -A^,
therefore a* = eV.
Expressing this equation in logarithms to base e, we have
a; log<, a = y l<^ge «
= y. by (147)
Substituting the values of x and y, we have
loga J^ loge a = loge JV,
therefore loga -^=q • log- JV. (149)
loge «
If a= 10, the base of the common system,
then logio I^ - j^-iy^ loga ^. (150)
From (150) it appears that every common logarithm may
be resolved into two factors, one of which is constant and
depends upon the base of the system employed, the other is
variable and depends upon the number itself. The constant
factor or multiplier which thus connects two systems of loga-
rithms, is called the modulus, and is denoted by M. Thus, the
constant , -~ is the modulus of the common system taken
loge 10 ^
relatively to the system whose base is e.
The modulus of the common system may, therefore, be
defined as the constant factor or multiplier, by which it is neces-
sary to multiply the Napierian logarithms in order to convert
them into common logarithms, and is equal to the reciprocal of
the Napierian logarithm of the Hase of the common system.
It will be shewn hereafter, in the chapter on the computation
of logarithms, that the Napierian logarithm of 10 is 2.302585... ,
therefore the modulus of the common system
• ^ I^iTlO " 2:302585:: " -4342944... , .
104 PLANE TiunoxoMKruv.
DcMotiiii; tlio coniinoa loyariihin ol a number by log, (I'^'O) may
bo written
log iV^=. 4342944 x log,, N. (151)
Hence we also find
= 2.302585 X log i\r, (152)
which enables us to finil the Napierian logaritlim when th«!
common logarithm of any number is given.
96. Since log^ N= Mloge N, and logb N=^ M' loge iV, wlu4-o
M and M' are the moduli, we have by division
therefore the logarithms of the same number in different systems
are proportional to the moduli of those systems. • '
97. Relation between the bases of two Systems.
Let X and y denote the logarithms of any number N to the bases
a and b respectively, that is,
let a;=log^ N, or a" = JV,
and y=zlogj^ N, or b'^=N,
therefore , a/>^—.jfV...
Expressing this equation in logarithms to base a, we have
»
and expressing it in logarithms to base 6, we have
- xlogj,a=y. '
Eliminating x and y from the last two equations, we obtain
logfc a log„ 6=1, (153)
M^hich expresses the relation between the bases of any two systems.
COMMON LOGARITHMS. 105
g8. Properties of Common Logarithms.
In the common system, the logarithms of all nuniborn which
are integral powers of 10, are immediately known. Thua
10'>=1, therefore log 1 =0,
10^ = 10, " log 10 =1,
10^=100, « log 100 -2,
lO^'^lOOO, « log 1000 = 3,
•
«kc., (fee.
Hence it appears that the logarithms of all numbers which
are not integral powers of 10 consist of either a fraction or an
integer and a fraction.
Thus, the logarithm of every number between 1 and 10, lies
between and 1, or is a proper fraction; the logarithm of
every number between 10 and 100, lies between 1 and 2, or is
1 plus a fraction ; the logarithm of every number between 100
and 1000, lies between 2 and 3, or is 2 plus a fraction, and so
on.
The integral part of a logarithm is called the characteristicj
and the decimal or fractional part the mantissa. Hence, if a
number is between
1 and 10, the characteristic of its log =0,
" =1,
" =3, ;
• •••
« =n-\.
Therefore the characteristic of the logarithm of a number
which has n integral places, is 7i - 1, or is less by unity than
the number of integral places in the number.
10 «
100,
((
n
((
100 "
1000,
((
H
I(
1000 "
10000,
((
K
((
• •••
....
IQn-l a
10",
<(
((
u
106 PLANE TRIGONOMETRY.
Again, since *.
10" = 1 = 1, therefore log 1 = 0,
10-^= t\7 =.1, " log.l =-1, •
10-*'^=tAu =-01, " log .01 = -2,
10-^ = TT5W--001, " log .001 =-3,
&c., <fec., «fec.
Therefore the logarithm of every number between 1 and .1
lies between and - 1, or is - 1 plus a fraction ; the logarithm
of every number between .1 and .01, lies between - 1 and - 2, or
is - 2 plus a fraction ; the logarithm of every number between
.01 and .001 lies between - 2 and - 3, or is -3 plus a fraction,
and so on.
Hence if a number is between
1 and .1, the characteristic of its log =1,
.1 " .01, " « « « ^'2
.01 " .001, « « « " ="3
.001 " .0001, « " « -' -1 "■'
~ > .
....■...•
....
10-» lO-Ci+i) « " « u ^ui^\
Therefore generally if a number has n ciphers -after the
decimal point, the characteristic of its logarithm is (tmT).
The negative sign is written over the characteristic to shew
that it alone is negative, while the mantissa, upon this supposi-
tion, is Always positive.
Hence, we have the following general rules for finding the
characteristic of the logarithm of any number. "" .
Rule I. — The characteristic of the logarithm of any number
greater than unity, is positive and less by unity than the
number of integral places in the given number.
-■#
COWMON LOGARITHMS. 107
Rule II. — The characteristic of the logarithm of a decimal
fraction, is negative, and numerically greater by anity than the
number of ciphers after the decimal point.
Thus, for the following; numbers, 14067. 521.64, 6.721,
.364, 04271, .0027, .0001672, the characteristics are respec-
tively 4, 2, 0, T, 2, "3, 1.
And conversely, if logarithms be given, having character-
istics 0, 1, 2, 3.... n, there are in the numbers to which these
respectively belong 1, 2, 3, 4....(w + l) digits; and if the
characteristics are 1, 2, 3....7^, there are in the corresponding
numbers 0, 1, 2 .... (»- 1) ciphers respectively after the decimal
point. , :• :
99. When the logarithm of any number is known, we can
at once write down the logarithm of any other number having
the same significanf digits, but differing only from the given
number in the position of the decimal point.
Let N be any number whose logarithm is known, then
10" X A^ is a number which has the same significant digits as
N, but with the decimal point moved n places to the Hght ;
N
again, -— or 10-" x iV is a number which has the same signifi-
cant digits as N, but with the decimal point moved n places to
the left. • . • ^ -
Now, log (10» X N) = log 10» + log iV, •
= H log 10 + log iV,
= n + log N' ; ". •
and log (10-« x N) = log 10-» + log N,
, . ., = - n log 10 + log Ny
— -n + \og N".
Hence, as n is an integer, the logarithms of 10" x iV and
10-" x N differ from the logarithm of iV in the characteristic
108
PLANE TRIGONOMETRY.
only ; therefore by giving the logarithm of N its proper char-
acteristic in accordance with the rules investigated in the last
Article, we can at once deduce those of 10" x. N and 10~»* x N.
Thus, log 2436000 = 6.386677,
log 24360 =4.386677,
log 24.36 =1.386677,
log. 2436 =1.386677,
"log .002436 =¥.386677.
IOC. In this system it is only necessary to register the
mantissa3 in the tables, for the characteristics can be determined
by the preceding rules.
These properties render the common tables less bulky and
more comprehensive than those computed for any other base,
and g've them an advantage over all others.
lOl. The arrangement and use of the common tables are
easily understood from the following portion of a table taken
from Chambers's Logarithms :
No.
1
2
3
4
5
9022
9921
0820
1719
6
7
8
9
D
90
4829
4830
4831
4832
6838672
9471
6840370
1269
8662
9561
0460
1359
9
8752
9651
0550
1449
18
8842
974 L
0640
1539
27
8932
9831
0730
1629
36
9112
.011
0910
1808
9202
.101
1000
1898
63
9291
.191
1089
1988
72
9.381
.280
1179
2078
81
Difif.
90
45
54
In this table we have the natural numbers between 48290
and 48329, and the mantissse are between .6838572 and
.6842078. The four left hand figures of the natural numbers
are placed in the column marked No., and the fifth figure at
the top of one of the other columns. As the first three figures
of the mantissse are the same for several consecutive numbers
LOGARITHMIC TABLES. 109
in the table, they are not repeated for each logarithm, but regis-
tered once only. Thus, the mantisssB of the logarithms of all
numbers from 48290 to 48305 inclusive, contain the same three
initial figures, viz. : 683, which are registered opposite the num-
ber 4829 and in the column under 0, while the last four figures
of each mantissa, are registered in the columns 1, 2, 3, &c., to
which they belong.
Thus, the mantissa of log 48296 = .6839112
«* log 48303 = .6839741
&c., ao.
In the second horizontal line and in the column under 6, a
dot is found in place of a figure. This indicates that the value
of the three initial figures has changed, and therefore both here
and all along the remainder of the same line, the three initial
figures taken from the horizontal line next below, are to be used.
The places of the dots are to be supplied with ciphers. Thus
the mantissa of log 48308 = .6840191.
In this manner we obtain from the tables the mantissas of
the logarithms of all numbers which contain, three, four or five
digits. The logarithms of all numbers between 1 and 100 are
generally tabulated separately. At the bottom of each page
there is a " table of proportional parts" by which the mantiss86
of the logarithms of all numbers containing six or seven digits
may be calculated approximately ; and conversely, the number
corresponding to a given logarithm whose mantissa cannot bo
exactly found in the tables, may be found approximately to six
or seven places of figures. . >
In the construction of the "table of proportional parts," it is
assumed that the increase in a logarithm is proportional to the
increase in the number. The proof of this principle will be
given in the chapter on the computation of logarithms. The
110 PLANE TRIGONOMETRY.
■ .v.
results given by it are only approximate, but the error is gener-
ally so small that it may be safely neglected, especially if the
numbers are large. ' • •
The last column marked D contains the difference between
two consecutive mantissse ; thus in the above table,
the mantissa of 48304 = .6839831
" 48305 = .6839921
Difference corresponding to 1 = .0000090
The significant digits only of this difierence, are registered
in column D.
Now, suppose the logarithms of 48304.1, 48304.2, 48304.3
(fee, are required, we have by the principle just stated
1 :.l
1 :.2
1 :.3
.0000090 : .0000009
.0000090 : .0000018
.0000090 : .0000027
&c., (fee, &c. '4»
HtMice. the mantissa of log 48304.1 = .6839840
log 48304.2 = .6839849
log 48304.3 = .6839858
&C., &C., •~' --TBC. '
The significant digits, 9, 18, 27, 36, <fec., of the proportional
parts found above, are registered at the foot of the page, oppo-
site to the whole difference 90, and in the columns 1, 2, 3, 4,
tkc, to which they respectively belong. The manner of using
this table will be easily understood from the following examples :
^tc. i.— Find the logarithm of 483136.
[n the table we find log 483130 = 5.6840640 '
Proportional part for 6= ■ M
Therefore ! ', t log 483136 = 5.6840694
;_ LOGARITHMIC TABLES. Ill
Ex. ^.— Find the logarithm of 48.29689.
In the table we find log 48.29600 = 1.6839112
Proportional part for 8 = 72
" « 9= 81
Therefore log 48.29689 = 1.6839192
Here we observe that since the local value of the seventh
digit (9) is one-tenth of that of the preceding digit (8), the
proportional part for the seventh digit is one-tenth of what it
would be if it Occupied the place of tho sixth digit, and there-
fore must be moved one place to the right, as shewn above.
For the same reason the proportional part for the eighth
digit must be moved two places to the right, as is seen in the
following example : '. ~
Ex. 5.— Find the logarithm of .48306357.
From the table, we have log .48306000 =1.6840101
Proportional part for 3 = 27
« »« 5 :- 45
•• * . 7- 63
Therefore log .48306357 = 1.6840133 ....
102. To find the Natural Number corresponding
to a given Logarithm.
If the mantissa is exactly found in the table, the first four
figures of the natural number will be found opposite to them"
in the column headed No., and the fifth figure at the top of the
page in the column from which the last four figures of the man-
tissa were taken. The position of the decimal point is deter-
mined by the characteristic, according to Art. 98.
If the mantissa is not exactly found in the table, find the
next less mantissa, and take out the first five figures of the
112 PLANE TRIGONOMETRY.
natural number as before. The additional figureB may be found
approximately by the proportional parts, as in the following
examples :
Ex. 1. — Find the number whose logarithm is 2.6841161.
The given mantissa is .6841161
Next less in the table is .6841089, and nat. No. = 48318
Difference 72, and nat. No. = 8
Therefore aH the digits are 483188, and since the character-
istio is 2, the required number is 483.188.
Ex. 2. — Find the number whose lopfarithm ia 2.6840700.
■'o-
The given mantissa is -6840700
Nest less in table .6840640, and nat. No. = 48313
Difference 60
Proportional part next less is 54, and nat. No. = 6
Residual difference 6, and nat. No. = 7
Therefore the digits of the number are 4831367, and since
the characteristic is 2, the required No. = .04831367.
103. Multiplication is performed by the prin-
ciple of Art. 91. •
^a;.— Find the product of .00098, .0761 and 41.
Let OS =.00098 X. 0761 X 41,
then logic = log .00098 + log .0761 + log 41.
log .00098 =1.991226
log .0761 =2'.881385 /
log 41 =1.612784
log X =3.485395
therefore «? . =.003057698
USES OF LOGARJTII.MS. 113
The student will observe that the characteristics are added
algebraically ; thus, in the above example we have 2 " to carry "
to the characteristics,
then 2 + 1 + ( - 2) + ( - 4) = - 3.
104. Division is performed by the principle of
Art. 92.
^x— Divide 18.792 by .0007834.
Let X =18.792^.0007834,
then log X ^\og 18.792 -log .0007834.
log 18.792 =1.2739730
log .0007834 = 4^8939836
log X =4.3799894
therefore x = 23987.74
Here we have 1 '' to carry " to the characteristic - 4, which
makes - 3. This subtracted from + 1 gives + 4, that is, the
characteristics are subtracted algebraically.
105. Although negative numbers have no real logarithms,
yet they may be multiplied or divided by means of logarithms,
if "We consider the quantities as positive, and prefix the proper
sign to the result, according to the rules of algebra. To indicate
that the natural number is negative, we append the letter n to its
logarithm; thus the logarithm of -43 is written 1.633468571.
Ex. Divide 5 by, - .029. ':
log 5 = 0.6989700
log -.029=1.4623980vi • . ,.
log -172.4131 =2.2365720n
therefore the quotient is -172.4131.
114 PLANE TRIGONOMETRY.
io6. If a Logarithm be subtracted from- lo, the
remainder is called its Arithmetical Complement.
M
Let . X ~ Yf »
then log X = log M- log N
= logif+(10-logiV^)-10,
but, by the definition, (10 — log N^) is the arithmetical comple-
ment of the logarithm of iV, and is written ar. co-log iV,
therefore log a? = log M+ ar. co-log iV - 10.
Hence, by means of the arithmetical complement the pro-
cess of subtraction of logarithms may be converted into that of
addition.
Ex. Divide .00815 by .000256.
log .00815= ";
ar. co-log .000256 = 13.5917600
log .00815= 3.9111576
11.5029176
subtract 10
log 31.83593= 1.5029176
therefore the quotient is 31.83593.
107. Involution and Evolution are performed by
the principle of Art. 93.
Ex. i.— Find the 7th power of .091.
Let a; = (.09iy,
then log a; = 7 xlog .091 . . •
= 7x2".959041
= a713,287, •
therefore a; = .00000005 1676. . . j^ :
TABLE OF LOGARITHMIC SINES, ETC. 115
Ex. ^.— Find the 7th root of .00074.
Let a; = (.00074)^
then log ic = — log .00074
T.8692317
_ 7 + 3.8692317 '
7
=1.5527474
therefore a; = .357065.
Here we add - 3 to the characteristic - 4 and + 3 to the
mantissa, in order to make the characteristic divisible by 7.
2\^
Ex, 3. — Find the value
-(^y
Let X = the value of it,
3 2 3
then log aJ = g log y- = g- (log 2 - log 7)
1367796
5 + 1.367796
, _ 5
= 1.273559, .
therefore a; = .18774.
io8. Table of Logarithmic sines, etc.
This table contains the logarithms of the natural sines,
cosines, &c., computed at intervals of 1' or 10", and is arranged
in precisely the same manner as the table of natural sines, &c.,
described in the last chapter, *
llf] PLANE TRIGONOMETRY.
As the sines and cosines of all angles, the tangents of angles
less than 45°, and the cotangents of angles greater than 45°,
are less than 1, their logarithms have negative characteristics
(Arts. 90 and 98). In order, then, to avoid the use of negative
characteristics and to secure uniformity in the entire table, the
logarithms of all the trigonometrical functions are increased by
10, before being registered, and the logarithm so increased is
called the Tabular Logarithm of the function, which is usually
denoted by Log.
As
Thus, sin 60° = -^ = . 8660254.
* log sin 60" =1.9375306.
(Tabular) Log sin 60" = 9.9375306.
Of course the real logarithm of any function is found from
the tabular logarithm by subtracting 10.
109. Since
cosec A = -: — 7 ,
sm A
we have cosec A sin -4 = 1
and log cosec A + log sin A = Q \
adding 10 to each of these, we have
!
(log cosec ^ + 10) + (log sin ^ + 10) = 20, i
or Log cosec A + Log sin A = 20,
hence Log cosec ^ = 20 - Log sin A. (1^4)
In a similar manner we find
Log sec ^ = 20 - Log cos A. 0-^^)
and Log cot ^1 = 20 - Log tan A. (1^6)
LOGARITHMIC TABLES. 117
Again, since
. sin A
tan A = 7
cos A
we have tan A cos A=am A
and log tan A + log cos A = log sin A
(log tan ^ + 10) + (log cos ^ + 10) = 10 + (log sin ^ + 10)
or Log tan A + Log cos A = 10 + Log sin Ay
hence Log tan ^ = 10 + Log sin ^ - Log cos ^. (157)
no. To find the Tabular Logarithmic Function
of a given Angle.
If the given angle is found exactly in the table, the required
quantity is at once obtained. When the angle is not exaCbtly
found in the table, the logarithm of the function is found
approximately by the method of proportional parts, as in the
case of the natural functions. (See Art. 84.)
Except near the lim^* ' , of the quadrant, the increase or
decrease of the logarithm of a trigonometrical function, is very
nearly proportional to the increase of the angle. This state-
ment will be proved in a subsequent chapter. ,,
The results given by the method of proportional parts, are
only approximate, but the error is in general so small that it
may be safely neglected. w
The application of this method will be easily understood
from the following examples :
Ex. jr.— Find Log sin 17° 18' 24".5. • .
From the table we have
Log sin 17' 18' 20'' = 9.473439 oy., ...r^ -
Log sin 17° 18' 30" = 9.473507 -
Difference for 10"= .000068 .
118 PLANE TUiaONOMETJtV.
Then we have
.10" : 4".r) :; .OOOOGS : .000030
hence Log sin 17° 18' 24".5 = 9.473469.
Ex. ^.— Find Log cot 42" 17' 53".
From the table we Hnd that the difference for 10" is .000042.
Therefore we have
10" : 3"::. 000042 : .000013
heuco Log cot 42° 17' 53" = Log cot 42° 17' 50" - .000013
= 10.041034 -.000013
= 10.041021
III. To find the Angle when the Logarithm of
the Function is given.
Ex. i.— Given Log sin A = 9.473469, find A.
From the table we find
Log sin 17° 18' 20" = 9.473439
and Log sin IV 18' 30" = 9.473507
Difference for 10" = .00006 8 ^
Hence it is evident that the required angle must lie between
17° 18' 20" and 17° 18' 30". «'
The given Log sin A = 9.473469
Log sin 17° 18' 20" = 9.473439
Difference = .000030
Then we have
.000068 : .000030 :: 10" : 4".5,
therefore ^ = 17° 18' 20" + 4."5 = 17° 18' 24".5.
LOGARITHMIC TABLES. " 119
Ex. ^.— Given Log cos ul = 9.893586, find A.
From the table we find
Log cos 38° 29' 30" - 9.893595
and Log cos 38" 29' 40" = 9.893578
Diflference for 10"= .000017
The given Log cos = 9.893586
Log cos 38° 29' 40" = 9.893578
Difference = .000008
Then we have
.000017 : .000008 :: 10" : 4". 7, •
therefore A = 38° 29' 40" - 4".7 = 38° 29' 35".3.
The chief point for the student to bear in mind here is, that
the proportional part for the sine, tangent and secant must be
added, while that for the cosine, cotangent and cosecant must
be subtracted for the reason already given in Article 84.
112. The method of proportional parts which has just been
used for finding the logarithmic functions of angles involving
seconds and fractions of a second, is sufiiciently accurate in all
cases, except near the limits of the quadrant where, from an
inspection of the table, it is seen that the differences for the
Logarithmic sines, tangents and cotangents of angles less than
2° or 3°, and for the Logarithmic cosines, tangents and cotan-
gents for the last 2° or 3° of the quadrant, are very variable,
and therefore the proportional part cannot be accurately found
by this method, since it is computed on the supposition that the
differences are constant for a difference of 1' or 10" in the angle.
The logarithmic functions of angles near the limit of the
120 PLANE TRIGONOMETRY.
quadrant, are computed with extreme accuracy by the following
process. A special table is formed containing for every minute
from 0" to 2°, the Logarithms of
sin A . tan A
and
which vary quite slowly and uniformly for the first 2°, and
therefore may be accurately found from the table for any inter-
mediate value.
Thiis, giving A the. values 30', 31', 32' &c., in succession we
have from the former of the above expressions,
Log sin 30' = 7.9408419
* Log 1800" = 3.2552725
Log b.n 30' - log 1800" = 4.6855694
In a similar manner we find
Log sin 31' -log 1860" = 4.6855690
Log sin 32' -log 1920" = 4.6855686
(fee, &c., &c.
These numbers are tabulated under the heading Log sin A
- log A" ; and in a similar manner is formed a column under
the heading Log tan A - log A". The cotangent being the
reciprocal of the tangent, a column is also formed under the
A"
heading Log j or Log cot A + log A".
tan ^
Since the cosine of an angle is the sine of its complement,
the column headed Log sin A - log A" at the top, is marked
Log cos A - log. comp. A" at the bottom, and when read up-
wards answers for the cosines of angles near the close of the
quadrant.
SPECIAL LOGARITHMIC TABLE. 121
Again, to find the tangent of an angle near 90°, we find the
tangent of its complement and then take its reciprocal, ao that
the column headed Log tan A + log comp. A", is formed by sub-
tracting the numbers found in the column headed Log tan A
- log A", from 20, according to (156).
The use of these special tables is easily learned from the
following examples :
Ex. l.—li A ■= 31' 17". 4, find Log sin A.
Here ^ = 1877". 4.
From the special table we find
Log sin A -• log A" = 4.6855689
Add log ^" = 3.2735568
Log sin ^ = 7.9591257
Ex. ^.— Given A = 17' 5".3, find Log tan A.
Here .4 = 1025". 3.
From the special table we have
Log tan A - log A" = 4.6855784
Add 'log .1" = 3.0108510
Log tan 17' 5".3 = 7.6964294
Ex. 5.— Find the Log tan 89" 47' 12". 8.
The complement of this angle is 12' 47".2 = 767".2.
From the table we get • ^ i;,m,vi;
Log tan 89° 47' 12".8 + log 767".2 = 15.3144232
Subtract log767''.2= 2.8849086
Log tan 89° 47' 12".8 = 12.4295146
122 PLANE TRIGONOMETRY.
Ex. 4.— Given Log sin A = 8.0794466, find A.
From the ordinary table we find that A = A:V 16" nearly,
therefore from the special table, we have
Log sin A - log A" = 4.6855645
but Log sin A = 8.0794466
therefore log ^" = 3.3938821
and ^ = 2476". 75
= 41'16".75.
Ex. 5.— Given Log tan A = 12.7478654, find A.
From the ordinary table we find that A = 89° 54' nearly,
therefore from the special table, we have
Log tan A + log comp. ^" = 15.3144251
but Log tan A = 12.7478654
therefore log comp. ^"= 2.5665597
and comp. ^ = 368".604
= 6' 8". 604
hence ^ = 89° 53' 61".396.
113. The best seven-figure tables hitherto published are
those by Dr. L. Schron, with an introduction by the late Pro-
fessor De Morgan. They contain the logarithms of numbers
from 1 to 108000, and of sines, cosines, tangents and cotangents,
to every 10" of the quadrant, with a table of proportional parts.
Of the other seven-figure tables published in England, may
be mentioned Hutton's, Chambers's, Babbage's and Shortrede's,
the last of which gives the log. functions to every second of the
quadrant.
The best American seven-figure tables are those of Stanley.
They give the log. functions to every 10" for the first 15°.
EXAMPLES. 123
The best American six-figure tables are those of Loomis and
Olney. The former gi\es the log. functions to every 10" of the
quadrant, with a table of proportional parts very conveniently
arranged ; the latter gives the log. functions to every minute
of the quadrant.
Loomis's tables are sufficiently extensive for all ordinary
purposes, and will be used hereafter in the examples of this
work.
Examples.
1. Find the logarithm of 256 to the base ^fS.
Let X = its logarithm ,
a? - ■ '
then 8^ = 256
to
or 22=2'
hence "^ = ^ ^^^ a; = 5 J.
2. Find the logarithm of 10 to the base
1^
3
Let x =
= its logarithm ,
then
(«•-«
and
flJ log ^ = log 10 = 1, (A
therefore
1 1
"^ . 1 - log 3
= - 2.0959.
3. The logarithm of a certain number to base 5 is n times
the logarithm of the same number to base 3 ; find n.
124 - PLANE TRIGONOMETRY.
If N denote the number and x and y the logarithms, we
have ,
5«' = 'N and 3i/ = iV^, •
therefore h^ ^^v
and X log 5 = 2/ ^og 3. (Art. 93)
But by the question x = ny, since x = logs ^ ^^^^ y = logg N,
therefore ny log 5 = y ^^o ^t
log 3
whence n = -, .
log 5 .
4. Given log i =1.698970, and log - =1.522878, find the
logarithms of |/3, f 2 and 1440^.
Log i = log 1 - log 3 = 1.522878
-log 3=1.522878
or log 3 = 1 -.522878
= 0.477121
therefore log ^3 = — log 3.
= 0.238560
Similarly we find
log f 2" = 0.100343 ;
log 1440^ = |- log 1440
= I log (10 X 2* X 32)
= 1^ (1 + 4 log 2 + 2 log 3)
= 1.263345. '
5. Find log 243 to the base 3*. irxr >i^; • . ; Ans. 7|.
EXAMPLES. 125
2
6. If the log of 9 is -, what is the base? Ans. 27.
o
7. Given log 98 = 1.991226, and log 112 = 2.049218, find
the logarithms of 50, .7 and 1750.
Am. 1.698970, T.845098, 3.243038.
8. Given log 1^ = . 096910, and log i =1.045757, find the
9
•logarithms of 2| and 2^.
Ans. 0.397940, 0.352183.
9. Shew that 5 log ^ + 3 log — - + log — = log 2.
10. Find the 50th root of 10. Am. 1.047126.
11. Find the value of (-)^*. Ans. .612295.
12. Find the 17th root of .071852. Am. .8565.
13. If a, 6, c be in geometrical progression, prove that
loga **> logb w> logo '* a,re in harmonical progression.
14. Shew that log . . oosec^ = log ^ sec ^.
©sin A Ocos/4
15. Find x from the equation 8* = 1000, having given
log 2 = . 301030. • ^71*. a; = 3.3219.
16. Given log 1.4 = .146128, log 144 = 2.158362, and
log 441 = 2.644438, find the logarithms of the nine digits.
17. Given log 2 = .301030, and log 3 = .477121, find the
logarithms of 135, 405, 3.24 and — ,
9
Am. 2.130334, 2.607455. 0.510545, T045757
18. Find the value of ^-^IlL^^il^lL . Am. .340653.
(19)* X (.061)2
19. Findlog2 6. '. Am. 2.584.
12G PLANE TRIGONOMETRY.
1
20. If 32«.5'«-i =23«+i shew that « = -— .^ -, s-
log 45 - log 8
21. Given xV =y^, and o? — y\ find a; and y.
9 27
22. Given 28«'.52a;-i =45«.3a'+i, find x
Ans. -.991.
23. Given Log cot 68° 21' 10" = 9. 598661,
Log cot 68° 21' 20" = 9.598600,
find Log cot 68° 21' 16".4.
Ans. 9.598622.
24. Given Log sin 37° 4' 40" = 9.780244,
Log sin 37° 4' 50" = 9.780272,
find Log cosec 37° 4' 47".
Am. 10.219736.
25. Given Log cos 75° 13' = 9.406820,
Log cos 75° 14' = 9.406341,
find Log cos 75° 13' 22".4.
Ans. 9.406641.
26. Given Log tan ^ = 10.572676,
find A, having given
Log tan 75° 1' 20" = 10.572622,
Log tan 75° 1' 30" = 10.572706,
Ans. A =-75° 1' 26".2.
RIGHT-ANGLED TRIANGLES.
127
CHAPTER VIII.
FORMULA FOR THE SOLUTION OP TRIANGLES.
RIGHT-ANGLED TRIANGLES.
114. The formulse for right-angled triangles have been
already given in Chapter III.
They are immediately derived
from the definition of the sine,
cosine and tangent. Thus,
from the right-angled triangle
ABC, we have
sin A = -, cos ^ = --, tan ^ = -r.
c c "
(158)
Special Formulae for Right-angled Triangles.
115. When A is nearly 90°, the first of this group cannot be
accurately computed from the tables, since the sines then^iffer
very little from each other; and for a similar reason, the second .of
the group cannot be accurately found when A is very small, or when
6 sTearly equal to c. Hence, an angle near 90° should be deter-
mined by its cosine, and a small angle by its sine or tangent. Special
TormX are therefore frequently necessary, especially in surveying
and astronomy, where great accuracy is required.
From (100) we have
2sin2 2-=l-cos4.
Writing 90°-^ for A, this becomes . . ,;^ ■
A
2Bm2 (45°_-)=l-8inil
=1 - - , by the first of (158)
c
a
128 PLANE TRIGONOMETRY.
•
whence . «"^ (^^°- 2~^=s]^17^' ^1^^)
■which may be used with advantage when A is near 90°.
Il6. From the second of (158), we have
sin" ^=1 - cos'^ A=l - —r. ,
whence sm ^= I —
l/(c+6) (c-b)
= ^ , -, (160)
b
which may be used instead of cos A=—, when A is small.
. From (100) we have
A
2 sin^ 2"=l""Cos -4
whence c — b— 2c sin^ -^ , (161)
by which c — b may be accurately found when A is small.
.117. From (83) we have s ;;
tan A — 1
=— — . , by the third of (158)
. a-b '■ '
a
which may be used instead of tan A=j-, when A is near 90°.
118. From (101) we have , v
. ^ |l-cos A ; ' •
tan — =^
2 Njl+cos^ • ;,
OBLIQUE-ANGLED TRIANGLES.
129
1-A
by the second of (158)
■J
1+-
6 c-b
whence
c+6
A
— &=a tan -x- ,
a
(163)
by which c - 6 may be accurately found when A is near 90^.
Since o» - b^—a^, we have from (163)
c^-b^
Q-b
a'
a tan
2
A
(164)
or c-\-b—acoi-^i
by which c-|-6 may be found when A is near 90*.
OBLIQUE-ANGLED TRIANGLES. i,%; ^
iig. The sides of a Triangle are proportional to
the sines of the opposite Angles.
In the triangle ABG^ let the sides opposite to tlie angles A,
B and G be denoted by a, h, c respectively. Draw CD perpen-
dicular to AB, produced if necessary.
Then, in the right-angled triangles ACD, BCD, we have by
the first of (158)
(Fig. 1) CD = b sin A, and CD = a sin B,
whence 6 sin il = a sin 5/
10-''
130 PLANE TRIGONOMETRY.
and (Fig. 2) CD = h sin A, and GD = a sin CBD = a sin B,
since CBD and ABC are supplementary angles,
whence 6 sin il = a sin 5.
Therefore, in both cases we have
h sin ^=a sin 5,
which may be converted into a proportion, thus,
a '.h '.'. sin A : sin B.
In the same way we may shew that
a : c : : sin ul : sin (7
and 6 : c : : sin J8 : sin C
These three proportions may be written as one, thus,
a ; 6 ;c ::siD il : sin ji5 : sin C,
or better thus,
^ 'L.^JL^. (165)
sin -4 sin ^ sin C
120. The Sum of any two sides of a Triangle is
to their Difference as the tangent of half the Sum
of the opposite Angles is to the tangent of half
their Difference.
, From (165) we have
a sin-il
6""siniS'
whence, by composition and division, * .
a + 6 sin il + sin 5
a - 6 ~ sin il - sin 5
^toi(4+|) (59) (156)
tan \{A - B)
OBLIQUE-ANGLED TRIANGLES. 131
Since ^+5 + C=180",i(^ + 5) = 90°-2- *
G
and tan ^( J + J5) = cot - .
Hence (166) may be written
tani(^-£) = ^cot|. (167)
Similar relations may at once be inferred between 6, c, B, C
and a, c, A^ G.
The last equation may be written thuB
»4 c
tani(i-5) = — ^cotg.
1+ —
a
Let — = tan ^
an assumption always possible, since a tangent may have any
value between and oc ; then we have
,.. «v 1-tan^ G
*^^i(^-^)=rTt^-^^'2 " ^ -
= tan(45°-e)cot|. (168)
In the last three formulae a > 6, therefore we may put
b
a
1 - cos (h G
then . ^h(^-^)-iT^^''°^2
'■'ir^-
= cos <6t .
+ COS <f>
tan^cot^. ' (168 6i5)
132
PLANE TRiaONOMETRY.
121. Geometrical proof of (i66).
Let ABC be any triangle of which the side BC is greater
than AC. With C as a centre and ACj the shorter of the two
sides, as radius, describe
a circle cutting AB in F^
BC in E, and BC pro-
duced in D. Join AD,
CF, AF, and draw FH
at right-angles to AB.
The angle FAD ia &
right angle ; the angle
AFC is equal to the
angle CAF; the exterior angle AC^ is equal to the angles
CAB, ABC, that is, A + B. But the angle AFC is half of the
angle ACD (Fuc. III., 20), therefore the angle AFC is equal
to ^(il + B). Again, the angle BCF is equal to the diflference
between the angles CFA and CBF {Fuc. I., 32), that is, the
angle BCF is equal to A - B ; but the angle FAF is half of
the angle ECF, therefore the angle FAF is equal to \(A- B).
Now, AD = AF tan AED = AF tan \{A + B)
and FH=AF tan FAF= AF tan ^{A - B),
and since ^5" is parallel to AD, we have
BD '. BE '.: AD '. EH
that is, a + 6 :a-6 :: il-^tan J(^ + ^) : ^^ tan \{A - B)
j: tan J(^ + B) : tan J(J - B)
therefore tan \{A -B) = ^5—-. tan i(A + B).
OBLIQUE-ANGLED TllUNGLES.
133
122. To express the cosine of an Angle of a
Triangle in terms of its sides.
From the angle C of the triangle ABC, draw CD perpen-
dicular to AB^ produced if necessary.
From Fig. (1) we have by Euc. II., 13,
BC'^ = A(P^AB''-2AB.AD.
But AD = ACQO^Ay
therefore BC r.AG' + AB'- 2AB.AC cos A,
that is, a'^ = 6^ + c' - 26c cos A.
Again, from Fig. (2) we have by Euc. II., 12,
BC^ = AC^-\'AB'-\-2AB.AD. ■ .
But ^i) = i4C co8(7^i> = ilC 008(180"-^)= -ilC cos J,
therefore BG^ ^^AC + AE"- 2AB.AG cos A,
that is, a^^b^ + c^- 2bc cos A.
Hence, in both cases, we have
a2 = 62 + c»-26ccos^. , (169)
In the same manner, we find ' i' . : ^^'
b^=:a'^ + c'-2accosB. :•, ; ; ./r(170)
c2 = a2 + 62_2a6cosC. . (171)
From the last three formulae, we have
b^ + c^ -a^
cos A =
cos B =
2bc '
a' + c'- 6"
2ac
(172)
(173)
134 PLANE TRIGONOMETRY.
123. The same results may, be obtained independently of
Euclid, as follows :
In Fig. (1) we have c = BD + AD,
= a cos -B + 6 cos A ;
and from Fig. (2) c = BD-AD,
= a COS B-b cos (ISO" 'A)t
= a cos B + b cos A ;
and similarly for the othei sides. Hence, in every triangle,
we have
a = b cos C + e cos B. '
b = c cos A + a cos C.
c = o cos B + b cos A.
(175)
Multiplying the first by a, the second by 6, and the third
by c, then subtracting the first result from the sum of the other
two, we have
a* = 6' + c'- 26c cos -4,
, 6« + c' - a» , ,
whence coSil = rr » as before.
12/,. To express the Functions of half of an
Angle of a Triangle in terms of its sides.
From (100) we have
A
2 sin'^ ^ = 1 - cos i4, .
2bc'b''-c'+a?
"" 26c * '
OBLIQUE-ANGLED TRIANGLES. 135
Ihc *
_ (a-b + c){a + b-c)
Wc • ^^'^^
This may be simplified by putting
whence -a + b + c = a + b + c-2a — 2(s-a).
a-b + c = a + b + C'2b = 2{s-b).
a + b-c = a + b + c -2c=2(8- c).
Hence (176) becomes, after dividing by 2 and extracting
the square root,
In the same manner we find from (173) and (174)
2 N oc ' ■• ■!
(178)
^'"^nI^^^^T- . "•^(^")
From (99) we have . ^i c, ; . \
2 cos'' — = 1 + cos ^,
-i+-^6r-' ^y(^^2)
2bo *
jb + cy-a'
26c *
{a + b + c) (-a + 5 + c)
26c *
136 PLANE TRIGONOMETRY.
_ 25 . 2(s - a)
267~'
Therefore, cob~:=^'^^^. " (180)
Similarly, <^o^o" = \|~^ ' 0^^)
-4=nI'-^^- • <^«^)
ah
Dividing (177) by (180) we have
^ A j(8-b)(s-c) ,,„„,
tan^=^ ^ „,„_,, \ '■ (183)
«(s-a)
and in a similar manner
125. To express the Sum and DilTerence of any
two sides of a Triangle in terms of the other side
and the opposite Angles. ' v
From (165) we have
a+6 sin ^4-8in B ' ", -
Jl ~ sin ' ^ V
BhiMA±B)coah(A-B) ,
" c > by (55) and (94)
- sin 2" cos -g
but ^+B+C=180°,
therefore -4+5=180°- C, and ^(^-|-JB)=90°-^;
and sin ^(^+5)=cos "2 , cos i(^+5)=8in ^.
OBLIQUE-ANGLED TRIANGLES. 137
Hence we have '^
;r> a+6 cos^cos K^-.B)
"7""' : G*
co8i(J.+ -B)co8 2"
~Zo3l{A-\-B)*
and «+»=''«7|(I+BF- * ^
In a similar manner we find »
Bin ^(A-E) ^g,^^
126. From (165) we have
t^-j-R-fc gin ^H-si n B+sin >: • -
a sin A
2 ai n ^(A4-B) cos ^(.1 -^)+2 sin ^U4-.B) cos hU-\-B)
2 sm 77 COB -X-
., sin \{A\B) Uos ^(^-■B)+cos \{A^B)\ ^
sin ^ COB g-
*
' A B ^ B
.- ; 2 COB ^ cos y cos ^ 2 cos -^ cos ^ ^^^^
sin 2" COS g- sm ^
In a similar manner are obtained the following, which may be
verified by the student :
-j^^_L/. 2sm-o sm -5-
-a-\-t>+c _ 2 2 ^jg9j
a .A
Bm«-
. -.,.^ -- 2
^ B .
^ y,^ 2 COS IT sm ■5-
^"^"^^ , ^ ^ (190)
COSTT
138 PLANE TKiaONOMETRY.
a+h-o 2 81112 cos 2
*:•■!('
a
A
■■"" cos 77
! '■
2
: (a+6+c), the last four
7
• S
cos 17 cos 77
8
2 2
a
. ^
sin 77
2
.5.0
Sm 77 Sm 77
8~a
2 2
a
. A
Bin 77
2
B .
8-b
cos 2 sin 2 ^
a
^
cos 77
2
. B
8-6
sin 2" cos 2
oc
^
cos 2
(191)
(192)
(193)
(194)
(196)
'^V • 8 — C
From these we may at once deduce those for -r, — ^— , &c.
Thus, writing 6 for a in (192) and (193) we have
C
a
cos
2 *'««2
h'
%-h
sin
. B
am 2
^ .
2 ''''2
. 5
am 2
(19G)
(197)
> &C. f &C;
The product of (193) and (197) gives (179), «S;o.
AREA OF A TRIANGLE. 139
127. To express the sine of an Angle of a Tri-
angle in terms of its sides. *i . r . ';
From (94) we have
sin il = 2 sin — cos — ,
= 2>J<-^^^^-sl'^.by(177)and(180)
= ^ Js(s-a) (s-b)(8-c). (198)
128. To express the Area of a Triangle in terms
of its sides and angles.
Let A represent the area of a triangle; then, from the
Figures of Art. 119, we have
£S. = iAB. CD, and CD = AC am A.
Therefore A = pcsin^. (199)
In a similar manner, we find
- A = |ac sin -B = ^a6 sin (7, " > . (200)
that is, the area of a triangle is equal to half the product of
any two sides into the sine of the included angle.
Again, from (199) ..' '
A = ^6csinil 'V
==^ . - Js{s-a){s-h)(s-c), by (198)
= >/s(s-a)(«-6)(s-c). (201)
c sin _o
From (165) we find b = . , which substituted in (199)
gives
_c' sin ^ sin jB c' sin A sin B
2lL^"^ " 2sin(^ + ^) • (^^^^
140 PLANE TRIGONOMETRY.
129. To express the Perpendicular of a Triangle
in terms of the side on which it falls, and the
Angles adjacent to that side.
Let p denote the perpendicular from C on the side c,
then pc = 2A,
and P='—* (203)
c sin il sin 5 , , ^ ,
° sin (^ -Hi?) ' ''y(202) (2(H)
If the triangle is right-angled at C, (204) becomes
p = c sin A sin JBf
-|-sin2il, ^205)
since sin jB = cos A.
Examples.
. .A , .b r*
sin»— sin'_ cos2~
1. In any triangle, shew that + = .^ *
a 6 «
From Art. 124 we have ■
2 _(8-b)(s-c)
t
sin' —
(^ abc
dn«- ,
2 (8~a)(s-c)
aba *
-^A . ^B
sm' — sin' —
therefore --+-_j_=JL^(,.„^.. J)
-, since 29^a + h + e.
ab
EXAMPLES. 1*1
s{s-c) 1^
ab 8
^ by (182).
2. In any triangle, prove that
c" cos (^ - J5) = 2ab + (a^ + 6') coa (A + B),
From (186) and (187) we have
c cos l(A -B)^ia + h) cos J(il + B)
c sin J(^ - 5) = (a - 6) sin J(^ + 5).
Squaring and subtracting, we get
cMcos«i(i-5)-sinH(^-^)} = (a + 6)'cosH(i + 5) ■
^ ^^ ^ (a - 6)« sin" i{A + i?)
c« cos (il - 5) = 2a6 {cos'* i(i + B) + sin' K^ + ^)}
+ (a» + 6^) {cos' K^ + ^) - «^"' i^"* ^: ^) i
or c'' cos (il - 5) = 2a6 + (a^ + 6') cos (^ + 5).
3. In any triangle, prove that ,t * :,
C
t (a + 6)' sin'* 2" * ("* " *)' ^^^ "2
cos (il -J?) '
From the result of the last example we have
c' 008 {A-'B) = 2ab - (a« + b') cos C,
since (i + B) and C are supplementary angles. ,
G C\
Hence c» cos (il - i?) = 2a6(cos' -^ + sin" g")
-(a« + 6»)(co8> 2-sin'2)
i: =(a + 6)»sin»|'(a-6)'cos«g,
142 PLANE TRIGONOMETRY. ^
(a + by sin^ I - (a - b)' cos» |
therefore c' = , . ...
cos [A - B)
4. If the sides a, 6, c of a triangle are in harmonical progres-
sion, prove that
cos' —
B sin A sin G
2 cos ^ + cos C *
Since a, 6, c are in harmonical progression
, 2ac
= . n\
a + c ^ f
As the sides of a triangle are proportional to the sines of
the opposite angles (Art. 119), we evidently have from the
above equation
. „ 2 sin il sin (7
sm B = .
sm ^ + sm (7
For (1) may be written
c a + c 1 /a
2a
_ 1 /a c \
'"2\a'^a)*
sin C l/sin-4 sin (7\
, . „ 2 sin ^ sin C
whence BUiB = —, — . i^ ,.-
sm ^ + sin G
TT o • -'^ ^ 2 sin -4 sin C
Hence 2 sm — cos — =
a .' 2 2 sin |(ii + G) cos i(^ " ^) *
sin ^ sin (7
COS 2 COS ^(A-G)
tx. e 1« 4 -^ sin .4 sin C
therefore rnr „ i
2 ^ »
2sin-cos J(^-0)
EXAMPLES. 143
sin il sin C ; • i .
■ 2co8^{A + C) COB ^(A-C)
sin A sin C . ,-_. i^
The student should notice the following transformations de-
pending on (165), which will be found very useful in the solu-
tion of problems. Whenever the sides of a triangle and the
sines of the opposite angles are homogeneously involved, and
occur either in the numerator or denominator of a fraction, or.
on opposite sides of an equation, we may substitute the one for
the other, as in the following examples :
5. In any triangle, prove that
(a" + 6' + c') sin -4 = a' sin -4 + a6 sin 5 + oc sin (7.
a^ + l^ + c^
Commencing with 7 , we have -.•(,/.-?
a^ + b^ •¥ c^ a a h h c c
ab " a b a b a b ' ,
sin' A sin« B sin» G
sin A sin B sin A sin B sin A sin B
1 /a sin il 6 sin J5 c sin (7\
sm
whence (a' + 6' + c') sin il = a* sin A + ab Bin B-^ ac sin C,
6. Shew that the area of a triangle ABG
^^^^ "^^ sin'^-sin'^C •
From (199) we have
A =ibc sin il
J ,x,j 2\S^^^ sin J5 sin C
■K^
144 PLANE TRIGONOMETRY.
7. If the angles of a triangle ABC be bisected by the lines
AD, BE, CF, meeting the
opposite sides in the points
Z>, E and F, shew that the
area of the triangle ABG is to
the area of the triangle DEF,
as (a + 6) (a + c) (6 + c) : 2a6c.
Here we have ^
^ ^ sin ^^i) sin^gjg sin ^2) (7
sin DAG ~ sin ^i)5 * sin DAG '
BD AG
^AB'DG^ ^y<^^^>
BD^ AG_
^DGAB
BD AB c
whence
and
or
DG~AG~~b
BD^DG 6 + c
BD ^ G
a b + c
BD
c
therefore BD = --^^, and i)C^ —,
b+c' b+c*
similarly EG —/' AE —,
a + c\ a + c*
a+0 a+b
Let A denote the area of the triangle ABG, and 8 that of
the triangle DEF,
then A = J6c sin ^, whence sin ^ = —
be
and fin J5 = — , and sin C= -- .
^ . ad '
EXAMPLES. 145
Area of AEF= \AE . AF sin A, by Art. 128
A fee
(a + 6) {a->rcy *
Area of BFD = \BD . BF sin 5
~(a + 6) (6 + c)*
Area of CiE^i) = JCi> . C^ sin C
Aa6
(a + c) (6 + c)*
Therefore the area of the three triangles, AEF, BDF, CED,
t he ac ah v
" \(a + *) (a + c) "^ (a + 6) (6 + c)^('a + c) (6 + c)j'
and therefore
(ho etc oft \
^""(a + 6)(a + c)"(^+6) (6 + c)"(a + c) (6 + c)j
A . 2a6c , ^
(a + 6) (6 + c) (« + c)*
or A : 8 :: (a + 6) (6 + c) (a + c) ; 2a6c.
Examples.
In any triangle, right-angled at C, prove that
A ITTc \, b'-a' ...
1. cos 7r = *.|-7; — C0S2il = £5 2'
2 > 2c 6* + a2
/ «x 2a6 . /i m (a + 6) (a -6)
2. 008(^-5) = -^ tan(.l-^) = ^ ^^ .
3. ton 2A = jr ^7T—\- ^^ 2il-sec 2B = j—'
(6 + a) (6 -a) ^-a
4. fLti^ J2 Bin (il + ^S"! c= V2»2» cosec 2^1."
c
11
146 PLANE TRIGONOMETRY.
5. A = {c^ sin 2i4 = — tan A = \{a + b + c) (a + b - c).
In any triangle ABG^ prove that
^\Ti{A-B) a^-l? tan5_a' + ft'-c»
^in (^ + i^) ~ c» * t^:^^"a'-62+c'*
versin .1 «(^-& + c) 7? ,«f /. ^
7. ; — n = 77 J ^ • COS B + sm Ji cot C ■— — .
versin i» 6(-a + o+c) o
A ^ B s-c A B 8-b
8. tan — tan — = . tan jr cot — = .
2 2 a 2 2 8-a
^ A ^ B BO
*^°2'-*^''2 c "^*2-''°*2 a
•7. ■
-4^ B a- b ' A 8 - a'
tan — - tan — . cot —
40 ^ J
10. COS ^ + COS ^ = 2 sin'' — . tan — tan — tan 77=0.
c 2 2 2 2 s*'
\, ,A B C a" ,A B G
11. cot — + cot — + cot — = — — cot — cot 77 cot — .
cot- + cot- ^
12. ^ .^ = - . 6 COS 5 + c cos C = a cos (B - C).
cot 77 + cot —
13. a cos ui + 6 cos jB + c COS (7 = 2a sin jB sin C.
14. (6 + c) COS ^ + (a + c) cos -B + (a + 6) COS (7 = a + 6 + c.
-4 B G
15. (« - a) tan — = (s - 6) tan — = (s - c) tan — ,
6 COS C - c COS 5 6 +
16. ; — ' — = • ■ . ., . f.
c
17. cot -4 + cot 5 = — cosec J5. ,. . ■. -^ - \
EXAMPLES. 147
18. c^={a + byHin' ~ + {a-by cob''^,
19. ^ (a' + 6' + c^) " be cos A +ac cos B + ab cos C.
i4 J5 C
20. »(co8 il + cos -S + JOS C)=>'a cos' 7^-1-6 cos' ^ + c cos' — .
21. (6' - c') cot A + {c*' a') cot 5 + (a' - 6') cot (7 = 0.
22. cos -4 + cos -B + cos C =« 1 sin B sin C.
s
ABC
23. (6 - c) cot — + (c - a) cot — + (a - 6) cot -^ = 0.
^, sin il + sin C cos C - cos ^1 ,B
cos A + cos G sm il - sin C 2
sin(i4-Jg) a'-6' a
sin(J?-C)~6'-c'' c' V .:, ^ .: , -.
26. a' sin 2B + 6' sin 2^ = 2a6 sin C.
a cos ^ cos C + cos 5 ,
6 cos 5 cos (7 + cos A
a' - 6' 6' - c' c^ - c^
28. — r— sin C + — 5- sin il + ,„ sin ^ + 4 sin \{A-B)
c' a'* 6^ "^
BinJ(^-C)smMC-^) = 0. • ^' .'■
ABC
cot--+ cot-+ cot-
^^' ~ B abc
sec - sec 2- sec 2- ,/';;(_ ^
^ ^ c M •■■•••-'•■I - '-'^ ■' ":
cos»2 cos»^ ^^ ^
31. cot ^ - cot -4 7 — coseo (7.
^i^Aiv-
148 PLANE TRiaONOMETRT.
A ^. ^ b + c
32
. Bin (- + B) cosec g = —
A S C
33. In any triangle, if cot — , cot -, cot - are in arithmeti-
A C
cal progression, shew that cot — cot — *= 3 ; and if cot A^ cot B,
cot G are in arithmetical progression, shew that the squares of
the sides are also in arithmetical progression.
34. If tlie line GE which bisects the angle (7, of any tri-
angle, meet the base in jE', shew that ■
< „ lah G ^ . „^ a + 6 ^ G
GE = r- cos TT, tan i!AC/ = r- tan — , .
a+6 2 a-0 2
and cos 5 - cos -4 = 2 cos -^ cos AEG
35. In a triangle ABGy if sin Ay sin J?, sin G are in har-
monical progression, so also are the versed sines.
A H C
36. In any triangle ABG^ if tan — , tan — , tan — are in
Z Ji ^
arithmetical progression, so also are cos A, cos J?, cos G,
37. If p, q, r denote the lines dravn from the angles of a
triangle bisecting them and terminated by the opposite sides,
shew that , ,,
A B &
cos^ cos^ ^_i 1 i
jp q r ~ a b G*
38. In any triangle prove that ,^ j
A =«(»-o) tan — . *^*- '• -' v.
5s > >
39. A = OOB TT COS -jr- cos 75" . ' K-
$ « A a
EXAMPLES. 14f9
sin A sin B sin C
s2(cps ^ + COS ^ + COS (7 - 1)
^^' ^" ^^rr+^ni? + sinC
■ ' , , siif ^ sin B
42. A«i(^»-6«)^-(jr-^. ,,.:
,:^.hi
■'•■p-^i
CTa&a(cot^+ COt B) ' ■■'■•'
^^' ^ ^ 2ca cGsec^l; ' :- "■•:^-; xvi^ ^ ^'^^i'r' •■<*^' '^'
.'.1 r.-i-; ■.' .{.,'J I. ..■/.■>^\ii, ■ • l....i; 'u'JfiJ "••t-i'.'; -ti':' ~i -I: .<•!•.'■ •x'
,. ,-. -.,^1.-., .. .X-. oip,i5.i;j ft ^<> vJ /^\ ,{; ><''i*.-."*{
An A : ■ I. ^",d.j H'.hI>< , •-UV-.uV-'I
^^' ^-A{cotA+cotB+cotCy '" '
46. A= J ^ ^'
4(cOt -„■ + cot — ■ + cot — ) , n?v,!.,j 5.: -jju: ui ;»''-
. ^ "" 2(cosecTTcosec B + cosec (7) ' ,. ... .
49. A = i(«^ «i^ 2^ + *' «i^ 2^)- ^"''^ ^^^^^^^^f^'-=" ''^ " '
■••■ • -■ :• ABC ''^^'i^^
50. A =«» tan - tan - tan -. (>) t?
v> .■■}
51. In any triangle, if pi be the perpendicular from tlie
angle A upon the side a, shew that "^
a + b + c b^ sin + c^ sin B
P' = B d= 6 + c
. cot - + cot —
l'">0 PLANE TRIGONOMETRY.
52. If perpendiculars be drawn from ^he angles of a triangle
bo the opposite sides, shew that the sides of the triangle formed
by joining the feet of these perpendiculars are a cos -4, 6 cos B
and c cos C and that its area = ^ab cos A cos £ sin 2(7.
53. If the sides of a triangle are 3^4 and 5, shew that the
cotangents of the semi-angles of the triangles are 3, 2 and 1.
54. If perpendiculars be drawn from the angles of a triangle
to the opposite sides, shew that the products of the alternate
segments of the sides thus made = abc cos A cos B cos C.
55. If jOj, p^y p, be the perpendiculars drawn from the
angles A^ B^ C oi a. triangle ABC to the opposite sides a, b, c
respectively, shew that
^ p^ sin A +p^ sin B-^p^ sin C
Sin B bin U
56. In any triangle shew that
2«*
(sin ^ + sin 5 + sin C)» = -t-(« cos ^ + 6 cos 5 + c cos C).
(aainA + b sin J? + c sinC)« -= (a" +b^ + c«) (sinM + sin^'iS + sin^C).
57. In any triangle, if — = . )^ — J^ , shew that a», 6», c»
c Bin {B-C)
are in arithmetical progression.
58. In any triangle shew that the distance from the middle
of the base (c) to the foot of the perpendicular drawn from the
1 y-, . , 1 -J. - ■, ' c tar A - tan B
angle C to the opposite side, is — ,
2 tan A + tan B
*
69. In any triangle shew that ^^"^ ^^ " ^^ = ^Iz^
, sin C c*
60. In any triangle prove that
a sin (^-C) + 6 sin (C - ^) + c sin (A - 5) = 0.
RIGHT-ANGLED TRIANGLES. 151
CHAPTER IX. ^
ON THE SOLUTION OF TRIANGLES.
RIGHT-ANGLED TRIANGLES.
130. The various cases of right-angled triangles are solved
by (158)-(164). ' .Vr
Ex. ^.— Given ii = 65" 17' 20'' and c = 17.216, to find the
other parts. (See Fig. Art, 114.)
From the first of (158) we have
' a >: e sin ^, '
or in logarithms log a «= log c -1- Log sin ^ - 10.
log c = 1.235932
Log sin il = 9.958290
loga-L194222
therefore a - 1 5. 6395.
Again, from the second of (158) we have _- , ., :■ .
, , ; . h = c cos Ay
or ,,,,,,,, log 6 = log c + Log cos ^ - 10. i, ,i
Log = 1.235932
^ J. .., Log cos -4 = 9.621221 : »> ,.,.',;;^
; '; -i - ^ log 6 = 0.857153
therefore 6 = 7 197
and ' 5 = 90°-^ = 24M2'40''.
152 PLANE TRIGONOMETRY.
Ex. ^.— Given a = .1799 and 6 = .2465, to find the other
parts.
From the third of (158) we have
tan A= — y , 1 ,
OP Log tan ^ = log a - log J + 10.
Loga=L255031
*, ' log 6=1391817 i' -
- - Log tan ^ = 9,863214
therefore -4 = 36° 7' 2r'.5
and 5 = 63°52'38".5.
From the first of (158) we have
a
sin^* ''/'; ^^^ '^'^' '•'■' ^''^■•
o^ logc = loga-Logsin ^ + 10.
<M »^ ' loga = r265031 ^rM^iiiii^.,/
Log sin .4 = 9.770496
log c= 1.484536
therefore : «-. c = . 305 16 6.
mir:.-> ,•■•
Examples.
3. Given c = 332.49 and a = 98.399, to find the angles and
*lie ^ase. Am. A = \T 12' 51", 6 = 317.6.
4. Given 6 = 374 and B = 52° 40' 18", to find the other parts.
An8. a = 285.2, c = 470.34.
5. Given A = 25° 18' 48" and a = 8.5623, to find 6.
Am. 6 = 18.1028.
6. Given a = 48 and 6 = 36, to find A,
Am, ^ = 53° 7' 48". 4.
RIGHT-ANGLED TRIANGLES. 15
7. Given c = 197.01 and a =196.64, to find A.
By (159). log (c - a) =1.568^02
log 2c = 2.595518
2)16.972684
Log sin (45° -4) = 8.486342
2
4
therefore \ 45°-^ = T 45' 21".7
1. . . . . ■.
and ,, \ =43M4'38".3
, . ; . ,1 = 86"29'16".6. -.
• l.!'.l ' f
\/ '• D J
8. Given a = 984.1 and 6 = .32 14, to find A and c.
By (162). log (a - 6) = 2.992919
log (a + 6) = 2.993203
Log tan {A - 45°) = 9.999716
therefore , . , A- 45° = 44° 58' 52."7
and " ' il = 89°58'52".7.
By (163). log a = 2.993039
Log tan 4 = 9-999858
k - V '
'■;^A yr.
■'•h'.<i ;'
log (c- 6) = 2.992897
c-6 = 983.78
therefore " ' c = 984.1014.
9. Given a = 4064 and A = 89* 58' 20", to find b and c.
By (163). loga= 3.608954
:>.!^ ; t^ Log tan ^= 9.999789
log(c-6)= 3.608743
c-b= 4062.027
154 PLANE TRIGONOMETRY.
By (164). loga= 3.608954
Log cot ^ = 10.000211
log(c + ^/)= 3.609165
c + 6 = 4065.98
therefore h = 1.9765 and c = 4064.0035.
10. Given b = 100.56 and c = 100.64, to find A.
4ns. .1 = 2°17'5".3. r
»
11. Given c = 270 and A = 18' 40", to find b.
Arts. 6 = 269.9960197.
12. Given A = 89'' 20' 40" and c = 10, to find a.
., J, , -; ; Ans. a = 9.9993674.
OBLIQUE-ANGLED TRIANGLES.
131. The relation investigated in Art. 119, between the
sides of any triangle and the sines of the opposite angles, fur-
nishes us with two equations involving the three sides and the
three angles, viz. :
sin -4 sin B sin (7 .it ,'
^ ~ -^ :'~^tJ -■''..
We aho have
^ + ^ + C = 180'.
It is evident, then, that if any three of these quantities,
except the three angles, be given, the other three m&j be iound ;
for on substituting the three given parts we would hf.ve three
equations, containing only three unknown quantities, which can
therefore be found by solving the equations. *"" ' '
If the three angles only are given, we can find the ratios,
but not the magnitudes of the sides, from the equations
a sin A a sin A
b sin jB ' c . a C *
OBLIQUE-ANGLED TRIANGLES. 155
Hence it is evident that one side, at least, or what is equivalent
to one side, must be given, in order to solve any triangle.
The only cases then which can occur in oblique-angled tri-
angles are the following :
(1) Given one side and two angles.
(2) Given two sides and an angle opposite to one of them.
(3) Given two sides and a» included angle. Me,/
(4) Given the three sides.
We now proceed to the solution of the different cases.
Case 1. Given one aide and two angles^ or a, B and C, to
find the other parts.
First Solution.
We have " ^ = 180'' - (5 + C)-
From (165) we have '
a sin B • -n ^ a k u-
J _ __ — =aam. B cosec A^ «' »*
sin il
or in logarithms
log & = log a + Log sin B + Log cosec A -20.
a sin C . ^ „ J
n„f1 c=—. — r- = « sin C cosec ^.
^^J. log c = log a + Log sin C + Log cosec A - 20.
Examples.
1. Given a = 512.24, ^ = 54° T 35" and C^7V V 41", to
find by c and A.
A = lSO''-{B + C)
= 180' - 125° 9' 16" = 54° 50' 44"
loga= 2.709473 loga= 2.709473
Log sin 5= 9.908652 Log sin C= 9.975744
Log cosec A = 10.087457 Log cosec A = 10.087457
log 6= 2.705582 log c= 2.772674
' ■ 6= 507.67 c= 592.58
156 PLANE TRIGONOMETRY.
2. Given c = 71.984, ^ = 61", and C = 58° 14', to find a
and 6. _ Am. a = 73.8838, 6-74.05.
3. Given ^=35° 42', i5 = 7^" 27' and c = 142, tolind a
and 6. ^m. a = 89.47, 6= 149.05.
132. Second Solution.
Given c, A and B. to find ihe other parts.
=180- (A +JB).
By (186) and (187)
cos A(^+i^) . G * ..- .. . ,
' sm —
and . ....sinH^-i?) .in ^^ - 7?)
sinA(.l+^) C *
cos 77 ,.
then a = ^(a + 6) + J(a - 6) and b = ^{a + b)- J(a - 6).
Examples.
1. Given c = .4781, A = 70" and £ = 60° 40', to find a and h.
i{A-£) = i°iO', ^ = 24° 40'
logc= T679518 ■ logc= r67951S
Logcos J(^- JS)= 9.998558 Log sin J(^ -i?)= 8.910404
G C
Log cosec - = 10.379512 Log sec -=10.041555
log (a + 6)= 0.057588 log (a -6)= 2.631477
a + 6= 1.14179 a -6- .04280
therefore a = .59229 and 6 = .54949. , ^
2. Given c = 100, ^ = 72" and B = 30°, to find a and ft.
^ws. a = ^7.23, 6 = 51.118.
OBLIQUE-ANGLED TRLA.NGLES.
157
' . 1'
133. Case II. — Given two sides and an angle opposite to
one of thenij or a, b and A,
From (165) we have
sin J5 = — sin ^,
a ,
or in logarithms
Log sin B = log b - log a + Log sin A^
C=180°-{A + B),
then
and (165) . ,
sin A
or log c = log a + Log sin C + Log cosec A - 20
sin (7 .
c = a — — r = a sm C/ cosec A^
Since the sine of an angle is equal to the sine of its supple-
ment, the equation
sin ^ = — sin ^
a
U ..^,:>^
cannot without other conditions, determine whether B is less
or greater than 90°.
Hence this is called the ambiguous case, for it is evident
that there may be two triangles which will fulfil the conditions
of the problem. The ambiguity, however, does not always
exist, as we now proceed to shew.
Ut When A < 90\
Let CAE be the given angle A. Take AC equal to 6, diaw
CD perpendicular to ^ '
AE and denote it by Fic.l.
p. Then from the
right-angled triangle
ACD we have
p — b sin A.
158
PLANE I'KIGONOMETRY
Fic.2.
With C as a centre and radius equal to a, describe a circle.
Now, if a > JO and a < 6, the circle will cut ^^ in two
points, B and B\ on the same side of -4, and there will be two
solutions, since each of the triangles ABC, AB'C fulfils the
required conditions.
li a > p and a > 6, there will be but oiie solution ; for,
although there are two
triangles ABC, AB'C,
the latter is excluded by
the condition that the
given angle A is acute,
while CAB' is obtuse,
and there remains there-
fore but one triangle
which satisfies the con-
ditions of the problem.
lia = p there will be but one solution, and the triangle will
be right-angled at B^ as is evi-
dent from the figure.
li a < p there will evi-
dently be no solution^ for the
circle will neither intersect
the opposite side nor be tan-
gent k) it.
If a = 6 there will be but one solution, and the triangle
ABC will be isosceles.
FiQ.3.
2nd. When A > 90°.
When the given angle is obtuse the other angles must be
acute, hence there will be no ambiguity, and a must be always
greater than 6. (^«*c. L, 18.) ' \^
OBLIQUE-ANGLED TRIANGLES. 169
Examples.
1. Given a = 925, 6 = 1256 and ^ = 30° 25', to find 7?, C
and c.
p = b sin A
or logp = log b + Log sin A - 10.
log 6 = 3.098990
Log sin ^ = 9.704395
log j9= 2.803385
therefore j9 = 635. 9
Since a > p and a < b, there are two solutions.
To find B we have
log 6 = 3.098990 "
Log sin ^ = 9.704395
ar. comp. log a = 7.033858 (Art 106)
Log sin 5 = 9.837243
therefore i?= 43" 25' 43".5
and 5' = 136-34'16".5
Hence, (Fig. 1) ACB = 180° - (^ + i? ) = 106" 9' 16".5
and ACB' = ISO'- {A + B')= 13° 0' 43". 5. . .
To find c, we have
loga= 2.966142 loga= 2.966142
Logsm AGB= 9.982504 Logsin^C5'= 9.352483
Ijog cosec A = 10.295605 Log cosec A = 10.295605
logc= 3.244251 ' . logc= 2.614530
,T - c= 1754.9 . ';. a c = 411.65
2. Given a = .5412, 6 = .308 and ^4 = 36° 60' 44", to find
£ and c (Fig. 2.) Ane. 5 = 19" 67' 16", C-.76519.
I no PLANE TRIGONOMETRY.
3. Given ^ = 107" 40', a = 5.32 and 6 = 3.58, to find J9, C
and c. Ans. B - 39' 52' 52", C = 32° 27' 8", c = 2 996.
4. Given A = 27° 44', a = 17 and 6 = 40.25. to find the other
parts Ans. Impossible.
6. Given A = 52° 19', a = 325 and h = 333, to find the other
parts.
Am. i5 = 54° 10' 56" or 125 49' 4",
C = 73°30' 4" or 1° 51' 56",
c = 393.756 I or 13.3668.
134. Case III. — Given two sides andj>li6 included angle,
or a, b and G.
First Solution.
To find A and B, we have from (167)
tan UA - B) = r cot —
2^ ' a + 6 2
or, Log tan \{A-B) = log (a - 6) - log (a + 6) + Log cot — ,
2
then ^ = J(^ + ^) + J(^-5)
and B ^l{A-^B)-l{A'B).
■ To find c we have from (165)
a sin G ' .
sm A
or log c = log a + Log sin G - Log sin A.
Examples
1. Given il = 176, 6 = 133 and C = 73°, to find the remain-
ing parts.
. Here we have a + 6 = 309, a - 6 = 43 and — = 36' 30' r \
2
OBLIQUE-ANGLED TRIANULES.
IGl
By (1G7).
log (a -6)= 1.G33468
ar. CO. log (» + &)= 7.510042
Log cot ^ = 10.130791
By (165).
loga= 2.245513
Log sin C= 9.980596
12.226109
Logsinil= 9.954216
Log tan ^{A -B)= 9.274301
therefore 1{A - li) ^-Wdd' 2".7
and 4(^ + i;) = 53'30'0"
hence ^-64° 9' 2". 7
.5 = 42°50'57".3
logc= 2.271893
tlierefore c= 187.022
2. Given a= 16.86, 6 = 9.60 and 6^=128° 4', to iind the
remaining parts.
Am. A = 33" 34' 40",
7?=18°21'20",
C = 24.
When A and B have been found by the above method, c
may be found by (186) or (1 87-), from which we have
cosJ(^+i?)
c=-{a-\-h)
co^ ^ (A -JJ)
and
^ ' sm ^{A-B)
135. Second Solution. — Given a, b and C, to find the
remaining parts. .
When the logarithms of a and h are given, which often
occurs in the computation of a series of triangles, we may find
A and B by (168) or (168 his).
• ' ' Example. * ^'
Given log a = 2.245513, log 6 = 2.123852 and C = 73', to
find the remaining parts. (Same as Ex. 1, Art. 134.)
12
162
PLANE TRIGONOMETRY.
By (168).
log 6= 2.123852
loga= 2.245513
By (168 his).
log b =-- 2.
loga= 2.245513
log 6- 2.123852
Log tan ^= 9.878339
^ = 37' 4'39".7
45°-^= 7''55'20".3
Log tan (45° -$)= 9.143510
Log cot ?= 10.130791
Log COS <^= 9.878339
«^ = 40° 54' 54"
^ = 20° 27' 27"
2 Log tan ^= 9.143510
* 2
Log cot 2" = 10.130791
Log tan i(^- 5)= 9.274301 Log tan J (^ - ^) - 9.274301
^(A - B) = 10° 39' 2".7 1{A-B) = 10° 39' 2".7
the same as by (167).
These methods are quite as short in practice as that of the
preceding Article. The latter, Lowever, should not be used
when a is nearly equal to 6. (Art. 115.)
136. Third Solution. — Given a, b and C, tojindc directhi
without finding the angles A and B.
From (171) we have
c2 = a2 + 62-2a6cosC.
This, however, is not adapted to logarithmic computation. In
the following forms it is easily computed by logarithms :
From (100) we have
then
C '
cos (7 = 1 - 2 sin'' — ,
Q
c' = a'^ + 52 _ 2a6 + iah sin'* —
(a - hf + iah mxi' ~
OBLIQUE-ANGLED TRIANGLES. 163
4a6 sin' —
- ('^ -'')'[' '-rrrw'
iab sin' —
Let ^^^'^-- (a-b)'^ •
or
then we have
/-f . C'
2 vao sin -
tan0 = r— , (206)
a-
c2==(a-6)2(l + tan2(9)
= (a-6)'sec'^
and c = (a -6) sec (9. •' ' ' (207)
When a is very nearly equal to 6, the denominator a-b
will be very small and 6 will be near 90"; in that case the fol-
lowing form will be preferable : ^
c2 = a'-l-6'-2a6 cosC
= a' + 6'-2a6(2 ^'|-1) by (99)
= (a + 6)' - 4a6 cos' —
2
4a6 cos' —
I ' . ■ ■ .^
/ 'iao cos- —\
Now since the first member of this equation is positive, th(^
iab cos'
2
second member must be so likewise, therefore ^ ^u must
be less than 1.
Hence we may assume
iab cos' —
sin' B - . ,y,
{a + by
164
PLANE TRIGONOMETRY.
2jai
C
cos
or
then we have
and
sin ^ = '
(a + 6) '
= (a + by cos" e
c = {a + b) cos $.
Example.
(208)
(209)
Given a = 176, 6 = 133 and C = 73", to find c. (Same as
Ex. 1, Art. 134.)
By (208) and (209).
log a = 2.245513
log 6 = 2.123852
log (a + 6) = 2. 489958
Log cos ^=9.781934
2)4.309365
log Va6" = 2.184682
log 2 = 0.301030
Q
Log COS -^ = 9.905179
2.390891
log (a + 6) = 2.489958
log c = 2.271892
c= 187.022
The remaining angles may
now be found by (165).
Log sin ^ = 9.900933
. ^ = 52° 45' 12".
137. Case IV. — Given the three sides, or a, b and c, to find
A,B,C. .
First Solution.
By (198) we have ,,
in J. = — is/s(s -a) (s- b) (s -c),
sm
be
(210)
■which may be i. ed when A is not near 90°. Similar formulae
of course exist for sin B and sin C
OBLIQUE-ANGLED TRIANGLES.
165
Second Solution.
From Art. 124 we have
si.-2=s/ f^
B_ ks-a){s-c )
''''2-Nl ^c '
2 \ ab
Third Solution.
COS
COS
2
B
C
COS —
2
Is (a - a)
t
Fourth Solution.
tan -- =
2
8(8 - a)
tan
tan
B
1 (8 -a)(8 - c)
\l 8(8-
,(8-h)
in which
C^_ Us -a)(s- b)
2"\/~7(s-c)
.'i^
(211)
(212)
(213)
When all the angles are required, the last group will be
the most convenient, since only four different logarithms are
required from the tables. Tlie first group is to be preferred
when half the angle is les8 than 45°, and the second when it is
greater tlian 45°. The third group is accurate for all angles.
106
PLANE TRIGONOMETRY.
• Examples.
1 Th(i sides of a triangle are 13, 14 and 15 ; find the anglea
L^t a = 13, 6 = 14 and c = 15 ; then
s = 21, 8-a = 8, s-6 = 7, «-c = 6.
To find A.
From the first of the last group, we have
log(«-6)= 0.845098
log(s-c)= 0.778151
ar. CO. log 5= 8.677781 (Art 106)
ar. CO. log {8 -a). 9.096910
2)19.397940
A
Log tan -= 9.698970
therefore
and
To find B,
2" = 26° 33' 54''
^=63' 7' 48".
therefore
and
log(s-a)=: 0.903090
log(«-c)= 0.778151
ar. CO. log 5= 8.677781
ar, CO. log (s - 6) = 9. 154902
. 2)19.513924
Log tan-- 9.756962
- = 29M4'41".6
5 = 59° 29' 23".2.
OBLIQUE-ANGLED TRLA.NGLES.
167
Tojmd C.
log(s-a)= 0.903090
log(«-6)= 0.845098
ar. CO. log 8= 8.677781
ar. CO. log (« - c) = 9.221849
2)19.647818
Log tan -= 9.823909
therefore
and
Verification
C
= 33°41'24".4
C = 67''22'48".8.
^ + ^ + (7 = 180^
The following transformation of (213) will facilitate the
computation when all the angles are required.
Multiply the numerator and denominator of the second
member of the first of (213) bj^s - a, and we have
tan
1 Us -a
-a) (s - b) {s ~ c)
8
Put
#-
-a) {s - b) {s - c)
= r.
then
also
8
tan- =
2 8-a
^ B r
tan — =
2 8-b
. C T
tan- =
2 «-c
(214)
2. Given a = 1468, 6 = 1359 and c = 1263, to find the angles.
Ans. A ==67" 58' 51", ^ = 59° 7' 4", ^=52° 54' 5".
1G8 PLANE 'VRIGONOMETllY.
3. Given a = ^56, 6=1 and c = 7, to find A.
Am. ^ = 115° 22' 37".
4. Given a = 5, 6 = 4.037 and c = 3.9575, to find A.
Ans. ^ = 77° 25' 12".
5. Given a = 25, 6 = 30 and c = 20, to find i4, 5 and C.
Am. A = bb° iC 18", 5 = 82° 49' 8", (7 = 41" 24' 34".
6. Given a = 39, 6 = 35 and c = 27, to find A, B and C.
Am. A = 7(>' 45' 21", 5 = 60° 52' 33", C = 42° 22' 6".
138. Given tioo sides and the included angle of a triangle^
to find its area.
From Art. 128 we have
A = |6c sin A = \ac sin B = |a6 sin C.
Ex. i.— Given a = 30, 6 = 40 and (7 = 28° 57', to find the
area.
loga=^.477121
log 6 = 1.002060
Log sin (7 = 9.684887
ar. CO. log 2 = 9.698970
log A = 2.463038
A = 290.427
JUx. 3. — The sides of a triangle are 103.5 and 90, and the
inelided angle is 100°, find the area.
Am. 4586.75.
139. Given two angles and the included side of a triangle, to
fii.d tlie area.
From (202) we have
c' sin A sin B
2 ' Bin{A'VB)'
EXAMPLES. IGO
Ex. 1.— Given A = 80% i?=:60° and c = 32 feet, to find th( .
area.
Log sin ^= 9.993351
Log8inj5= 9.937531
Log cosec (A + B) = m 191933
2 logc- 3 010300
ar. CO. log 2= 9 698970
logA= 2.832085
A= 679.33 square feet.
Examples.
1. In the ambiguous case of triangles, given a, b and A,
shew that if Cj, Cg be the two values of the third side,
Ci + C.2 = 26 cos Af CjC2 = b^~ a\
and that the area of both triangles = 16^^ sin 2 A.
From (169) we have
a^ = P ■\-c^ - 2bc cos A *■
or c2-26cosyl.c + 62_^2^0 (1)
If Ci and Ca be the two values
of c in tliis equation, we have
by the theory of quadratics,
Ci + C2 = 2b cos A and c^Ci = b^~ a\
Now, AI) = '-l±^' = b cos A,
therefore q + Cj = 2b cos A ,
Again, c,c, =^ABx AB' ^AP'. {Euc IIJ , 36.)
^AC-PC^
^b'^-dK
By solving (1) for c we have
c = 6 cos vl ± sjoir - h^ sin^ A. (2)
170 PLANE TRIGONOMETKY.
Hence
Ci = 6 coH ^ + Jn"^ - P sin'' A and Cj = i cos ^ - \lci^ - h^ sin' A.
Area ABC = Ihc^nxnA = U'-'sinylcos^ + ^6sin^ sld^ - b'^ sin''* A.
Area AB'C= hbc^nmA = Jji^gi^vlcos^ - ^isin^l Jd^ - b'^ sin*'' A.
therefore the area of both triangles = b'^ sin ^ cos -4
From (2) it is seen that when a = & sin ^, c = 6 cos A. whirh
gives only one value for c and the triangle is right-angled at B.
2. Given the perpendiculars from the angles of a triangle
upon the opposite sides, to find the sides and angles.
Let Pa, Pb) Pc denote the perpendiculars on the sides a, b, c
respectively ; then by (203)
2A 2A 2A
Pa = -^, Pb-J-. Po=--*
2A , 2A 2A
or o, — ) = , c = f
Pa' Pb' Pc
Substituting these values of a, 6, c in (172) we have
4A'' 4A2 4A^
. Pl Pl Pl
2 ^^
'Pb Po
Pl"^ pl ' pi _ PlPc + PaPb - Pb Pc
2 ^pfiPbPe
PbPo
which determines A,
Then & = -^» ^--^' ^^'
sin A . sin ^
EXAMPLES. 171
3. Given the sum of the sides a and h = k, the side c and
the angle C, of a triangle, shew that
a = k cos'* ^ , h = k siv? ^
, . ^ . n/A^c^ C
where sin ^ = ± — sec - .
k A
From (171) c' = a^ + i^ - 2a6 cos G
= a2 + 2a6 + h' - 2ab (1 + cos C)
Q
= J^ - iah cos^ — ,
C
therefore 4a6 = (J^ - c^) sec'^ —
A
which is possible since {a + 6)^ > 4a6,
then 1 - 7 rrr, = 1 - sin'* 6 ' '
a-h
or r = cos
a + 6
2a
+ 6
1 + cos ^,
or -?- = 2 cos'* ~r
k 2
and a = A; cos^ — -
2
Also r = 1 - cos 6
a +
OP 7- = 2 sm^ -^
A; 2
and h = k sin' — .
A
172 PLANE TRiaONOMETRY.
This prohlem may also be easily solved by (18C) and (187).
Thus, from (186) we have
cos 1{A -B) = - con i(A + B)
G
h .
Hence the angles are known.
Fiom (187) we have
a — o = c »» , ■ .
cos 2
Ilonce the sides a and b are known.
4. The sides of a triangle are 13, 14 and 15, find tho area.
Ans. 84.
5. The sides of a triangle are 13. 20 and 21, find the perpen-
d cuLar on the longest side and the area.
Ans. Perpendicular =12, area = 1 26.
6. Tho sides of a triangle are 25, 6.l and 52, find the area
and the angle opposite the shortest s?de.
Ans. Area = 624, angle = 28" 4' 21".
7. The sides of a triangle are 137, 111 and 124, find the
area. Ans. 6510.
8. The angles of a triangle are 70°, 60" and 50°. and the
perimeter is 150, find the sides. (See formula 192.)
Ans. 64.81. 50.51, 44.68.
9. The sides of a triangle are 85 and 75 and the included
angle is 75°, find the other angles.
Ans. 57° 10' and 47° 50'.
EXAMPLES. 173
10. The ratio of two sides of a triangle is 7:3, and the
angle they contain is 6° 37' 24", tij^d the other angles.
Ans. 168° 27' 25".4 and 4° 55' 10".6.
11. The aides a, 6, c of a triangle are as 8 : 10 : 16, find the
angled. Ana. J5 = 55° 46' 16".
12. Given two sides of a triangle and the difference of their
opposite angles, to solve the triangle. (See formula 166.)
13. In any triangle prove that tan - = ^ ^ -, whore
d is the difference between the sides a and c, and p is the per-
pendicular upon the side h, and that tan B = ; — - .
c-b cos A
14. If a, y8, y be the perpendiculars from the angles of a
triangle upon the opposite sides, prove that
a? be
15. In the ambiguous case, if A and 8 be the areas of the
two triangles, prove that ; ' ' • , i ; 1 »\
A'^ + 8'''-2A8cos 2^1 = ^, (A + 8)'
a, b and A being given.
16. In the ambiguous case, if c^, c^ be the values found for
the side c, when a, b and A are given, prove that
'i
c* + cl- 2c ^c^ cos 2 A = 4^2 cos^ A
and ^! ~ ^8 = 4^ *^^^ ^ "^^ ~ ^^ sii^^ A. ,
17. In any triangle, given a, B and the sum of the other
two sides (equal to m)j to solvo the triangle. • '
. C m-a B
Ans. tan — = cot — .
2 m + a 2
18. If a, )8, y be the distances from the angles A, i/, (7 of a
174 PLANE TRIGONOMETRY.
triangle to a point P within it, from which the sides subtend
equal angles, find the sides and the area.
Ana. The sides are J{jS^ + y^ + ^y)
si {o? + -/ + ay)
and the area is -— - (a/3 + ySy + ay).
19. In a triangle, given C, c and ab = m\ to find the sides
a and b. Ans. a + b — c sec 0,
a-b = c cos 0,
, . /, 2w C , . ^ 2m . G
where tan v = cos — and sin d> = sm — .
c 2 c *>.
20. In a triangle, given C, the perpendicular from C=p
and a + b = wi, to find c.
Ans. c = m tan — , where tan = — tan — .
J p 2
^^ 21. The perimeter of a triangle is 100 rods, ^ = 102° 51' 30",
J5 = 25° 42' 45" and (7 = 51° 25' 45"; find the sides.
Am. a = 44.51, 6 = 19.8, c = 35.69.
22. The perimeter of a right-angled triangle is 24 rods, and
one of the angles = 30° ; find the sides.
■ Ans. 5.072, 8.784, 10.144.
23. Given the base a, the vertical angle A, and the difference
of the other two sides = c?, to find the sides. (See formula 187.)
24. Two adjacent sides of a parallelogram are a and b, and
the included angle is 6 ; shew that the diagonal drawn from this
angle is ' . . : . , ■
sj{a'^-^b'^-\-2abQose), • ',
and that the other diagonal is ^
EXAMPLES.
175
25. The sides of a triangle are 17, 25 and 28 ; find the area
of the triangle formed by joining the feet of the perpendiculars
drawn from the angles to the opposite sides. (See problem 52,
Chapter VIII.) Arts. 24:^V4V
26. In a right-angled triangle, the sum of the hypothenuse
and base is 2986, and the angle at the base is 52°, find the
perpendicular. Ans. 1456.37.
27. Two sides of a triangle are 356 and 294, and the angle
opposite the latter side is 51° 27', find the other side.
Ans. 316.309 or 127.4079.
28. The perimeter of a triangle is 128, the angle C = 28° 4' 21"
and the perpendicular from C on c = 49.92, find the side c.
• Ans. 25.
29. Given 6 = 39, c = 51 and B = 115°, to solve the triangle.
. An^. Impossible. Why 1
30. In a triangle given (7, c and d^ - b^ = 7ri?, solve the
triangle. •. '^^ - n
, Ans. sin {A - B) = — ^ sin C.
When will there be two solution*? ; ' :
■.^.;Vr^'-:.' M!-.,
iX
176 PLANE TRIGONOMETIIY.
CHAPTER X.
APPLICATION OP TRIGONOMETRY TO SURVEYING, NAVIGATION
AND ASTRONOMY.
140. In this chapter we will give some examples shQ;wing
how the formulae of Trigonometry are applied in the solution
of some very interesting and useful problems in Surveying,
Navigation and Astronomy. Horizontal and vertical angles
are most conveniently measured with a theodolite. This
instrument is composed essentially of two graduated circles
having their planes perpendicular to each other. When the
instrument is in use, one of the cix'cles is placed in a horizontal
position by means of spirit levels. On this circle horizontal
angles are measured ; on the other are measured vertical angles
whether of elevation or depression, that is, whether the object
is above or below the horizontal line passing through the centre
of the axis of the vertical circle. A telescope is attached to the
axis of the vertical circle, by means of which a clearer view of
the object may be had. The sextant and reflecting circle are
employed to measure angles in any plane whatever. They are
the only instruments which can be conveniently used at sea for
measuring the altitudes of the heavenly bodies above the horizon.
For the descriptions, adjustments and modes of using these in-
struments, the student is referred to treatises on surveying,
such as Gillespie's Land Surveying, or Sims's Treatise on Mathe-
matical Instruments.
Their construction and adjustments will, however, be best
learned by a careful study of the instruments themselves. We
V
HEIGHTS AND DISTANCES.
177
shall therefore suppose that the manner of adjusting and apply-
ing them to practice is known, and proceed at once to give a
collection of problems which they enable us to solve.
141. To find the distance 0/ an inaccessible object upon a
horizontal plane.
Let it be required to find the distance from A to an
object C situated on the
opposite bank of a river.
Measure a base AB, whose
length we will denote by c.
Measure also the horizontal
angles CAB, CBA; then
(Art. 131)
AC_
AB
sin B
whence
or in logarithms
AC = c
sin (7'
sin B
»• 11
:A'i'-
(■■ '
i V
sin {A-\rBy ^•'>--
log ^C = log c + Log sin B - Log sin {A + B).
This problem may also be solved without the use of any
instruments for measuring
angles, as follows :
Having measured fhe base
AB as before, measure any
length Ad along AG and an
equal one Ag along AB; then
measure dg. Similarly mea-
sure Bm and Bn, equal to each
other, and then measure mn.
Bisect dg in h and join Ah, then since Adg is an isosceles
triangle Ah bisects the angle A, and we have
13
178
PLANE TRIGONOMETRY.
similarly
81112 =
. B
sin :r- =
Id''
mk
Bm
gd
inn
'2 Bm
Hence the angles A and B are known by the aid of the tables,
and the solution may now be completed as before.
142. To find the height and the distance of an inaccessible
object standing on a horizontal plane.
Let DC be the ob-
ject. Measure any
base AB, and denote its
length by c. Measure
the horizontal angles
CAB (a) ABC (13) and
the angle of elevation
CAD (y). Then from
the
have
triangle ABC we
AC^c
sin
B
; = C'
sin P
sin (7 "sin(a + /3)
Ih the right-angled triangle ACD we have AC^ just found,
and the angle CAD (y) to find DC.
Hence DC = AC i&nCAD
sin /8 tan 7 ~
=
Or thus :
At any convenient point
A measure the angle of ele-
vation CAD (a); then mea- .
sure a base AB (c) directly
from the object, and at -5 -g
measure the vertical angle
sin (a 4- j8)
HEIGHTS AND DISTANCES. 179
ABD (/3). Then the angle ^Z) B = a - /?, and from the triangle
^^i) we have by (165)
c sin )8
AD=:
sin (a - yS) *
then in the right-angled triangle ADC
DC = AD Bin CAD
c sin a sin /3
and AC = AD COS CAD
c cos a sin j8
sin (a - /8)
•■'-:'»,
i-;;o
143. To find tlie distance between two incuccessible objects on
a horizontal plane. » ...
Let G and D be the two
objects. Measure the base AB
equal to a, and the angles CAB^
DAB, ABC, ^i?A' which put
equal to a, )8, y, 8 respectively.
Then from the tria^^le ABC we
have by (165)
*n An sin^5C smy
AG = AB —. 77v-9r=a
sin ^C-S sin(a + y)*
sin ^5i) sin 8
similarly, ii) = ^5 ^^^j^ = « ^1^(^ + 8) '
Then in the triangle ACD, the sides AC, AD and the in-
cluded angle CAD = (a - (3), are known, therefore CD can be
found by the methods of Arts. 134-136.
180
PLANE TRIGONOMETRY.
144. Three inaccessible objects A, B, C (not in the same
straight line), situated on a horizontal plane and at known dis-
tances from each other, are visible from another point P on
the plane, and the angles
A PC, BPC are given; to
find the distance from P to
each of the objects A, B,C.
Let A, B, C be the
three inaccessible objects,
and P the point of obser-
vatioa. Let the observed
angles APG, BPC be de-
noted by a, )8 respectively
PAChye,QXidiPBChy<l>.
Since the sides of the
triangle ABC are known, J^
the angle C can be found
by (179) or (182)
^ + <^ + a + ;8 + C = 360°, , ., , .,
therefore K^ + <^) = 180''-i(a + )8 + O). '
From the triangles APC, BPC, we have by (165)
6 sin ' - '^ a sin <h
, and PC = — ; — ^ ,
PG =
therefore
sin a
sin a sin a
sin ^ b
sin /3
*,.'irt .,'1 i-t
then
sin^
= tan »/r, suppose j
sin ^ - sin ^ 1 - tan i/r ^' ^i
sin ^ + sin 6 1 + tan xj/^
^:i,*'i
HEIGHTS AND DISTANCES.
181
^l^ence '^^, = tan (45« - ^), by (59) and (81)
tan ^(^ + c')
therefore tan | (,^ - ^) = tan (45° - ip) tan ^(<^ + 6). (2)
From (1) and (2) 6 and <^ become known, and therefore by
(165) AP, BP and GP can be found.
When e and </> are supplementary, ^ = 45°, and the solution
fails when the point P is without the triangle, in which case
the quadrilateral AGBP is inscriptible in a circle. , .
145. To determine the height and distance of an inaccessible
objecty by having given its angles of elevation observed at three
points at given distances
from each other, and in
the same straight line.
Let DE be the object,
A, B, C, the three stations
of observation: let AB = af
BC =- 6, and DF - x; and let
the observed angles of eleva-
tion at Aj B, C, be a, ^, y,
respectively :
Then AD = x cot a,BD = x cot l3,CD = x cot y.
Draw i>/ perpendicular to ACy then in the triangles ABB,
BCD, we have by Fuc. II., 12, 13,
,. . . x^cof^ a = a'^ + x^cot^ ^ + 2a. Bf
a;2 cot*'' y = 52 + a;2 cot^ )3 - 26 . 5/:
Eliminating Bf and solving for x, we have
i i.
>=\|^
ab (a + b)
cot^ y-{a + b) cot^ /i + b cot^ a '
And thus AD, BD, and CD become known.
182
PLANE TRIGONOMETRY.
146. A flagstaff of kriown length, standing upon the top of
a tower of known height, subtends at the eye of a spectator on
the same horizontal plane as the
tower, a given angle ; to find tJie
distance of tlie tower.
Let AB = a, the height of
tower ; BC = h, the length of the
staff*; P, the position of the ob-
server, at which the staff" sub-
tends an angle BFC = fi.
Let AFB=^e, then APG==e + /3.
AF = a cot e and AF=(a + h) cot {6 + ^),
whence . a tan (^ + /8) = {a + h) tan 6 X
and tan 6 = ^ cot ^ ± ^ A'' cof^ ^ - 4a (a"T X)
2{a + h) '
therefore AP=\ { A cot /8 + Jh^cot^ ^-ia (a+T) } ,
from which we see that there are two values of AF, unless
K^ cot'* ^ = ia {a + h).
If a segment of a circle be described on BC containing an
angle equal to yS, it will in general cut the horizontal line in
two points, F and Q, corresponding to the two values of AF.
When A cot ^ = 2 J a (a + h), the circle will touch the hori-
zontal line, the points F and Q will coincide, and ^ will then
have its maximum value, so that tan $= \ 7 .
\a-hA'
and then / ' AP= J a {a + h), \Euc. III., 36)
Or we may proceed thus :
From the centre E draw ED, JS^iJ perpendicular to CB and
HEIGHTS AND DISTANCES. 183
AQ; join CJE and EP ; the angle CFD = CPB = CQB = ^.
In the right-angled triangle, CDE^ we have
h h
DE tan )3 = -, or BE = — cot fi^AH,
and CE sin ^8 = ^, or 0^ =\ cosec /? = ^P.
Pir= JEP'-EH^
= JEP'-AD^
-J
— cosec* B - (a+ — )
4 ' ^ 2
therefore ilP^^fi'- PZr=i{7i cot )8- ^/t^cot^^- 4a (a + A)}
and iig = ilfl' + Pir=j{Acot^+ Jh^cot''fi-ia{a + h)}
the same as before.
-/;,.;.
147. In Surveying, the course or bearing of a line, is the
angle which it makes with the meridian passing through one
extremity, and is reckoned from the north or south
point of the horizon, toward the east or west. N
Thus, if I^S represent the meridian and the
angle B'PQ be 30", then the bearing of PQ from
P is 30'' to the east of north, or as usually read
" north thirty degrees east " and written N. 30° E. ^
The reverse bearing of a line is the bearing
taken from the other extremity. Thus the bearing S
of P^ from ^isS. 30* W.
184)
PLANE TRIGONOMETRY.
In Navigation, the course of a ship is generally referred to
the points of the Mariner's Compass which consists of a circu-
lar piece of card board attached to a magnetic needle, and is so
balanced that it can move freely in any direction.
The circumference is divided into thirty-two equal parts
called points, and each point into four equal parts called quarter
ooints,
The points are read as follows, beginning at the north and
going east : " north, north by east, north-north-east, north east
by north, north east," and so on as seen in the figure.
The angle between two adjacent points is W° or 11° 15'.
A quarter point is therefore 2° 48' 45", a half point 5° 37'
30", and a three-quarter point 8° 26' 15".
LEaiGTH OF A DEGREE OP LONGITUDE. 185
148. To find the length of a degree of longitude on any
'pa/rallel of latitude.
Let V be the pole of the earth, C the centre, EQ an arc of
the equator containing one degree, and
AB ^rv arc of one degree on any parallel
of latitude. Join CE^ CQ and draw
AD and BB perpendicular to the raalus
PC ; then it is evident that the angle
ADB is equal to the angle ECQy and
therefore
CQ : BB :: EQ : AB.
But regarding the radius CQ as unity, DB is the sine of the
angle BCF or the cosine of the angle BCQ, that is, the cosine
of the latitude : hence
1 : cosine of the latitude : : EQ : AB
and therefore AB = EQ cos. lat.
From which it follows that similar portions of different parallels
of latitude are to each other as the cosines of the latitudes. «. ,
Ex. — Find the length of a degree c )ngitude at Toronto,
latitude 43° 39' 24".
The equatorial radius of the earth is 3962.8 miles, but in
consequence of the flattening in the direction of the polar
diameter it is only 3956.514 miles at Toronto: and consider-
ing th5 earth a sphere with a radius of 3956.514 miles, the
length of a degree of longitude on the equator would be 69.054
miles =EQ. " • '■vv'.'. ',-■-■■'■■•'.•. ^ -- ■- •'■.:'.■:•-■ ,■ ;-i'^r:-
Hence ' ? - log ^^= 1.839190
'"' Log cos lat. = 9.859432
Vr«'i
,..,, .n ■ . . log ^5= 1.698622 : ,
therefore .4^ = 49.96 or 50 miles very nearly.
186
PLANE TRIGONOMETRY.
As a degree in longitude makes a diflTerence of four minutes
in time, it follows that fifty miles east or west on the parallel of
Toronto is equivalent to four minutes difference in time.
149. To Jmd the MoorHs distance from tlie Earthy a/nd her
diameter. ; .
Let PF be the earth's axis, JEQ the equator, and A and B
the positions of two observatories on the same meridian and
whose latitudes ACQ, BGQ are known. Draw the radii, CAs
CBy and produce them to meet the celestial sphere in Z and Z' ;'
*
through A and B draw AH, BH at right angles to CZ and GZ'
respectively, then Z and Z' are the zeniths, and AH and BH
are the horizons of A and B respectively ; let M be the moon,
and when she is on the meridian of the observatories, let her
altitudes MAH, MBH be measured with a sextant, theodolite
or any other suitable instrument. Join AB, then regarding
the earth a sphere in order to render the problem as simple as
possible, we have in the isosceles triangle ACB, the sides AG^
GB and the angle AGB, the sum of the latitudes given, hence
AB and the angles GAB, GBA can be found. The angle HAB
= HBA=\iqM of ACB, and since MAH and MBH are deter-
mined by observation, the angles MAB, MBA are known; hence
AM and BM can be found. Finally, in the triangle ACMf
PARALLAX.
187
the sides AM^ AG and the included angle MAC are now known,
therefore MG can be found, which is the distance sought.
From A draw AN tangent to the moon and join MN, then
ANM is a right angle and the angle MAN the moon's apparent
Bemi-did.meter can be determined by observation,
therefore MN= AM sin MAN,
whence her radius is known and therefore her diameter.
■ Parallax.
150. The parallax' of a celestial body is, in general, the
apparent angular displacement which is produced by viewing
the body from two different points. The term is used in
astronomy to express the difference of altitude or zenith dis-
tance of a celestial body when seen from the surface and the '
centre of the earth respectively.
Let G be the centre of the earth, P the place of an observer
on its surface, M a celestial
object seen in the horizon, M'
the same seen at the zenith
distance M"PM\ and M" the
same object seen in the zenith.
Now it is evident that when
the object is at M" it will ap-
pear in the same direction
whether viewed from P or G ;
there is then no displacement
and therefore no parallax.
If the object be at M\ its apparent direction is PM' -while
its true direction is GM\ and the angular displacement TM'G is
the parallax due to the zenith distance M"PM' or to the altitude
MPM.
It is evident that the angle PM'G increases sis the body
188 PLANE TRIGONOMETRY.
approaches the horizon, and when it is in the horizon as at M
the parallax PMC has its maximum value, and is then called the
horizontal parallax which is, in fact, the largest angle which
the earth's radius subtends at the object.
Parallax increases the zenith distance and consequently
diminishes the altitude. The angle M"FW is the apparent and
M"CM'the true zenith distance of the object when at M'.
Hence, to obtain the true zenith distance from the apparent,
the parallax must be subtracted; and to obtain the true altitude
from the apparent, the parallax must be added.
It is evident from the figure that the efiect of parallax is
wholly in the vertical circle passing through the observer and
the body, the azimuth or angular distance from the meridian of
the observer is therefore not affected.
151. To Jind the parallax at any altitude when the hori-
zontal parallax is given.
Let P=(7i/P, the horizontal parallax,
p = CM'F, the parallax in altitude,
Z' - M'PM", the apparent zenith distance,
R = CM= GM\ the distance of the object,
r = (7/*, the earth's radius,
then in the triangle CPM' we have
^mCM'P ^ CP^ ,;,;.,!. ■ ■;■ .;, ■, ... .
sinCPJ/' ~ CM' ,v.-5 . . .., „
sin jo r . • ;
or -: — ^= ^=smP,
sin Z R ,, , ,
whence i') ^/_ . sin p ::= sin P sin <^'. ,.^ , ,.. (215)
In the case of the sun and planets, the horizontal parallax
is so small that we may, without sensible error, use the more
convenient formula
p"=P' Bin Z'. . ,;.;>!.:. •
PARALLAX IN ALTITUDE. 189
Jjf the true zenith distance M'CM" = Z he given, instead of
the apparent, we have '
sin CM'P CP
I i
sin GFM' CM
sinp r . _
or -: — 7-^ r = — = sinP; (a)
sm {^+p) R
hence
sin (Z+p) + sin p 1 + sin P
sin (Z + p) - sin p~ 1 -sin. P^
tan (-+;?) p
or —4 tanM45° + --),
tan-
Z P Z
whence tan (— +^) = tan2 (45° + — )tan--. (216)
This equation males known ^ Z+p^ from which we obtain
Z
p by subtracting — . ■ ' . ^ -, ■ ^- , -) yi'.^^.^//''h t1^ '■■h:;t;;^ ^^'f
Another solution of equation (a) will be given in a subse-
quent chapter. • . . . .* .
^ ^ . .:■■■.■■■.■' ',;,><;.■■} --;■■.;; ,\.^,^i ^'.kX--
152. To find the augmentation of the MoorHs semi-diameter
on account of her altitude above the horizon.
The apparent diameter of the moon is the angle which her
«lisk subtends, and is not the sa'- ■ for all points on the earth,
on account of their different distances from her.
Supposing the moon's distance from the centre of the earth
to remain constant, her distance from the observer must dim-
inish as her altitude ajaove the horizon increases, and therefore
her apparent diameter must increase. This increase of the
apparent diameter is called the augmentation of the diameter
due to altitude.
190 . PLANE TRIGONOMETRY.
In the figure of Article 151, let 8 denote the moon's apparent
semi-diameter at i/, and 8' the augmented semi-diameter when
at M\ then we shall evidently have
8:8':: PM' : CM' or CM.
8 PM' sin PCM' sin Z'
B'~CM sin CPM'~ am Z*
sin Z
slKZ'*
whence ^' = ^^^^ (217)
.. Examples.
1. To determine the distance of a ship at anchor at (7, a base
line ^5 of 100 rods is measured along the shore, and the angle
ABC was observed to be 83*' 18' and the angle BAC 32° 10' ;
find the ship's distance from B. Ans. 58.96S rods.
2. From the decks of two ships A and B, 880 yards apart,
the angle of elevation of a mountain C due north of both ships
was observed at each; at A the angle was 35° and at B 64°;
required the height of the mountain above the surface of the
sea, the deck of each ship being 21 feet above it.
. , . • , . , , , Am. 942.75 yds.
3. A tower subtends an angle of 39° at a distance of 200
feet from its base, what ejiigle will it subtend at a distance of
350 feet from its base 1 Ans. 24° 49' 53".5.
4. There are two monuments whose heights are 100 feet
and 50 feet respectively, and the line joining their tops, when
produced, makes with the horizontal plane on which they stand
an angle of 37°; find their distance apart.
Ans. 66.352 feet.
5. To determine the distance between two inaccessible rocks
C and Df in the sea, a base line AB of 670 yards was measured
EXAMPLES, ^ - - 191
on the shore and the following angles were measured at the
extremities of the base: BAD^^O" 16', BAC = Q7° 56' ; ABC
= 42° 22', ABD^IW 29'; find CD.
Arts. 1174.26 yaxdg.
6. The hypothenuse of a right-angled triangle rests on a
horizontal plane, and is 100 feet long; one of the angles is 36°
40' and the inclination of the triangle to the horizontal plane is
60° ; find the height of the right angle above the plane.
Arts. 41.48 feet.
7. An object stands on the top of an inaccessible hill, and
from a certain point on the horizontal plane the angle of eleva-
tion of the top of the hill is 40°, and that of the top of the
object 61°. Going back 100 yards in a direct line from the
object, the angle of elevation of the top of the object was then
33° 45' ; find the height of the object. Ans. 46.663 yards.
8. The angle of elevation of a tower 100 feet high, due north
of an observer, was 50° ; what will be its angle of elevation after
walking due west 300 feet? Ana. 17° 47' 50". 5.
9. "Wishing to find the distance between a battery at B and a
fort at Fj which cannot be seen from the battery in consequence
of the ground between the battery and fort being covered with
a forest, distances BA and AC to points A and C where both
the fort and battery were visible, were measured, the former
being 2000 yards and the latter 3000, and the angle BAF was
found to be 34° 10', FAC 74° 42', and FCA W 10'; find the
distance between B and F. Ans. 5422.3 yards.
10. The elevation of a balloon was observed to be 20° bear-
ing N.E., and by another observer 4000 yards due south of
the former its bearing was found to be N.b.E.; required its
height. Ans. 511.24 yards.
11. Coming ift from sea, at a certain point P I observed
192 .PLANE TRIGONOMETRY.
two headlands ^ and 5, and inland at C a tower which appeared
between the headlands : the distance between the headlands was
5.35 miles, and the distance from A to the tower was 2.8 miles,
and from B to the tower 3.47 miles. I observed the angles
APC, BPC, the former being 12° 15' and the latter 15° 30';
required my distance from each of the three objects.
A-,v6. ^P=11.257; CP=12.7523; i?P= 11.034 miles.
12. Bequired the distance of the three objects Aj B, C from
the point P situated within the triangle ABC, from the follow-
ing data: ^5 = 267 rods, ^C = 346 rods, BC = 209 rods, angle
^P(7=128°40', angle^P^ = 9r 20'. . •
Aiis. ^P= 248.854; J?P = 91.134; CP= 130.81 rods.
13. While sailing along a coast a headland G was observed
to bear N.E.b.N. ; having run E.b.N. 15 miles to B, the head-
land bore W.N.W.; find the distance from the headland at each
observation. Ans. 8.499, 10.81 miles.
14. ^ and B are two points lying N. and S., and 50 rods
apart ; what must be the distance of a third point C from each,
that it may bear N.b.W. from B and W.b.S. from A f
Ans. 9.754 and 49.04 rods.
15. The courses of two ships are N. and E., and their rates
of sailing are equal, the bearing of the former -from the latter
was E.N.E, ; but after each had sailed ten miles the bearing of
the latter from the former was S.S.E, ; find the distance be-
tween the ships at the first observation. Ans. 7.653 miles.
16. The elevations of two mountains, in the same line with
the observer, are 9° 30' and 20°; on approaching two miles they
both have an elevation of 38° 15'; find their heights and the
distance between them. ' ' ' ' * >^>*-^!' '--'■>■ v>;i.
An^. The nearer, 747.77 yds. ; the more remote, 2380 yds.
The distance between them, 2070.48 yds. " '
EXAMPLES. • 11)3
17. From a ship sailing N.W., two islands appeared in
sight, one bearing W.N.W., the other N, and after sailing
eight miles farther the first bore W.b.S., and the other N.E. ;
required their bearing and distance from each other.
Ans. S. 58° 40' 50" A W.; distance, 12.95 miles.
18. Two ships sail from the same port at the same time,
the one due N. at the rate of six miles per hour, the pther on
a course N. 60° E. at the rate of ten miles per hour for two
hours, she then tacks to cut the other off or to overtake her ;
how far must she sail to do it, and on what course ?
Ans. 23.75 miles, on a course N. 46" 49' 35" W.
19. From two stations on the deck of a ship 100 feet apart,
the bearings of an object on shore were N.E. and N.N.E., and
the ship's head was N.b.W. ; find the distances of theobject
from each station. Atis. 145.177 and 217.27 feet.
20. A cape C bears from a headland II, W. ^ S. 4.23 miles;
how must the cape bear from a ship which runs in towards
the headland on a course N.b.W.^ W., until the headland
is 2.3 miles distant from the ship? ; Atis. W.H^f.W. :
21. A ship was 2640 yards due south of a lighthouse, and
after sailing N. W.b.N. 800 yards, its angle of elevation was
5° 25'; required its height. Ans. 191.94 yards.
22. A tower, 65 feet high, subtends an angle of 60° at the
eye of an observer standing on the same horizontal plane as
the tower ; find his distance from it, his eye being five feet
above the plane. ,. ,. , Atis. ii.S ieet.
23. A lighthouse standing on a rock, is observed from two
points in a line with it, and one quarter of a mile apart ; from
the nearer point the elevations of the top and bottom are 52°
14' and 48° 38' respectively, and from the more remote point
14
194 PLANE TRIGONOMETRY.
the elevation of the top is 16° 28'; find its height and eleva-
tion above the sea.
Ans. Height, 60.82 feet; height above sea, 445.2 feet.
24. From the top and foot of a lighthouse 58 feet high,
standing on a cliff by the sea coast, the angles of depression
of a ship's hull measured from the visible horizon are 5° 47' and
5° 8'; find the ship's distance, the diameter of the earth being
taken at 7926 miles. Ans. 1688.7 yards.
25. At noon, a column in the direction E.S.E. from an ob-
server, cast a shadow the extremity of which lay in the direction
N.E. from him ; the elevation of the column was found to be
46°, and the length of the shadow 80 feet ; find the height of
the column. Ans. 61.23 feet.
26. From the top of a hill, two telegraph posts standing on
the horizontal plane below, were seen in a line with the observer ;
their angles of depresyion were 30'^ and 45° respectively and
their distance from each other 176 yards; find the height of
the hill. . ■ ! • An^. 240.416 yards.
27. From a ship a lighthouse bore N.NvE. ; after sailing
E.b.S. seven miles it bore N.W.b.N. ; find its distance from the
ship at each station. Ans. 5.953 miles; 8.257 miles.
28. A flagstaff 24 feet long, standing on a cliff by the sea
shore, subtends at a ship an angle of 38' ; the elevation of the
cliff is 14*; find the height of the cliff and the ship's distance
from it. Ans. Height, 508 feet ; distance, 2038 feet.
29. From the top of a hill, a vertical pillar 220 feet high,
standing ou the horizontal plane below, subtends an angle of
1° 12', and the depression of its top i& 12° 20'; what is its dis-
tance, and the height of the hilH J-
; i. . ^-^ Ans. Distance, 9977 feet; height, 2401 feetw !
EXAMPLES. 195
30. If B denote the earth's radius, h the height of a moun-
tain and B the dip of the horizon from its summit, show that
6 6
A = 2j5 sin^ — sec = ^ tan ^ tan — .
31. A tower standing on a horizontal plane is surmounted
by a flagstaff; from a certain point on the plane, the tower
subtends an angle /8, and the flagstaff an angle a ; from another
point c feet nearer to the base of the tower, the flagstaff sub-
tends the same angle a ; shew that the height of the tower is
c tan (3
l-tan;8tan(;-^^) • ' '^'■^-' -''
32. From the bottom of a tower 100 feet high, the angular
elevation of the summit of a hill was 32°, and on retiring 180
feet from the foot of the tower its top is seen to be in a straight
line with the top of the hill; find its height.
.J,' Ana. 901.5 feet.
33. A person walking along a straight road observes the
greatest angular elevation of a tower to be a, and from another
straight road he observes the greatest angular elevation of the
tower to be p. The distances of the points of observation from '
the intersection of the two roads, are a and b ; shew that the
height of the tower is .^ : ,_ . , ,, . ^
I
a'-b^
'cot'"^ a-cot^ /3*
34. Three observers -4, B, C situated in the same straight
line, A and C being each at a distance of 1000 yards from B,
find at the same instant the angular elevations of a balloon to
be at A 48" 10', at B 54° and at G 60° 30'; find the height of
the balloon. Ans. 2169.05 yards.
35. A person itanding on a horizontal plane observes that
196 PLANE TRIGONOMETRY.
the top of a telegraph post standing on the same plane, is in a
line with the top of a building which stands on a hill at a greater
distance than the post ; the distance of his station from the foot
of the post is b and the angle subtended by the height of the
building is /?. He then moves to a station farther off from the
foot of the post by a distance a, and in the same lino with the
former station and the foot of the post, and he there observes
that the angle subtended by the building is the same as before,
and that the top of the post is in a line with the foot of tlie
building. Shew that the height of the post is J ab + b'\ and
the height of the building
a (a+ 2b) tan ^
a - 2 Jab + b^ tan /8
36. The angular altitude and breadth of a cylindrical tower
are observed to be a and /8 respectively, and at a point c feet
nearer to the tower they are y and 8 ; shew that its height is
, and its breadth
cot a - cot y
. /5 . 8
sin 2" sin -
sm---sin-
37. A tower, 51 feet high, has a mark at the height of 25
feet from the ground ; find at what distance the two parts sub-
tend equal angles to an eye at the height of 5 feet from the
ground* , ., .4 ws. 160 feet.
38. The angular elevation of a tower at a place A due south
of it is a, and at a place B due west of A, and at a distance o
from it, the elevation is /3; shew that the height of the tower ig
c sin a sin ^
. ' V'{sin(a+^)sin(a-/3)} * ' -; '
EXAMPLES. 197
39. A ship sailing on a S.S.W. course bore due south,
and the angle subtended by the ship was 20' 15", and her
length was known to be 160 feet ; find her distance.
Ans. 1.94 miles.
40. From the top of a mountain three miles high the true
depression of the horizon was 2° 13' 27"; required the diameter
of the earth, supposing it to be a sphere. Ans. 7952 miles.
41 At noon, on the shortest day of the year, the shadow
of a perpendicular post was seven times as long as its shadow
at noon on the longest day ; find the latitude, the sun's declina-
tion being 23° 28'. Ans. 38° 27' 47".5.
42. When the sun's declination was lb" T 12". 5 N. the
shadow of a perpendicular post was to the height of the post
as 5 to 3 ; find the latitude.
Ans. 74° 9' 23" N., or 43° 54' 58" S., according as the shadow
fell N. or S, of the post.
43 The length of the shadow of a perpendicular object was
4 feet, arfd its longest shadow when sloping was 5 feet ; find the
sun's altitude. Ans. 36° 52' 11".5.
44 If a ship sail from a certain place 174 miles eastward
on a parallel of latitude, then due south 5°, and then west-
ward on a parallel of latitude 194 miles, and reach the same
longitude ; required the latitude of the place arrived at, sup-
posing the earth a sphere 7912 miles in diameter.
Ans. 48° 43' 22" N.
45. A ship sails westward on a parallel of latitude 125
miles which are found to be equal to 2° 30' of longitude; find
her latitude. C / \. •.. .■& Ans. 43° 36' nearly.
46 An object 12 feet high, standing on the top of a tower,
subtends an angle of 1° 54' 10" at a point 250 feet from the
base ; find the height of the tower. Ans. 160.85 feet.
198 PLANE TRiaONOMETRY.
47. A statue 10 feet high, standing on a column 100 feet
liigh, subtends at the eye of an observer in the horizontal
plane on which the column stand.s, the same angle as a man
() feet high standing at the foot of the column ; find the dis-
tance of the observer from the column, his height being 6 feet.
Ans. 121.095 feet.
48. A flagstaff 8 feet high, standing on the top of a tower,
subtends an angle of 57' 17". 75 at 100 yards from the foot of
the tower; find its height. . Ans. 232 feet.
49. A tree leans towards the north, and at two points due
south at distances a and 6 respectively from the base, the
angular elevations of the tree are a and ft. If 6 be the incli-
nation of the tree, and h the perpendicular height, shew that
tan = r — — -, h =
h cot a- a cot )8' cot ^ — cot a '
50. Four inaccessible objects A^ By C, D are situated in
the same straight line, and visible from only one point E. The
distance between A and B is 20 chains, and between C and 1)
12 chains; the angle AEB=20\ AEC = U\ and yl^i) = 50°;
find the distance between B and C. Ans. 16.48 chains.
51. If the three segments AB, BC, CD of the straight line
in the last problem, be represented by a, x, b respectively, and
subtend at E angles a, ^, y respectively, shew that
sin ^ sin (a + /8 -l- y)
a? + (a + h)x = ah
sm a sin y
cosec' )8 /cot a + cot )8\ /cot y8 4- cot y\
ah V a + x /\ 6-t-a; /'
52. A vertical pillar standing on a horizontal plane appears
of the same breadth all the way up, to a spectator standing at
a given point on the plane; shew that at any point of the pillar
EXAMPLES. 199
whose angular elevation is ^, the radius is a sec tf, where a is
the radius of the base.
53. From the top of a mountain the angles of depression of
two stations on the plane at its foot are observed to be a and /?,
and the diflference of their bearings is y. If a be the distance
between the stations, shew that the height of the mountain is
a sin a sin B . . „ . sin 2a sin 2B „ y
- — -. ^r 2 > where sin" 6 =-- — r— ,-7 ^r- cos'' j- .
sm (a + ^) cos 6 sin- (a + y8) 2
54. The boundaries of a tract of land are described in a
deed as follows: " Commencing at the intersection of two roads,
thence on a course N. 52° E. 21.28 chains; thence S. 29° 45' E.
8.18 chains; thence S. 31° 45' W. 15.36 chains; thence to the
point of beginning." Find the bearing of the last course and
the area of the tract of land.
Ana. Bearing, N. 61** W.; area= 19 ac. 2 r. 36 p.
55. Calculate the bearings of the last two courses and the
area from the following field notes : Commencing at a post,
thence N. 45° W. 20 chains; thence K 18° E. 12.25 chains;
thence E. 12.80 chains; thence N. 32° E. 6.50 chains; thence
S. 45° 30' E. 13.20 chains; thence 14.75 chains; and thence to
the point of beginning 16.30 chains.
Ans. S. and S. 65° 25' W.; area = 59 acres.
56. The boundaries of a tract of land are: A£, W. 25
chains; BC, N. 32° 15' W. 16.09 chains; CD, K 20° E. 15.50
chains; BU, E. 25 chains; BF, S. 30° E.; and FA, S. 25° W.
to the point of beginning. A line is run from A cutting off 70
acres 1 rood 33 perches from the west side; find the second
point in which this line cuts the boundary.
Ans. The side DF, 18 chains east of D.
57. A ship sailed from latitude 51° 24' N. as follows: S.E.
40 miles; N.E. 28 miles; S.W.b.W. 52 nules; N.W.b.W. 30
200 PLANE TRIGONOMETRY.
miles; S.S.E. 36 miles; S.E.b.E. 58 miles; find her bearing
and distance from the point of departure. • ' '^« • -
Ana. Bearing S. 25° 59' E. ; distance = 95.87 miles.
58. A ship sailed 320 miles on a parallel of latitude from
longitude 81° 36' W. to longitude 90° W, ; in what latitude
was she? Ans. 56° 30' 47". 3.
59. Twilight ceases when the sun is 18° below the horizon,
find the latitude of the places at which twilight lasts all night
at the time of the summer solstice, the sun's declination being
23° 27' K ■ ' Ans 48° 33' N"
60. The meridian altitude of a star is 64° 10', and its
depression below the horizon at midnight is 28° 30' ; find the
latitude of the place and the declination of the star.
Am. Latitude, 43° 40' N ; declination, 17° 50' N.
61. At two places on the same meridian, one in latitude
59° 31' 30" N., the other in latitude 33° 56' S , the moon's
meridian altitude was observed to be, at the northern station
43° 47' 40", and at the southern station 41° 21' 40"; find the
moon's distance from the northern station, the earth being re-
garded a sphere whose radius is 3956 miles
Am. 238020 miles.
62. When the moon's horizontal parallax is 57' 32", and
the apparent altitude 50° 40', what is the parallax in altitude?
•^ --^... .... .v: ,,.. :. . Am. 36' 28".
63. Find the augmented semi-diameter of the moon when
her true semi-diameter is 15' 42", her horizontal parallax and
apparent altitude being the same as in the last question.
Am. 12".41.*
64. When the moon's horizontal parallax is 58' 10" and her
true altitude 50°, find her parallax in altitude.
.-■v- Am. 37'52''.6>
INSCBIBED CIBCLE.
201
■ >\ I'
: r i c CHAPTER XI.
CIRCLES INSCHIBED IN AND CIRCUMSCRIBED ABOUT A TRIANGLE,
POLYGONS, AREA OP A CIRCLE, ETC.
152. To find the radius of the Circle inscribed
in a Triangle. . „ , ,.
The centre is in the intersection of the three lines bisec-
ting the angles of the
triangle. Let the in-
scribed circle touch the
sides of the triangle in
the points D, E and F.
Join OD, OE, OF; the ,
angles at D, E and F
are right-angles. {Euc. B"
III., 18.)
Let OD=^OE=^OF= r, the radius of the circle ;
then area of triangle £0C = iBG. 0D = -^,
br
area of triangle COA = \AC.OE^ — ,
■;f ■'■'.'■■
area of triangle i 05 = i-4^.<5^= -g i
therefore, by addition, ..^ ; . :^ j^ ;;,^ .
(a -h 6 + c)^ = area of triangle ABC = A.
; ?
Put, as before, a + 6 -1- c = 2»,
:\ ^V;'v*^»
th^n
A
r== — .
8
(218)
202
PLANE TRIGONOMETRY.
Tiiat is, the radius of the inscribed circle is equal to the area
of the triangle divided by half the sum of the sides. Different
forms can therefore be obtained for the radius by employing the
various expressions already given for the area of the triangle.
Again, from the triangle BOG we have by (204)
BC sin OBD am OCD
0D==
Bin (OBD+OCD)
or
. B , G
asm. — sm —
2 2
sin ^{B+G)
. B , G ^
a sm -- sm --
2 2
A "V
cos^
From the figure we have
AE=AF, BD=BF, GD^GE;
AE+BD+GD=s,
■■ AE+a=s,
, AB=s-a=::AF. \
£D=s-h = BF.
GE=s-c = GD.
hence
or
therefore
Similarly
or
From the right-angled triangle AGE, we have
.■ OE=AEtsinOAE
A
2'
B
2'
Similarly
r=(s-a) tan
r=(a-6) tan
'hi
, r={s-c) tan
G
Multiplying the last three equations together, we have
0»
A B
r^ = (s- a) {s -b) (s- c) tan -~ tan - - tan
A" A B G
=— tan — tan — tan —
« 2 2 2 ... A .,
Bsr^s tan — tan — tan — ,
2 2 2 '
, A B G '-.
r=s tan - tan — tan — .
■ 8- .1 _, .J
G
(219)
(220)
(221)
■?'!";' ;"Mt.
(222)
ESCRIBED CIRCLES.
203
From the figure we have , . ! . i
AO=AE sec OAE ' ^
, . A s-a
= (s-a)sec-- = — ^,
iv:.'.7
(223)
Similarly
be
s
ac
s
ah
A
cos ~-=
2
£
cos-=
C
cos -=
cos -
2
c
s-a
.be.
B0--
s-6
. ac.
C0=
= (-«
ah.
(224)
153. To find the radii of the Escribed Circles.
The escribed circles are the three circles which touch one
side of a triangle and the other two produced, and are exterior
to the triangle.
''V■<.i■:^
:'.{ i\i
>\1K_
?ii)vi
TietABC be a triangle, bisect the exterior angles at B and
C by BOi, COi, then Oi, will be the centre of the escribed circle
204 PLANE TRIGONOMETRY.
which touches BC and the other two sides AB, AC^ produced.
Join AOi which will also bisect the angle A, and draw O^D,
OiE, OiF, to the points of contact 2>, U, F; then the angles at
2>, E and F are right angles.
Let OiD = O^F=OiF=ri, the radius of that circle which
touches the side a, and let ra, r^ denote the radii of those that
touch the sides b and c respectively.
The quadrilateral ABO^^C = the tria.ngha AO^C, AO^B,
\ * = the triangles uijBC, BO^G,
therefore, the triangles A 0^0, A O^B = the triangles ABC^ BO^C^
A
therefore r, =
In like manner we find r, =
and '.'i .« r3 =
■ ^vA ''; S - C
(225)
Bisect the angle ACS by CO, then is the centre of the inscribed
circle, and OCO, is a right angle; heuco the angle 0501=90°
sLiidBC0^==9QP~~, ;: ^ '- > Ci
From the triangle JBOOi we have by (204)
^ _- sin OBOi sin BCO,
^ Bm(CBO^+BGOi)
■'\, . . ■
, ^ ; • asm(90''-|-)8in(90<'-^)
- :. , n.. V,^. sin {180°-i(JS+0)^
ESCRIBED CIRCLES.
205
that is
ri =
Similarly
B C ^
a cos — cos —
2 ^
A
, cos-
COS — COB ~
2 ^
r- =
,^»
r« =
5
cos-
^ B
c COS — cos —
2 ^
008-
J226)
i '.:(':(
.v/ (82^ r-f'-.'
1. .
From the figure we have , ^ .;;„ „ j^.
AE=AG+CE=AC+OD,
AF=^AB+BF=AB+BD.
But AE=AF,
therefore 2^^=2^J'=^5 + ^.C+BC
■ >' ■•'
=a4-&+c,
"-■■■■ ,''C!/ ■ ;i . -'--''■■ '•-
and AE=AF=^a'\-h-\-c)=s. ^
Also CD=CE=AE-h=s-h,
and BD='FB==AF-c=-s-c.
. t
^
»'■■
;•.! <;,v
D/;. or ,|vil
(227)
From the right-angled triangle AO^E we hav«
Oi^=^-E^tanj&^Oi
or
Similarly
r, =5 tan — . if Ha •> '
^ 2
JO
r^=s tan — . \- rm 'h~J)^
(228)
C
r«=» tan —
.^ ft:"S^Qi.>
206
PLANE TRIGONOMETRY.
find
From the right-angled triangles AEOi, CDOu BDOi ^e easily
A ' ' '
AOi =AE sec EAOj^ =s sec — .
BO^=BD sec DBOy_ = {s'-c) cosec -.
COj =DC sec DCO^={s-h) cosec -.
(229)
Similar expressions may be easily deduced for AO^, BO2 , &c.
From (228) we find
^ A
^^ 2 B ^ A
,, or t*! tan —=r2 tan — ,
tan
B
that is,
BF=AEf and therefore AF^BH^^a.
154. To find the distance between the centre of
the Inscribed Circle and that of one of the Escribed
Circles.
. ■ ■ ^'■, ^ . ■ ' '. ■ ,
From the figure we have '* l' t ; '; ,
00i=AOj,-AO
A , A
=» sec — - (a _ a) sec — , by (229) and (223)
2
A
-a sec
Similarly 00^=^ sec — ,
00^=0 sec
O
■i} (230)
CIRCUMSCRIBED CIRCLE.
207
155. To find the radius of the Circle circum-
scribed about a Triangle.
Let ABC be a triangle and the centre of the circum-
scribing circle whifth is found geo-
metrically by Euc. IV., 6.
Draw the diameter BOA' and
join A'C, then the angle BA'C:=
angle BAG^ because they are in the
same segment, and BGA' being in a
semi-circle, is a right angle. Denote
the radius BO by R,
then
OP
whence
A' B sin BAV = BG, >
2B sin ^ = a,
f ■ ^ -■ a
i?=
2 sin A
h '
2 sin^
' "■©■■ ^■
by (165). 2 sin (7*
Or we may proceed as follows:
Draw OD perpendicular to BCj
then , OB Bin BOD ^BD,
r.' -'yt.u'
a
,'j.
or
whence
^ sin -4 = — ,
a
, as before.
(231)
2 sin A
To express R in terms of the sides and area we have
R-=
a
2 sin ^
abc
2bc sin A
abc
"IK
by (199).
(232)
• 1 1
208 PLANE TRIGONOMETRY.
From (180), (181) and (182) we have ■ \ \
A B C As
cos ~ cos — cos — = , which combined with (232) gives
2 2 2 abc
JJ=— sec -- sec -— sec — •, (233)
4 2 J ^ ,
156. Relations between the radii of the InscribiBd,
Escribed and Circumscribed Circles. ,
The product of (218) and (225) is
''• t'l..
From (225) we have
111 s—a-]-8—h-\-8—o
I 1 = -
ri ra r^ A
3s-(aH-6+c) 5 1
(235)
A A r
From (222) and (228) we have
, ^ ^ Cf ^ ^^ JB ^ O,
ri4-»*2+**3 -*•=«(*»« ^+tan -4-tan --tan - tan - tan ~).
Dividing by (233) we obtain ' '^ l^ '"'\
ri+r2+r3-r ^^, A B C ^''''' A . B C ^ *•**
—— — =4(8m- cos - cos -+C0S - sm - cos - ..
A B . C . A . B . C-'
. +COS - cos — sin ^-sin - sm - sin -)
^^. A B-\-0 A . B+C
=4 (sm - cos --—fees - sin — -—)
A A
• id =4(sin'' — •+cos'^— )=4, \"yi m 'A ;^^^t]^<v/X
2 2 . ,
therefore ri+rg+rg -r=4iJ. (236)
From (231) we have
a a
i2=
"2 sin ^ ^ . ul A*
f.'-> 4 sm — cos —
- ' 2 2
INSCRIBED AND CIRCUMSCRIBED CIRCLES.
209
i A
whence a=Ui sin ^ cob --, which substituted in (219) gives
2 2
r=«4Ji sin — am - sin —
J J J
=12 (cos 4+ cos 5+ cos (7-1) by (112 his).
In a similar manner we obtain from (226)
,. . ^ ^
ri=»4i? sm — cos — cos — ,
2 iB ^
„ A . B G
r3=4E cos - sm - cos - ,
„ A B .
r3=iB cos — COB — sm — .
J ^ «
(237)
(238)
157. To find the distance between the centres
of the Inscribed and Circumscribed Circles.
Let be the centre of
the inscribed, and Q that
of the circumscribed circle. -
Draw O-B, and QF perpen-
dicular to il5. Let 0(?=D. ;
The angle AQF=C, there-
fore FAQ = 90" -C, and
FAO=-i
hence 0^^=90° -C- g =i(^ " ^)' *^^ ^^^^
. A
sm-
By (169) we have
OQ^=AQ^-i^AO^ - 2AQ . AO cos OAQ
Ot
2)2 =1224- ■
r^ 2Ercos^(.B-C) -
but from (237) we have
B ,
:.: 4i?r8in-Bm-
(a^
8m'
sm-
m
210 PLANE TRIGONOMETRY.
^ . ^ . Cf
4ifrBin— sin-- .
2 2 21tr COB MB - u)
therefore D^=B^ + ^ " ^^
sin - sin -
=.^..2iJ.^2Lil^=i?.-2iJn
. ^1
sin-
(230)
Otherwise as follows. From (a) we have
A A
(2)2-222) sin2 ^^^^ _2Br sin — cos UB-C)
or ^^(2)2-222) (l-cos ^)==r2-222r cos i(B+0) cos \{B-C)
=r2 -22r (cob ^+cos C).
In a similar manner we find
^(2)2-222) (1— cos JB)=r2 -22r (cos ^+cos C),
the difference of which is
^(1)2 -222) (cos ^- COB J5)=- 22r (cos;^ -cos B),
whence D^=B^-2Br.
158. To find the distance between the centres
of the Escribed and Circumscribed Circles.
Let Oi be the centre of the escribed circle touching the side a
(Fig. of last Art.), and OiQ=Di, then since AOi= ^^ , we have
sm-
from the triangle OiAQ
r? 222r,coB^(B-C)
"^1"-" '^ A .A
sin2 — sin —
B G
^ 422ri cos - cos -
but from (238) we have '^^= ^ t
sin2- sin-
INSCRIBED AND CIRCUMSCRIBED CIRCLES.
211
therefore
Dl=B' +
B
4Rri cos - C08 —-
1 2 2
sin
A
2Rr, coH ^jB-C)
. A
sin
Similarly
D2=B2+2iBr2.
(240^
The sum of (239) and (240) gives
D'^+X>i-|-i>3+I>3=4B=+2B(ri-fr2+»'a-r)
(241)
==12E% by (236)
159. To shew that the distances between the
centres of the Escribed Circles and the centre of
the Inscribed Circle are bisected by the cu-cum-
ference of the Circumscribed Circle. ,, • (
Let Q be the centre of the circum-
scribed circle of the triangle ABC
Bisect CB in D and draw DE perpen-
dicular to CB, and join AE ; then the
angle GAB is bisected by AE, and the
centres of the inscribed and escribed
circles are in AE ; let them be at O
and Oi respectively. Join EB and
OB.
Since the angle GBE is equal to the
»ngle GAE, we have
EBO=GBE+GBO=UA+B),
EOB=OAB+OBA=^{A+B),
EO=EB=DB sec DBE
and
therefore
{Euc. I. , ;-i2j
but by (230)
therefore
_ a A
0E=^— sec — -
2 2
A
00 1=0' sec -,
00i=20E.
212
PLANE TRiaONOMETRY.
We may here notice, for the sake of problems, that if OQ be
joined and produced both ways to meet the circumference, we have
hy Euc. III., 35,
{B-{-OQ) {B-OQ)=EO . AO
a A
.in-
ar
2ar
^ . A A 2 sin ^
2 sin — cos —
2 2
.2Br,
whence
OQ2=i22-2i2r.
i6o. To find the Perimeter and Area of a Regular
Polygon of any number of sides, which is inscribed
in or described about a Circle of given radius.
Let AEB be an arc of a circle whose centre is 0; AB a
side of the inscribed regular
polygon of n sides ; OE at right
angles to AB, and therefore bi-
secting it; CD a tangent at E^ A)
and meeting OA and OB pro- c
duced in C and D; then CD is '*
a side of the circumscribed regular polygon of n sides.
TT
•TT
Let ^ = r. The angle AOB = — , and therefore AOF^-
n
n
IT
AB = 2AF= 2r sin AOF= 2r sin - ;
n
IT
therefore perimeter of inscribed polygon = 2wr sin — . (242)
n
Area of the inscribed polygon = 7i x triangle AOB
AO.BO
= n-
sin AOB
= inr^ sin .
n
!VV
(243)
AREA OF A CIRCLE. 213
Again, CD = 20 E = 2r tan COE = 2v tan - ,
therefore
perimeter of circumscribed polygon = Inr tan - . (244)
Area of the circumscribed polygon = n x triangle COD
= w. OE. CE^nr" tan -. (245)
From (243) and (245) we have
area insc ribe d polygon ^ area^triangle_^0^ ^ ^^
area circuiii^ibed polygon area triangle COE OE'
^% = co.'-. (246)
OA' n •
i6i. Circumference and Area of a Circle.
. The circumference of the circle is evidently intermediate
in length to the perimeters of the inscribed and circumscribed
polygons, hence the circumference of the circle lies between
that is, between
IT ""
2wr sin — and 2wr tan — ,
IT "^
sin — tan —
71 ^
27rr .^— and 27rr '
IT
n
n n
Now, let the number of sides be indefinitely increased, then
is very small, and when 7i = oc,^ = 0, therefore (Art. 74)
sin — tan —
!L = 1 = 1, when- = 0.
n n
Hence when n is infinite, the perimeters of both polygons be-
tween which the circumference of the circle lies, become 27rr,
therefore the circumference of the circle = 27rr.
214 PLANE TRIGONOMETKY.
Again, the area of the circle lies between
^nr sin and nr''' tan — ,
27r , TT
sin tan —
or between nr^ and irr^ — ,
Zir IT
n n
each of which becomes irr^ when n = cc .
Therefore the area of the circle which is always intermediate
in magnitude to the areas of the two polygons
= Tr^. (246 bis)
Hence, if 6 be the circular measure of the angle of a sector
supH as AOE in the last Figure, then the area of the sector
A0M=^6r\
162. To find the Angles and the Area of a Quadri-
lateral inscribed in a Circle.
Let AB=^a, BC=b, GD=c, DA=d; join ^
AG: then
Area il^Oi)=area ABO-\-ADC
=^ab sin B-\-icd sin D
=^{ab-\-cd) sin B. (1)
We have by (169)
AC'^=a^-\-b^-2ab cos B=c^-{-d^-^2cd cos £
since cos .D= — cos B, (Art. 39)
therefore cos B= ■ ,
2{ab-^cd) '
hence 2 sm^ — =1 — cos 5=^^
2 2{ab+cd)
(2)
and 2 cos" — ==l4-co8 5= — ^— ^^ ^^ ^, ^3)
2 ^ 2(abi-cd) * ■ . ^^^
therefore tan^ — —
EXAMPLES.
B (c+d)2 -(a-b)
215
2 (a+6)2-(c-d)
2
2
(247)
" {a-\-h-c-\-d) {a-\-h-\-c-d)
{s -a){s- h)
'='{s-c){s-d)'
where 2s=a-\-h-\-c-\-d.
Multiplying (2) and (3) together we easily find
8inB=-T^l/(«-«)(«-^)(«-^)(^-'^)
(W-\-cd
which substituted in (1) gives
Area ABGD=V (s -»)(«- &) (« - c) (« - <i). (2*8)
If v/e substitute the value of cos B, in the expression for ^OS
we readily find
],_ {ac-{-hd)(ad-]-hc) ^
ab-\-cd
If i?! denote the radius of the circle, we have
AG _, kob+cd) (oc+bd) (ad-jrhc) .g^g)
Examples.
L In any triangle prove that
(1) acot^ + &cot5+ccotO=2(i2+r),
(2) a cos ^+6 cos B+c cos — 412 sin A sin B sin C.
From (1) we have
a cos J. 6 cos B c cos (7
acot^+6cot5+ccot0=r-r^-+-^— + ^.^ ^
_ lJL_ (cos ^+co8 5+cos (7), by (165)
sin A
= 2J2(1+^), by (231) and (237)
.; , =2(E+r).
216 PLANE TRIGONOMETRY.
From (2) we have
, , , „ _ a sin ^ C08 A , bsiriB cos B csinC cos C
a cos ^+6 cos J5+C cos (7==:- 1 1 —
. sin A sin B sin G
a
(sin 2^+8in 25+8in 20)
^ sin A.
= 4R sin ^ sin jB sin C. by (112).
2. If a, y8, y be the distances of the angles of a triangle from the
centre of the inscribed circle, and a, b, c the sides opposite to them,
prove that
a?a + ^h + '/c=' abc.
In the Figure of Art. 152, let A =a, BO =^ a.nd GO =y;
,, r r r
thb.1 a= J, ^=— ^, y = ^,
sin- sin- sin-
hence
sin"'' — sin'^ — sin'^ —
2 ia ^
„ „/ a A , b B c C\
=2r { -; 7 cot — +-: — - cot — H- — ^ — - cot — I
Vsin A 2 sin jB 2 sin (7 2/
2ar» / J[ B , G'
sin J.
(^cot- + cot- + cot-j-
But r cot— — AE, r cot ~ = BF, r cot — = D0,
« 2 2
therefore r (cot — +cot -+cot --)—AE-{^BF-{-DG=s,
2 2 2
2a 2a
hence a'a + S^b + '/c = — r rs == -: r A,
' sm -d sin A
2a be .
. — Bin A =zabc.
sin A 2
3. Prove that the radius of the circle which circumscribes the
triangle O1O2O3, formed by joining the centres of the escribed
abc
circles of a triangle ABG, is ^-r- , or equal to the diameter of the
Examples. 2l7
circle which circumscribes the triangle ABC; and that the area of
the triangle O1O2O3 is — — , where r is the radius of the inscribed
circle. (See Fig. of Art. 153.)
In the Figure we find the angle BC0i=l(A-\-B)=AC02,
CA02=:^UB+C) = BAO.,, £OiC = i(B+C?), AO,C = UM-C).
From the triangles AO^C, BO^C we find by (165)
cos — a cos —
CO^ — ^ ■ and (70i=— 7-,
cos - cos -
B A B A
cos — COS — a cos'^ —4-6 cos'^ —
2 2 2 9
therefore Oi O2 = a — ^ + 6 p = ^ p
cos — cos — cos — cos —
2 2 2 2
a^ 6c 2.9 — g - 6 c
V~ g(3 - g ) . s{s - 6) ~ /(i-a) (s-6) ~ C '
6^7^^ \ ^~~ «^ 2
b n
Similarly Q1O3— p and O2O3— ' . .
Let i?x denote the radius of the circumscribing circle,
A
^ ^ a coaec — •
then B,^ "^^'^ 2
i.l^-
2 8in£0iC „ A ^ . A A
2 cos — • 2 sin — cos —
2 2 2
a abc
= 2B.
sin A 2A
Area of the triangle OiOgOa^i O1O3 . O1O2 sin OgOjOa
6 c ^
sin-- sm-
218 PLANE TRIGONOMETRY.
A
. cos —
he 2
sin -sin -
"'^ by (219)
2r
abc
= 4^(ct-f *+c).
4. The sides of a triangle are 13, 14 and 15, find the radii of the
inscribed and circumscribed circles. Ans. 4 and 8^.
5. Find the radii of the inscribed, circumscribed, and of an
escribed circle, when the triangle is equilateral. •
^^. g/3, -/a, -»/3.
J5 Q
6. In the Fig. of Art. 152 prove that J. 0=::c sin — sec — .
it 2
B C
7. In the Fig. of Art. 153 prove that J.Oi =:c cos — cosec — .
<" 2
8. In any triangle prove that r= p . ,
. , cot — - cot —
. 2 2
A B C , A , B O
and Ti cot — =»*2 cot — =r3 cot — =r (cot — +cot — +cot — ).
A A 2i 2i jL VL
9. In the ambiguous case of triangles, when a, b and A are given,
shew that the circles circumscribing both triangles are equal, and
that the distance between their centres is
\/{a^ cosec^ A-.b^).
10. Prove that the perpendicular from an angle of a triangle
on the opposite side, is a harmonic mean between the radii of the
adjacent escribed circles.
11. In the Fig. of Art. 153 prove that OA . OOj =4i2n
12. In any triangle prove that {ri - r) (rg - r) (rg - r) = 'kRr^.
13. A circle is described passing through the vertex -4 of a tri-
angle and touching the base BC a.t its middle point; prove that its
,. . 2(6«4-c'')-a'' -
radius is — -;; — : — -; — «
86sinO • ,
EXAMPLES. 219
14 The sides of a triangle are 13, 20 and 21 ; find the radii of
the inscribed, circumscribed and escribed circles, and the distance
between the centres of the inscribed and circumscribed circles.
Am. r=4f , R^IO^ , ri=9, r2=18, r3=21, andZ) = iv/65.
15. An inaccessible tower standing on a horizontal plane sub-
tends an angle of 36° at each of three points on the plane, whose
distances from each other are 29, 35 and 48 yards ; find its height.
Am. 24^}/ 5-2y/'Q yards.
16. In any triangle prove that
(1) A = JBr(8in^+sin5+sin C);
ABC
(2) =r« cot — cot -^ cot — ;
(3) =222* sin il sin 5 sin (7;
(4) =iJSMsin2^+sin25+sin2(7).
17. In any triangle prove that ' .
(1) B = i\j . ^ "'' . ..
\ sm A BUI B am C
2(sin ^+8in JS-f sin 0) ' •
«^c / A , B, , C'
B sin ^ + sin 5+ sin (7
r 2 sin .4 sin -B sin C? * I
^ ahc !,. '<'■..■■ ..'■.. ■■ ■ ■■' .' ' '-i
(5 22?rrr: -— — .
a+o+c • ,
• ■ ^ . , •
18, In any triangle prove that
c sin A sin B
(1)
sin ^.-f-siin 5-|-sin G '
'^ ahc
(^ = i i;. ■ u <»i^ ^+8in B f sin C).
{a-\-b-]rc)'
19. If the radii of the escribed circles be in arithmetical pro-
gression, the tangents of the semi-angles of the triangle are in arith-
metical progression ; if in harmonical progression, the cotangents of
the semi-angles are in arithmetical progression.
220 . PLANE TRIGONOMETRY.
20. If p, q, r denote the lines drawn from A, B, C, bisecting the
angles of a triangle ABC, and terminated by the circumference of
the circumscribed circle, prove that
p cos — -irq_ cos — -f-r cos — =: a-\-o-\-c.
2t M A
21. A circle is described about a triangle ABC, and another
triangle is formed by joining the points of bisection of the arcs
subtended by the sides of ABC ; prove that the sides of the new
triangle are
a Ah Be G
-cosec-, -cosec-, -cosec — ,
, , . A A B C
and that its area = — cosec — cosec — cosec — ,
8 2 2 2 *
where A is the area of the triangle ABC.
22. In the Fig. of Art. 155, if BO meet AC m E, prove that
i2:=^Oco8(^-C)sec-B.
23 Perpendiculars are drawn from the angles A, B, C oi oxi
acute-angled triangle to the opposite sides, and produced to meet
the circumscribing circle ; if the produced parts are a, ji, y respec-
tively, prove that
— +-5-+— = ^(tan ^+tan B-ftan C).
a fi y
24. Perpendiculars AD, BE, CF are drawn from the angles of a
triangle ABC to the opposite sides, meeting one another in G; if B
and R' be the radii of the circles which circumscribe the triiingles
ABC a.n.d DEF respectively, and / the radius of the circle inscribed
in the triangle DEF, prove that
(1) AG = 2B cos A ; (4) DG = 2R cos B cos C ;
R
(2) EF = JB sin 24 ; (5) Area ABC — — {DE^-EF-i-FD) ;
(3) B-'LEi (6) r'=i2i2co3 4co8£cosC;
(7) AG-YBG^CG-'l{R^r),
Shew also that the circle which circumscribes the triangle DEF,
bisects the aides of the triangle ABC and the lines AG, BG, CG.
EXAMPLES. . 221
25. Prove that
tan — cot — c tan — cot —
,^ , A rvi ^ 2 2 2 2
tan-+tan— tan-+tan~
' a h e ^, A B C^
26. In the Fig. of Art. 153, if r and r' denote the radH of the
circles inscribed in the triangles ABC and O1O2O3 respectively,
prove that
r , A B C^^ A ^ B ^ a
— = (cos — 4-008 — + C03 — ) tan — tan — tan — .
r' 2 2 2 2 2 2
27. If the middle points of the sides of a triangle be joined with
the opposite angles, and J?i, iJg, -R3, <fec, , be the radii of the circles
described about the six triangles so formed, and ri, r2, ra, &c., the
radii of the circles inscribed in the same, prove that
^ ^ ^ ^ ^ ^ ,111111
BiRiR^-RzEiR^, and — + — + — = — + — + — .
^•i ^3 ^5 ^2 Ti ra
28. If a and a' are homologous sides of two similar triangles
described, one about and the other within a circle, prove that
a =4a sin — sm — sin — . , ,, ,
a Z a
29. If a, (B, y denote the distances from the angles of a triangle
to the points of contact of the inscribed circle, prove that
30. If a', b', c' denote the distances between the centres of the
escribed centres of a triangle, prove that
a'h'c' . , . « . ^ :'.•
r^rgrg = — — sin A sm B sin G.
o
31. With the same notation as in Example 2, prove that
(1) r=i"^(a + 6 + c); (2) -^ + i- + i-= l_(i + i?)
' abc ' ^ aa? bjS^ c-/ abc r'
<«' "'(t-7)^^(7-^)->'(M-)=»-
222 • PLANE TRIGONOMETRY.
32. In any triangle prove that r=-
rira + rira+rarg
33. The radii of the escribed circles of a triangle are 16, 48 and
5:^ ; find the sidesL Ana. 25, 51 and 52.
ABC
34. In any triangle prove that rirjra =r^ cot^ — cot^ -- cot^ --.
^ ^ 2
35. If r be the radius of the inscribed circle, and ra the radius
of the circle inscribed between this circle and the sides containing
the angle A, shew that
1 - sin ^
ra = r ^ = r tan^ (45° - -).
1 + sin —
36. The sides of a triangle are in arithmetical progression, and
the distance between the centres of the inscribed and circumscribed
circles is a geometric mean between the greatest and least; shew
that the sides are as
|/¥-l : i/6~: 1/5^ + 1.
37. In any triangle prove that
(1) ^=2-2(sin2 ^ + 8in2 ~ + sin2 -);
(2) ah + ac + bc=rri+rr2+rr3+rir2+rir3+r2ri ;
(3) ^j^^ Jr^+r^Ur.+r^U r^+r,) ^
38. If Pj, Pj, p, be the perpendiculars from the angles of a
triangle on the sides a, b, c respectively, prove that
111111
(1) -_+_+_- = _+ — +_j
Pi P* P» r, r, r,
„ 'Zrr.rtr.
m Ji= — ^-^; .
P1P2PS
2r,r3
.. 1111
(4) -+ =-;
Px P, P» ••«
p* pl pl be ac ah '
P^Pb PiPs PiP» a' * c
EXAMPLES. • 223
39. If r,, ra, r, be the radii of three circles which touch one
another externally, shew that the area of the triangle formed by
joining their centres is K' (r, + r, + r^) r^rtr,.
40. If and Q be the centres of the inscribed and circumscribed
circles of a triangle, and if r^ , r^ , r^ be the radii of the circles
which circumscribe the triangles BOO, AOG, AOB respectively, and
Ba , Bb , Be the radii of the circles which circumscribe the triangles
BQC, AQC, AQB respectively, prove that
a h c _ abc '"'oTb^c _ B
~Bl'^~B'b^~B^~~W' *^ abo ~^'h^c*
where B is the radius of the circle which circumscribes the triangle
whose sides are a, h and c.
41. Shew that there is only one point within a triangle from
«rhich, if perpendiculars be drawn to the sides, circles can be in-
scribed in each of the three resulting quadrilaterals ; and if ri , r2, rj
be the radii of these circles, and r that of the inscribed circle of the
triangle, then
(1.1) (i-i)+(l-M (i-MH^-i) (--i) = ^. -
42. Shew that the area of the triangle 0x0^0^ (Fig. of Art. 153)
= All+ r + 7 + T \t
\ -a + o + c a-o+c a+o-cf
where A represents the area of the triangle ABC.
43. ABCD is a quadrilateral inscribed in a circle, shew that :
AC am A=BDamB. ,'i
44. Shew that the perimeters of an equilateral triangle, a square
and a hexagon, each containing the same area, are as
t/27 : t/'16 : */12. . ", v •
45. In the Fig. of Art. 152, prove that . . • '
AO^'-i-BO^ + CO^zzab + ac+bo- ^^° -.
a-\-b-\rc
46. Prove that the area of a regular hexagon inscribed in a circle
is geometric mean between the areas of the inscribed and circum-
scribed equilateral triangles.
224 PLANE TRIGONOMETRY.
47. In a regular polygon of n sides, of which a side is 2a, prove
TT
that Bi-r = a cot — , where B and r are the radii of the circum-
scribed and inscribed circles respectively.
48. If A, Ai, A2, -^j denote the areas of the four circles which
touch the sides of a triangle, A being that of the inscribed circle,
prove that
1111
r=z= + ■
Va Va[ VA2 VA3
49. The external bisectors of the angles of a triangle are produced
to meet the circumference of the circumscribing circle ; shew that
the area of the triangle formed by joining the three points thus
R
obtained is ~ (a + b + c).
4t
50. li r^, r^, r^ denote the radii of the circles inscribed in the
triangles BOC, AOC, AOB, in the Fig. of Art. 152, prove that
— + — + — = 2 cot — + cot — + cot — ).
Ta n r^ 4 4 4'
51. The square of the side of a pentagon inscribed in a circle is
equal to the sum of the squares of the sides Qf a tegular hexagon
and decagon inscribed in the same circle.
52. Shew that the areas of the circles in Euc. IV., 10, are as
6+/ 5": 2. ■ .
53. If through any point within a triangle three straight lines
be drawn from the angles A, B, C, meeting the opposite sides in
D, Ej F respectively, prove that
OP OE OF
AD^BE^GF' '
54. If a, )8, y be the angles which the sides of a triangle a ibtend
at the centre of the inscribed circle, prove that
4 sin a sin )8 sin 7=sin ui +sin f+sin (7.
INVERSE TRIGONOMETRICAL FUNCTIONS. 225
CHAPTER XII.
INVERSE TRIGONOMETRICAL FUNCTIONS.
163, If sin 0=a, then 't) is an angle whose sine is a; but instead
of writing this description of 6 at full length, the following notation
Id employed to express this relation :
^=3in~' a.
Similarly, if cos (p=b and tan i{)=c, then ^=cos~' h and i/^=tan~' c.
The student must be careful to remember that the index - 1
does not here imply an algebraic operation, but merely expresses the
relation between the angle and its functions as above enunciated.
The functions sin~^ a, cos~^ b, &c., are called inverse trigono-
metrical functions, from the nature of the notation which is anala-
gous to that employed in Algebra, where x~^ is the inverse of x.
Since there are several angles which have the same sine, several
which have the same cosine, &c. , it follows that the above equations
are more correctly written as follows : '
sin~' a=mr+( - 1)" 0,
COS"' b=2mr±<j),
tan~' c=n7r+i/;,
where n is any integer, either positive or negative.
Any relation between the trigonometrical functions may be ex-
pressed by the inverse notation. Thus, we know that
6 sin e
tan -— =•
2 1+cos^
1 • 1 , ... ^ , / sin \
which may be written --:=tan~' ( :; ) ,
^ 2 \l+cosdr
.and therefore fl=2tan-i| ),
Vl+cose/'
which is read thus,
6 is equal to twice the angle whose tangent is
l+cos 0/ : . ■ • il
sin ': ' '•
l+cos
16
226 PLANE TRIGONOMETRY.
164. Oiven «tn""' a and sin~^ 6, to find svn~^ a ± «tn~' 6 and
coar-i a ± cos~^ b.
Let 0=8in~' a and ^=sin~' 6,
then Bin 0=a and sin ^=&,
hence cos fl=-i.v/(l — sin'' ^)=±v/(l -»')>
and cos <p=± s/(l - sin^ (j,)=^s/(l - 6»),
Bin (0+0)=8in 6 cos 0+cos sin <j>
=±av/(l-6«)±6v/(l-a«),
therefore (9+0=sin-'{ ±av/(l - 6'*)±6n/(1 -a')},
or sin-' a+sin"' 6=sin-' { ±a>/(l - 6*) ±b\/{l - a'^) } . (250)
In a similar manner we find
Bin-« a - sin-' 6=sin-'{ ±a^{l - b^)l^by/{l - a^) }. (251 )
cos-' a±co8-' fc=cos-i{a6=F v/(l - a*) (1 - b^) }. (25::)
165. Oimn tan~^ a and tan~^ b, to find tan''^ a±tan~'^ 6,
Let 0=tan~' a and ^=tan— ' 6,
then tan d=a and tan ^=&,
tan d±t8i.n <t>
tan (0±0):
ITtan^ tan^
a±b
~lTab *
(a+b \
- ),
tan-»a±tan-i 6=:tan-'(— 5=^). (253)
Xn a similar manner we find
cot-J a ± cot-i 6=cot-' ( ^^^^-^ ) . (254 )
V 64-a / '^ • ■' ' ■
166. Here we may remark in regard to (253), that since there
are several angles whose tangent is a, several whose tangent is 6,
and several whose tangent is ~ , it follows that the sum or dif-
1-rao
ference of any two of the angles whose tangents are a and b respec-
EXAMPLES. 227
tively, is o»|uaJ to flome oiui of thu angles whose tanguut is ^ -
a - l> ...
or respectively.
Si)nilar roniuvks ui)ply, of course, to the other formiilio of this
ohapter, and in faot t(» all fonnula; containing the inverse functions.
Examples.
1. Prove that
tan-'{(v/2'-<l) tan e} - tan-'ICv/^ -1) tan //}=tan-« sin 20.
By (263) wo have
tan-'{(v/"2+l) tan^}-tan~'j(v/2' -1) tan ^}
((^"2+1) tan 0-(v/2 -1) tan ^1
=tan~' •{ zz: : r
{ H-(\/2+l) (^2 -l)tan2f/ J
/ 2 tan 6 \ . ^
=tan~'i )=tan~' sm 2t/.
\l+tan2 6>/
2. Given versin"' versin-' (1 - 6)=versin-i — , find x.
a' a
Since the versin=l — cos, the above becomes
X hx
cos-' (1 ) - cos-' 6=cos-' (1 ),
a a
or cos-' \ (1 - -)hWa - J>') (1 - (1 y) ] =co8-' (1 - -), by (25i>)
Cancelling cos"' and solving the equation for x, we have
M"Vrs)- . ^ V
a cos 6 a — sin ^
3. Prove that tan-' ;; :— --tan"' — =f ;,
1 - a sm <p ,, cos <p
a cos 6 1 . * - ^^ ^
Let tan-' -—-=x • and tan-' =y, "
l-asin0 cos^
sin X a cos ^ , sin y a- sin ^
then =; :— 7 and = — - ,
cos as l-asm^ cosy cos^
whence sin x=a cos (^ - x) and a cos y=sin (^+i/),
i-
1 ,.■«,-
228 PLANE TRIGONOMETRY.
twice the product of which is
2 sin X cos y =2 sin (<»+i/) *50« (^ - *)
or .sin ix+y) +sin (x - i/)=8in (2<i - a;4-i/)+sin (aJ+y),
whence sin (x - t/)=8in (2^ - a+y)
or ac - y =2«4 - x+y,
that is :b - y =<>
a cos a ~ siu c/>
or tan-i : tan~> =^.
1 - a sm <> cos ^
.>: a,fr -y , a--o,
4. Prove that tan"' -=tan-' — — -■+tan-' -— ~
4-tan-' ^-+ . ... tan-' ^-t-tan"' -.
a,aj+l ani^-i + l On
where a, , a,, . . . . <(» are any quantities whatever.
By (253) wo have
X 1 , <*i«-y
tail"' tan~' — r=tan"' ,
y «! «i y ■*- »
1 1 a» - a,
tan-' tan-' — =tan-' — ,
1 1 a, - a,
tan-' tan-' — =tan-' ,
1 , 1 , ttn - <'f'n— 1
tan—' tan-' — =tan-'
a
n— 1
On a„a„_i+l
by addition we have •
tan-' tan-' - =tjvn-' - — ^+tan-' — — ^^+... H-tan-' -5~ ^-1-,
y a« ftiy+a ajai+l a„o»_i4.1
therefore by transposition
X a,x-v a^-a, . an-On-i . 1
tan-' -=tan-' - — ^+tan-' — — \+... +tan-' ^ ' -f tan-'- ,
y ftiy+x a,ai+l a,ja»_i+l a,i
6. Prove that cot (fl+tan-i tan^ e)=2 cot 26. .• , , "
6. Prove that cos sin"^ cos sin"^ fi=±d.
\l — X |1+X ;,: V '
7. Provethatco8-ix=2Bin-^ l-^'=2co8-i 1-^,
EXAMPLES. — 229
8. Find the general values of cos"^ sin 6 and tan"^ cot 0.
Am. 2w7r±(— -0), nir+{—-d).
3 1
9. Prove that tan"^ — =2 tan-^ — .
10. Prove that
(1) sm-i-+sin-i-=-.
4 ,-.
(2) sec-i 3+tan-i 2V 2 =tan-i ( - - v" 2 ).
, 2 1 , , 12
S) tan-^ -=- tan-i -.
4) 2tan-i-+tan-i-=j.
(5) sin-i (_v/2+s/"2) + sin-i (- v^2^^V?)=|-
1 1 , 1 TT
(6) 2tan-i— +2tan-i— +tan-iy=:— .
11. If sin (tt cos 0)=co8 (tt sin 0), shew that 0= - — sin -.
6 ' I ^
12. If 2 sin — =co8 d, shew that 0=2 cos-^ . Icos — .
2 >| " f^,.
13. Find the values of
tan (tan-^ e+oot-^ 6) and sin (sin-^ — +cos-^ — ).
Ans. «, and 1, or
14 If tan {n cot a5)=cot (» tan «), shew that
mn ^ ^. 1 . . 4w
m and r being any integers
■ 15. If 2 tan-i a;=8in-i ^y, shew that ]/=j--j-^.
1
2
2'^^ _ PLANE TEIGONOMETRY.
16. Find the value of x in the following equations :
(1) tan-J2a;+tan-i3ic=:y. Ans. cc=-lor--.
4 6
(3) sin-i 2a;-sin-i>/3' x=sin-i cc ^m. a;=0 or +—
2 '
(4) 8ec-i--sec-i-+sec-ia-sec-i6=0. ^m. x=±a6.
(5) sin2cos-icot2tan-ia;=0. Am. a;=±l,or±(l±v/¥).
(6) ver8in-i(l+a;)-ver8in-i(l-a;)=tan-i2\/rr^.
1
AtiS,. aj=— or— 1.
(7) sin-i a;+tan-i a;=^ -4rw. a:=:J^::l^^.
(8) Bin (tan-i a;)+tan (sin-'a;)=m!r.
Am, x=—Vm* - 2m'i T 2v/l + 2m2-2.
-'-•■. . , '.
17. If sec 6>-cosec 0=2\/2", shew that <?=— sin-* ~.
18. Shew that 4 tan-i tp.n-i — =^
5 239 4
19. Shew that tan-i -+tan-i -H-tan-^ i+tan-* ~~.
20. Shew ihat ts \ (2 tan"* a)=2 tan (tan-* aH-tan-* a3).
21. Show that tan-i ( - ^^=sin- ' ( -^)^,
V a / \a+x/
22. Shew that tan-* '^ + '^~^ + ^^^..x ^ ^?5
v/3 - ^"2 V2 4'
23i Shew that cot-> 3+cosec-* s/?—.
4
DIVISION OF ANGLES. 231
CHAPTER XIII.
DIVISION OF ANGLES — SOLUTION OP EQUATIONS — AUXILIARY
ANGLES — ELIMINATION OP TRIGONOMETRICAL FUNCTIONS.
Division of Angles.
167. We shall here determine, a priori, how many values any
assigned trigonometrical function can have when determined from
any other function of the angle or of a submultiple of the angle.
168. Cfwen sin A, to find how many values sin — cam, ham when
expressed in terms of it, »
Let a be the circular measure of the least angle whose sine is
equal to sin A; then all the angles whose sines are equal to sin A
axe included in the general expression ^
W7r+(-l)"o. f^rt. 42.) " " '
A
Hence all the values which sin — can have when expressed in terms
of sin A are included in
rn7r+(-l)'»a>
. / nn+{-ira \
Now n must be of one of the forms 22. or 2X+1, since every
number is either divisible by 2 or divisible by 2 with a remainder 1.
Let n=2A, then
^^-^j=8in (A7r+--)=±Sin -,
according as A is even or odd.
232 PLANE TRIGONOMETRY.
. Let n=2A+l, then
Sin
(—2 )=sm(;i-+^-)
(IT a\ a
---^=±C08-,
according as A is even or odd.
A ■■
Therefore sin —, when expressed in terms of sin -4, has four
different values, viz. , + sin — and + cos — .
A
169. Given cos A, to f/nd how marvy wdues cos — cam have when
o
expressed in terms of it.
Let a be the circular measure of the least angle whose cosine is '
equal to cos A; then all the angles whose cosines are equal to cos A
are included in the expression 2w7r-j-a, and therefore all the different
A
values which cos — can have when expressed in terms of cos A are
o
included in *
- ' • ■ • 2n7r4-a ' : ^ ,.
Now n must be of one of the forms 3p, 3p+l, Sp+2, since every
number is either exactly divisible by 3, or divisible by 3 with a
remainder 1 or 2.
Taking n=Sp, we have
2r(i,7r4-a ,^ a. , o. . a
cos — - — =cos (2i)7r±-)=cos (±-);=cos -.
Taking n=3p+l, we have
2n7r4-a / 2Tr4-a\
cos — - = cos i 2^n-+ — W~)'=^
Taking 71=3^+2, we have
2n7r -fa / 47r + a\
_ 47r-j-a
cos — :;:: — =cos yipir^ — - — |=co8 — - —
/„ 27rTa\ 27rTa
=C08 I 27r — j=cos — - — ,
DIVISION OF ANGLES. 233
A
Therefore cos —, when expressed in terms of cos A^ has three
o
diflferent values, viz. :
a 27r+a , Qtt — a
cos — , cos — - — and cos — - — .
170. Given sin A, to determine how many values sin f J. can have
when exp, essed in terms of it.
Let a be the circular measure of the least angle whose sine ig
equal to sin A ; then all the values which sin ^A can have are in-
cluded in
sin|(w7r+(-l)«a},
where n is of one of the forms ip, -ijj+l, 4p+2, 4p+3, since every
number must be exactly divisible by 4, or divisible by 4 with a
remainder 1, 2 or 3.
If n=4p, which is even, - .
sin I {w7r+(-])'»a}=sin (3p7r+|a)=±sin fa,
according as p is even or odd.
If w=4j9+l, which is odd,
sin |{n7r+(-l)"a}=sin {3p7t+|(7r-a)}
=-j-sin |(7r-a),
according as^ is even or odd; and so on.
Hence we find that sin |-4, when expressed in terms of sin .4,
has eight different values, viz. , '
± sin I a, -t-sin |(7r-a), ±cos |a, ±8in|(7r-3a),
2r7r+e
171 • To find the number of vahies which cos has when
n
suAxessive integral vcdv^ are assigned to n.
Here r, being an integer, must be of the form mn+p where m
is or any integer, and j:> is or any integer less than n; that is,
r must be exactly divisible by n, or divisible by n with a remainder
which is 1, 2, 3 ... or M— 1, #
234
PLANE TRIGONOMETRY.
Hence giving r, the values, 0, 1, 2, 3 . .. n- 1, n, n+1, . . . &c.,
in succession, we have
when
r=0,
2r7r+^ e
cos =cos — ,
n n
r=l,
2rK+e 2Tr + d
cos cos ,
n n '
r~2,
cos =cos
» n
r=3,
2rn+d 67r +
cos =cos ,
n n
&c.,
&c., &c.
r=»-3,
2rTv+d Qir-e
cos =cos
,« n
r=n-2,
2r7r+d 4:Tv-e
COS =cos
n n
r=n— 1,
2rTr+d 2ir-e
cos =cos
n n
r=n,
2r7r+d d
cos =cos — ,
n n
r=n+l,
2rTT+e 2ir + e
COS ^=cos
n n
&c.,
&c. , &c
Therefore there are n and ordy n (liferent values of cos -— —
corresponding to the values 0, 1, 2, ... n— 1, of r; for the same
values of the function recur in the same order when r is succes-
sively made equal to n, n+l, &c. ^
2r-K-\-d
In a similar manner we may shew that sin has also n dif-
ferent values.
n
The preceding examples are quite sufficient to shew the mode of
prooofding in any assigned case,
SOLUTION OF EQUATIONS. 235
Examples.
1. Shew that sin ^, when determined from tan A, has two values.
2. Prove, a priori, that sin mA, when expressed in terms of
sin A, will have one or two values, according as m is odd or even;
and that cos mA, in terms of cos A, will have only one value, m
being in each case a positive integer.
\^
3. Prove that tan — -, when expressed in terms of sin A^ will
4 ,■•..,■,.■..
have four different values.
IT
4. If tan ^=sin 2^, find A. Ans. nir or ri7r+(- 1)" — .
4
• 5. If cos 0+cos 20+cos 3^=0. then will, r being any integer,
e^{n+{-iY5\ {(7+(-ir'i)^±i}^^.
Solution of Equations.
172. An equation in which the unknown quantity is a trigono-
metrical function of an angle, is, in general, readily solved by the
aid of the ordinary trigonometrical transformations. We shall here
illustrate the mode of solving a few easy equations, such as are most
frequently met with in Spherical Astronomy.
173. Given dn 6=cos (3 sin {6+ a), to find 6.
Developing the second member by (45) we have
sin ff=cos a cos ^ sin ^+sin a cos /3 cos 8,
or tan 0=cos a cos p tan 0+sin a cos /?,
sin a cos /? :' i* . ■
whence tan 6=- -.
1 — cos a cos p ^
Now it is evident that if ^ is not limited to any particular quad-
rant by the nature of the problem under consideration, there wilJ
be an indefinite number of solutions; for all the angles ", ^+180",
^+360°, 0+540°, &c. , that is, all the angles included by the expression
6+nir, have the same tangent. (Art. 44.) In practice, however,
only the first two values of 0, viz., 8 and 0+180", or those less than
3G0^, are considered ; tind the conditions of the problem are generally
such as enable us to determiae which of these is to be takl^.
236 PLANE TRIGONOMETRY.
It is evident then, that when an angle is determined by a single
trigonometrical function, there will be two values less than 360°;
but if the values of two functions of the required angle, which have
not the same sign — such as the sine and tangent, or the cosine and
cotangent — can be found from the problem, the solution is deter-
minate under 360°.
Suppose, for example, that the required angle is found by its
sine and tangent ; if the sine is positive and the tangent negative,
the angle will evidently be in the second quadrant or between 90° and
180°; if both functions are negative, then the angle will lie between
270° and 360°, and so on.
The solution of the last equation cannot be eflFected by logarithms ;
. a formula adapted to logarithms is easily deduced as foflows :
Put the given equation in the form
sin 6
=cos 13,
then
sin {d+a)
Bin {d+a)+Bin 6 1+cos/?
— — • — ■
sin (d+a) -sin d 1 -cos fi
tan(e+|-)
taai-
a a ■ B
whence tan (0f— )=tan - cof^ — -.
a
This determines ^+— , and therefore Q becomes known by de-
2k
a a B a
ducting — , thus 0=tan~i (tan --- cot" — ) - — .
2 i2 2 ^ ,
174. Given tan {6+a)=sin ^ tan d, to find 6,
Putting the given equation in the form
tan(e + a) . ^
— -=sm B.
tane *
we have, by composition and division,
tan(6/+o)+tan0 1 +sin j3
tan(^+a)-tan0~'i-sin,'3' •
f'
SOLUTION OF EQUATIONS. 237
whence — : =tan2 (45°+-),
sin a 2
therefore sin (20+a)=tan2 (45°+ M sin a
and <9=4 sin-' j tan' (45°+ 1) sin a ( - -|.
175* Given tarn, {a+6) tan 6=ian fi, to find 6,
Here we have
1- tan (a+e) tan d l~tanj3
1+tan ^a+e) tan 6 ""l+tan )3 *
cos (a + 2d) ,,^„ ,
whence * -^-^^ ^=tan (45°-/3),
cos n
therefore cos {a + 26) =tan (45° - /?) coi a
and 6=^ cos-' { tan (45° - /3) cos a J - -.
176. Criven x sin d=a and x cos 0=b, to find 6 and x.
By division we have
a
tan 0=7" >
which gives two values of 6, one less and the other greater than .180°,
and also two values of x from the equation x=a cosec 6.
Limiting the values of 6 to those less than 360°, the solution is
determinate under the following restrictions :
1st. When X is positive.
The signs of sin 6 and cos will be the same as those of a and h
rsspectively, and therefore the quadrant in which 6 must be taken
is determined.
2nd. When x is negative.
The signs of sin 6 and 908 are the opposite of those of a and b,
and therefore must be taken out accordingly.
3rd. WJien 6 < 180° or > 180°.
Under either of these conditions the equation tan 0=t gives
only one value of 6 (a and b being unrestricted aus to sign). Under
238 PLANE TRIGONOMETRY.
the former condition, x has the sign of a; and under the latter, the
opposite sign to that of a.
4th. When 6 is limited to acute values, positive »/ negative.
, Under this condition x will always have the same sign as 6,
Auxiliary Angles.
•
177' Ij^ t^6 solution of equations by logarithms it is necessary
to express the sum or difference of two quantities by means of a pro-
duct. This can always be effected by introducing the sine, tangent
or some other function of an angle chosen for that purpose. An
angle which is thus introduced to assist in trigonometrical calculation,
is called an auxiliary atigle, and is of great utility and extensive
ajjplication, particularly in Spherical Trigonometry and Spherical
Astronomy
Auxiliary angles have already been employed in a few examples.
(See Arts. 120 and 144)
Ex. 1. — Adapt jc=v/(a«±62j to logarithms.
nI-:-.-
(1) x—v'(ci''^+h^)=a
Assume tan 6=—, an assumption always possible, since the tau-
a
gent may be of any magnitude whatever, then
a;=av/(l+tan2 6)=a sec 6;
hence x will be found by the two equations
Log tan 0=log 6-log a+10;
log x=log a+Log sec 6-10.
(2) x=^{a^ - ¥)=y/(a+h) (a - 6),
which is in a form adapted to logarithms; or we may proceed as
follows :
x=a. 1 •
'i
.6.6
Assume sin 0=—, since — must be less than 1,
a a
•then a=:av/(l - sin* ft)=a cos 6.
AUXILIARY ANGLES. 230
Ex. 2. — Adapt x=a±\/a^+¥, to logarithma,
Tho equation may be written thus,
Assume tan ^=--,
a'
then «=» (l±sec ^)
l+COS ^ 1-C0S6
and — ct ~
cos (ft COS
a sin 6 1+cos <b , a sin 1 -cos d
cos ^ sm cos 1^ sin
=a tan 6 cot — and - a tan tan —
2 2
==& cot — and - 6 tan ^,
2 2
which are both adapted to logarithms.
Ex. S. — Adapt x=a±Va^-h^j to logarithms.
Assume sin ^=— , then we shall easily find
a ''
B 6
sc=:2a cos2 — and 2a sin2 — .
2, 2,
Ex. J^ — Given cos cos 8 cos h+sin <j> sin d=sin a, to find ^ in a
form adapted to logarithms.
Assume x sin 0=cos 6 cos h )
and X cos ^=:sin 6, i ^ ^
that is, tan ^=cot d cos ^ j (2)
then the given equation becomes
X (cos sin &+sin <j> cos d)=am a,
or x sin (0+^)=sin a :;.
and £c=sin d sec ^;
., , . , «\ sin a ....
therefore sm (0+0)=
sm d sec ^
=sin a cosec S cos ff, (3)
wrhich gives two values of (0+0).
240 PLANE TRIGONOMETllV.
Tn this example, let (I=-30' 22' 47".5; o=20" 10'; h=-lu\
I'm:! (,..
13y (1) Log COB (T= 9.035857
Log cos /i= 9.984944
. Logwsinfl =9.920801
Log « cos ^ = Log Bin <J = 9.703919n
Logtfen/?=10.216882w
e = 121° 15' 13*.
iy (3) Log sin a = 9.687843
Log cosec (5=10.296081u
Logcosff= 9.7l5023n
Log sin {(}>+e) = 9.698947
( 29° 59' 53^5
^'*' ""lor 150" C 6". 5,
therefore <l> = ]
lor -
28° 44' 53". 5
9r 16' 19". 5.
If we take the acute value only of 6, we have
ff = - 58° 44' 47",
which gives ^+ft = -29°59'53".5 or -150° (T 6". 5
and 0= 28° 44' 53".5 or - 91^16' 19''.6
as before.
Ex. 5. — Given tan l=:{coit a cos ft- cos 6) cosh- si/n a cos p sin /».,
to find h in a form adapted to logarithms.
Writing the given equation in the form
cos a cos B - cos 6 , . ,
tan A cosec a sec 3= : cos a - sm h.
sm a cos p
and assuming cos 0=cos a cos /3, since cos a cos p is always less than
1, we have - ^
cos 6 - cos d , . ,
tan A cosec a sec p=—. — cos h - sin h
sin a cos p
2wnJ(d+e)_BinK^-W)
=: ; -r QOB h - SlU h,
sm o cos /? •
AUXILIARY ANGLES. 241
Again, assuming cot <l>= --,-'— --^— \ wo Imve
Bin a COB li *
tan A cosec « sec /?=cot cos /i - sin /j,
cos cos /i - sin </» sin /i
sin '
cos (0+/l)
sin *
whence cob ((p+h)=t&n X coscc a sec /? sin (/>,
therefore /i=co8-» (tan A cosec a sec /:J sin f/>) -0.
*
Examples.
6- If *=^"^ ) shew that x=cos 20, where tan 0= -.
7. If a=v/a+6 + v^a-6, shew that
a «
a;=2v/a cos (45°--), where cos <?= .
^ a
. I +
8. Given
sin 6 s/(l+tan2 a tan*^ /?)+cos ^ \/(l -tan'^ a tan'^ |9)=tan a+tan /?,
find ^ in a form adapted to logaritiims. ... " , -^
A • //. \ si^ («+/^)
^rt«. sm (#+0)=-— , where cos 20=tan'^ a tan' j9.
V 2 cos a cos iTi
9. n tan (^+45°)+tan (d-i5°)=2 tan 00% find d.
Am. 30" or (3w+l) ^.
10. Shew that
sin ^+co8 ^+sin 5+co8 B=2 v/2"cos {i6''-l(A+B)}coa ^(A -B).
» 11. If cos (a+0)=cos a sin 0+sin /?, shew that " "
cos (0+J)=sec o sin /? cos 6, where tan J=v/"2sin (45°+a) sec a.
12. Given sin a(;=tan J cot w,
sin (a3+2a)=tan (J'cot w, find 33 and w.
. ^ , . sin (fJ' + rf)
^ns. tan (x+a)=— -)-^ — 7, tan a,
'O'C) .'r.l . ^ ^^"('^ ~'^)
• '. and cot w=siii x cot d.
- 17
,1
242 PLANE TRIGONOMETRY.
13. Given cot x sin /j=tan d cos /?-co8 h s'ti /?, find « in a form
Adapted to logarithms.
Aris. cot a5=cot h coaec <l> cos (0+/^),
where tan 0=cos /i cot d.
14. Given cos h=cot ^ tan d,
-cos (af/i)=tan tan 6, to find J^ and <t,
Ans. tan (/i+— )=cot ■— sec 2<pj
tan d=co8 h tan 0.
Quadratic Equations.
178. To solve the equation
x«+ax+6=:0, (a)
tc^ere 6 w essentially positive amd a either positive or negative.
By the ordinary algebraic method we have
-li^^sj^-t)'
*
which may be easily adapted to logarithms in a manner similar to
that of Bxs. 2 and 3 of the last Article.
The following method, however, will generally be found more
convenient :
,0 A ■
2 sm — cos -^ 2 tan - ~
^ 6 6 2 2 2
(1) Bin 6=2 sin — cos ~= , = .
22 ^.0 0*
CO88 -^+8in2 1- l + tan'^
2 2 2
whence tan^ — -2 coseo tan -^^+1=0. (225^
2 2 •:,.'.,■•:■ '■) . '
2 sin ^ cos ^ 2 tan ~
Bin 2 2 2
tan 0= =: =s _iBa«M^^ :: . >\ • 5 .0. •
CO8 ^ . ^ - ' •
COS — -Bm« -^ 1 - tan" —
2 2 2
" '-. ■;''r,>' (• ".,.'; ' ..- . -
whence tan« ^+2 cot tan ~-l=0. (256)
2 2 . "
■ QUADRATIC EQUATIONS. 243
In (a) let x=y\/h, where the radical is taken with the positive
sign, X and y having the same sign. We thus reduce (a) to
V
which compared with (255) gives
a A ' -
- 2 cosec ^=— — , y=tan -jr ,
■•/■,
or ' sin 0=- , and a;=\/6 tan — ,
a 2
which gives two values of 0, and consequently two values of «, Let
B be the smaller of these two values of 0, then all the values of
which have the same sine are . — t ,,
0, TT-e, 27r+0, Stt-^, &0.,
,: : ,-f'
-* ■■. .
and all the values of tan — are , ,. *
a
tan — , tan ^(tt-^), tan ^(27r+5), tan K3^~^)> **'•» ' '
^ (? <^ .
OV tan — , cot — , tan — , cot — , &c.
a a a 2 ^
Hence the roots (wj , Xj) of (a) are found by the formulas \
sin 0= , a5,=v/y tan — and x»=v/r cot — ,
a 2 ' 2
where (9 is always to be taken less than 90°, with the sign of its
sine, v/ 6 being regarded as a po^tive, quantity, and a either positive
OT negative. When 2v/6 is greater than a, sin is impossible, and
both roots are imaginary. .,
179* To solve the equation
a«+ax-6=0, (c)
where -h is essentially i^ative, a h«mg either positive or negatim.
Let x=y\/ h , then (c) becomes
. V
244 PLANE TRIGONOMETRY.
which compared with (256) gives
a 6
2 cot i>=-7^> y=tan -^ ,
or tan 0= , and x=\/b tan -^, i
which gives two values of ^ and also two of x. If ^ be the smaller of
the two values of 0, the values of which have the same tangent are
e, TT+e, 27r+0, 3n-+0, &c.,
and all the values of tan — are
2
Q d e e
tan—, -cot—, tan—, -cot—, &c.
2' 2' 2' 2' '
Therefore the roots of (c) are found by the formula)
tan 0= , «,=v/6 tan— and a5„=-\/6 cot—, '
a ' ^ 2 ' 2'
where 6 is to be taken less than 90°, with the sign of its tangent, the
radical with the positive sign, and a either positive or negative.
tSij Here tan (^ is always possible, therefore both roots are real.
' Ex. —Given x^ - 1. 7246x+. 72681=0, find x.
Here we have a=- 1.7246 and 6=. 72681. ' ■
log ( - 2)=0. 3010300/1 log v^= X 9307104
log v/ 6 =1.9307104 ^ e
Loff tan o q^aqi *7a
ar. CO. loga=9.7633116n n ** 2
Log sin (9=9. 9950520
0=81° 22' 3^
■^=40° 41' r.5
2
■
log
Xj=
= 1.8650282
-..■ ■.
Xl=
: .732872
Log
cot
"2"
=10.0656822
•
log
Xj=
= 1.9963926
x,=
= .991728
Cubic Equations.
180. Let the equation be transformed, if necessary, to another
which wants the second term, so that it may be of the form
x3-qfx-r=0; (a)
CUBIC EQUATIONS. 245
y
if x=— , this becomes
From (104) we have, by writing for JL, ,5
^1
3 cos 3^ ^
C083 - —cos <b 5 =0, ■> . ;
4 * . ,
, ; i ';,■■ v> ;..■, -.W' '.-''' *
which compared with (6) gives
cos 6
cos ^=1/ and as= ,
,3 lis"
7i«flf=_ or n=— ^ I—,
^ 4 2SJq
^-21^^n^ or cos 30=4n3r=^^ ^ ,
where the radicals I— and ^ I— are to be considered positive.
Sjq Sjq^
If 6 be the circular measure of the least angle whose cosine is
equal to cos 3(f>, then by Art. 171 the three values of cos ^ are
e 2ir + e ^ 2n-d
cos — , cos — r — and cos — - — ,
3 00
and therefore the three values of X are ' .
2 ll.cosi, 2 \l.coB?l±l and 2 JI . cos ?I^.
\J3 3 \/3 3 Nj3 3
r I27 T* cfl
Since cos 3^ < 1, — ^ — < 1, or — < -i .
2 \ 33 4 27
8
Ex. Given aj^- 4a-— =0, find x.
o
*^^' .444
.. V ' ^^ -4m. — :=. cos 10% - — =r cos 50°, -—-r: cos 70°.
-■ v/3 n/3 v/3
The other ^rms of cubic equations can be solved by Trigo-
nometry, but the solution is a matter more of curiosity than of
utility. We shall therefore pursue the subject no further, but refer
the student to the standard treatises on Algebra for a fuller eluci-
dation of this subject.
246 PLANE TRIGONOMETRY.
Elimination of Trigonometrical Functions.
l8l. The elimination of the trigonometrical functions from a
given number of equations, is frequently requii-ed in some of the
higher branches of mathematics, and is generally effected by the aid
of the various trigonometrical transformations given in the preceding
chapters. The following examples illustrate the mode of proceeding
in most cases: -
Ex. 1. — Eliminate d between the equations ■>
a sin d 6 cos d v. k
« . y
I'
=0. (2)
a sin2 6 b cos'^
Clearing of fractions gives us
,;,!,. \ bj' cos 6+ ay sin d=ab sia 6 cob 6 f . . f' (3)
bx oos2 d+ay sin^ 6=0. (4)
Adding bx sin^ 6 to both members of (4) we have
bx (cos« 0+sin2 6)={bx - ay) sin^
Ol , jt \/bx=i\^bx- ay sin 6 f
Vbx
whence sin 0=
s/{bx~ayy
n
'<•••>•
and :. ' ,,,.vi ■: cosg= ^/^\ ^ ,
y/{ay-bx)
wliich substituted in (3) give after reduction
s/bx y/(bx - ay) + s/^y s/{ay - bx)=ah.
Ex. 2, — Eliminate between the equations ...v .
y cos ^ - X sin ^=a cos 2^ (1)
- X cos (ft+y sin 0=2a sin 2^. » ' ) (2)
From (1) and (2) we have by division - . ^ i
«cos0+yBi n^ • : .,
" : .- — =2 tan 2a
y cos - X sin ^
'i' 1 • » : . <I '•'•11 . ■ ■
. ,-. . .. •T^.M. r '. V aj+y tan 4 tan ^
y - « tan ^~"r- tan» •
feLIMlNATION OF TRIGONOMETRICAL FUNCTIONS. 247
M^hence
but
3cc X
tah3 fj,A— tan* ^+3 tan ^ =0,
If V
(4)
tan3 ^ T 3 tan2 ^+3 tan <(> =F l=(tan <t>^l)\
Subtracting (3) from (4), and using first the upper and then the
lower sign, we get
whence
3(^-1) tan* <t>+-^ l=(tan <p - 1)',
{x-yf (tan 0-1)'
/ (3tan2 + l)^
(cc+y) (tan0+l)g
; / (3tan2 + J)^
Dividing (6) by (5) we have
and likewise
/3e + i/\ 'i /t an ^ + 1 \ ^
Vac - 1// Vtan ^ - 1/
whence
'.therefore
tan 01
, (x + y)* + (a;-t/)*
(x + i/)*-(x-i/)*
sin 0i-
(x+y) +(x-y)
i
cos 0!
n/2{(x+i/)^+(x-i/)*}*'
(x+y) -(x-y)
v^2.|(x+y)^+(x-y)^}*
s(n&
(x + y) -(x-y)
"Bin 20=2 sin cos 0— « |»
(x+y)'+(x-yr
which substituted in (2) give after reduction
(x+y)V(x-y)*=2al
(5)
(6)
, 1
248 PLANE TRIGONOMETEY.
•^ Examples. •
3. Eliminate 6 and ^ between the equations
sin ^=»/i cos 0+n sin ^
cos d=^n sin ^ - w cos tp.
> .: [.
Ans. m'+n'=l.
4. Eliminate 6 between the equations , ; .
cosec2 6=m tan d
sec* 6=n cot 6.
, . , Ans. (mny=(s/m+y/n)\
5. Eliminate B and (7 between the equations
a-h cos C-c cos jB=0
6 - c cos A -a cos 0=0
c - a cos B - 6 cos A=0.
Ans. a2=62+c«-26cco«X
6. Eliminate 6 and ^ between the equations
tan 0+tan ^=a ''
tan 6 tan ^ (cosec 25+cosec 2^)=6
cos (^+0)=c cos (6 - (j)).
Ans. o,=h +hc.
7. Eliminate 6 between the equations
sc=a(cos 6+coB 26) ■ ..
'«/=6(sin e+sin 2<^).
8. Eliminate a and ^ between the equations
a B
<f tan a=m, tan ^=ji, tan — tan — =c.
2 2 " '-i
^m. 4c(i+-)('i+')=(l-c*)».
9. Eliminate 6 between the equations
sec 6 - cos Q=m
coseo d - sin 0=n.
ELIMINATION OF TRIGONOMETRICAL FUNCTIONS. 249
10. Eliminate 6 and ^ between the equations
a sin^ e+h cos'' 6=c
h sin^ 6+ a cos^ 0=6^
a tan d-b tan ^=4).
1111
r ■.,- -4»W. - + - = - + —.
• a o c a
11. Elimiftate 6 between the equations ^ ....
, , , , y. aaind+h coBd=m ,
;, -. ■„', ;, a co&6-b aind=n.
Ans. a^+6^=m^+n^.
12. Eliminate 6 and between the equations • • ^
., - * , £c=a cos"* ^ cos"* ^ -. ' '. • V .'1
!/=& cos"* ^ sin*" ' ,• .„
z=c sin"* 0.
, , , " ' . ;• .' ^ JL JL J.
13. Eliminate 6 between the equations i-
.;.i : yf?: i:M. ; (a+6) tan (0-^)=(a-6) tan (0+^)
a cos 20+6 cos 2<^=c. ,...,*.,,.„ ;iii:j
I ^ns. a^ - 6^ +c'=2 oc cos 2(/>.
14. Eliminate ^ between the equations ' ' " '^^^
■ • '' ^' ••'"'' ' M sin (? -mcos0 =2m sin ^
. n sin 2^ - m cos 20=n,
Ans. (n sin d+m cos 0)'=2m(m+»j)-
15. Eliminate 6 and r between the equations
<
r=2acos20, a;=rcos0, i/=rsinfl.
16. Eliminate 6 and r between the equations
. r <wtti.»^ ^.--li .J io *'^7^' x=rcose, 2/=r sin ^.
-4m. (x2+|/2) tan-> — =:o-.
as
r ""'^ ^ ' " ■ ■ v>^-v: .......
250 ' PLANE TRiaONOMETJiY.
■iU.X *■ 1
CHAPTER Xir.
ON THE COMPUTATION OP LOGAEITUMS.
We will here prove all the formulf* necessary for the computa-
tion of Napierian and Common Logarithms ; but before commencing
this chapter the student shpuld read carefully Articles 88-97, of
Chapter VII. .> .--
The Exponential Theorem,
182. To expand a" in a aeries of aacending powers ofx,
<i« = {l+(a-l)}*
Lu{(a^l)-i{a-iy+l(a-iy-i{a-l)*^&c.}x
+terms in a'^+terms in x^+ &c. '«»■—' - •
This shews that a" can be expanded in a series beginning -mth
unity and proceeding in ascending powers of x.
Let the coefficient of x, which is
(a-l)-Ka-ir+K«-l)'-K«-l)*+&c., ■ ■
be represented by J., and the coefficients of x'^, x^, &c., by J?, 0, &c.
respectively, than we have
a* — l+Ax+Bx^+Cx^+Dx*+&c.,
where A, B, G, &c., are independent of x, and therefore remain un-
changed however we may change x.
Assume also
ITow, since a'+v^a' a^ , and . ... ^^^ ,, ..s>,au*n!^i M •
a'-^=l+A{x+y)+B(x+yy+C(x+yy+kc.f
the last series is evidently equal to the product of the two former ;
therefore we have
l+A{x+y)+B{x+yy +C(x+yy + &c.
=(l+Ax+Bx'^+Gx^+&c.)(l+Ay+By^+Cy^+&c.},
THE EXPONENTIAL THEOREM. 251
Expanding both members of thig equation we have
1+Ax+ Bx^+ Cx^ + Dx* +&c.\
Ay+2Bxy+ZCx'^y+4.Dx^y + &c.
-Bl/2 +3(7x1/2 +Gi)a2y 2 +&C.
Cy^ +U)xy^ +&C,
1+Ax+Bx^ + a»' +• Dx* -f.Stc.\
Ay+A^xy+ABx-y+ACx^y +&c.
By"" +ABxy''+B^xhr-t&c. -
. i. \>l , » . S-' ,1, Oy3 +ACxy^ +&C.
JOy* -ir&C.
Cancelling the terms common to both series, we have
2Bxy+3Cx^y+iDx^y + &g.\ (A''xy+ABx^+ACxhj+ &c.
+3Cxy^+eDx^y'i+ &c.\ = \ H-ABxy^+B^xY + &c;
' ■i'AChyy^+ &<p.
+4Djct/3 + &c.
Equating the coeflScients of xy, x^y, x^y, &c. , W3 hav&
2B^A\ OP B=4~,
, "..) 1.2 ' .,{
ZC=:AB, or' (7 = ^' ,
1.2.3 ' r r Tt
4I>=^(7, OP J3 ^ ^* / ^'*' '
1.2.3.4*
&c. '^ &c.
Therefore a»=l+^at+ + + -h&o.^
L2 1.2.3 1.2.3.4 ^
where '' ' ^=(a-l)-i(a-l)!^+|(a-)3-&c. '':.« -J^v
Since this result is true for all values of x, take x such that
1
Ax=^, or «=7, then ^ ^
y 1.2 1.2.3 1.2.3.4
=2.718281828459.... =e,
henoe J a=e^ and J.=loge a.
252 PLANE TRIGONOMETliy.
Therefore we have finally
a« =Tl+(log, a) Y + Goge ay —^ + (log, a)' —^ + Ac. (257)
which ia called the Exponential Theorem.
IS asati, we have
e* =1+35+ + + + &o. (258)
1.2 1.2.3 1.2.3.4 ^ ^
The Logarithmic Series.
183* To express loge (1+ar) in a aeries of ascending powers ofx.
From the last Article we have
log, a= A . : - •
= (a-l)-K<»-l)'+K«-l)'-i(a-l)' + &c.,
in which write l-¥x for a and we get
log, (l+a5)=x-^+| -^ +1 - &a (259)
In the last seriefl write -x tor x, then we have
x* a.3 X*
log, (l-a!)=-»- 2 - 3 - 4 - *c- (260)
184. To prove the Logarithmic Series indepen-
dently of the Exponential Theorem.
A«iume Ax-¥Bx^+Cx^ +Dx* + — =loga (1+a)
and Ay+By^+Cy^+Dy*+.... =loga(l+y)
By (subtraction we have , . ,,
il(x-i/)+B(»a-ya)+C(x»-i/»)+....=loga(l+x)-loga(l+y)
-~"'V
c,5 =loga(l + ^
We may consider -~ as a simple quantity, and therefore
/ X ""1/ V
loga ( 1+^j — ) may be developed in the same manner as log„(l+x).
THE LOGARITHMIC SERIES. 253
Thus,
Therefore
A{x- y)+B{x^ - 1/») +C{x^ - 1/3) + . . . .
Dividing hy x-y we have
A+B(x+y)+Cix''+xy+y^)+. ...=a{ ±-)+B^^,+G^~^,+ ....
Since this equation is true for all values of x and y, it must be
true for x=y ; hence, writing x for y, it becomes
A
A+2Bx+ZCx^+....
1+x
=ui(l-x+a2-£c='+.... ).
Equating the coefficients of like powers of x, we have
A=::A, 2B=-A or £=--,
V. ■, I.
A '■ ■ '
-O -
SC=A, or C= -, ,
4D=-^, or I>==-T»
4
AQ^ • &C. ;
Therefore log„ (1 +x)=:A {x- -+—-—+ ....).
fv.
'ijv
2 3 4
Dividing by x we have
1 CT 35 CT "'^
- l0ga(H-x)=^(l--+- --+.... ),
X J o 4
but ' ■ - log« (l+x)=loga(l+x)« , (Art. 93.)
,, , 1-x l-3x + 2x«
=log„(l+l+ — +-^— +
by the Binomial Theorem.
^54 PLANE TRIGONOMETRY.
Therefore
Since this equation is true for all values of x, it must be true
when «=0, hence
^=Iog,.(l+l+-^ + j|^ + -^ + ....)
=loga e=- , (Art. 97)
log, a
=the modulus of th i system whose base is a. (149)
Therefore we have generally
If axsef we have
«' X^ 05*
loge (l+a;)=«=x -■^ + ^-~r"^"** *" t)eforo.
^ O 4:
185. To deduce the Exponential Theorem from
the Logarithmic Series.
From (261) we have, by writing A for
loge a*
a 3 3 ^=l+x, and raising both members of this equation
to the power -— — , we have ■ *
Ax
. y y' «n/ , y* a ;y'+2xy'
^ 1.2.^» 1.2.ui» L2.3.i» 1.2.3^2 + •»
/; ^* '■'
md when x=^, this becomes
THE NAPIERIAN BASE. 25.^
Restoring the value of A, we have
ov = 1 +(log, a) I + (log, ay^ + (log, a)' ^^ + &a'^
i86. The Napierian Base.
The sum of the series
which we have denoted by e, is the base of the Napierian system of
logarithms. This base renders the logarithmic series simpler than
any other base would, as is evident from a comparison of (259) and
(261). Napierian logarithms are sometimes called natural loga-
rithms, because they occur first in the investigation of formulee for
their calculation. The Napierian system is used for the most part in
the higher Analysis, and, with the exception of the common system
which is universally employed in arithmetical and trigonometrical
calculations, it is the only one which we shall ever have to use.
137. The Napierian Base is incommensccable*
For, if possible, let e=— , where m and n are integers, then"
m _L._?_ 1 '1
n" "*" '''1.2 1.2.3'*' '"1.2.3. '..». 1.2.3... (n+1) "^•"
Multiply both members by 1.2.3... n, and we have
1.2.3...(»-l)m==2.2.3.4...n+3.4.5...n+... "J
(1_ 1_ )
' ' . ^ ^Wl"*'(n + l)(n + 2)"*'-7'
but the former member being integral, the latter must also be so,
1 1
•which is impossible since — - + ; -7- — + ... is. greater than
n+1 {n + l){n + 2)
t ..11
and less than the sum of the geometrical series — - + --
fi+l n+1 (n+iy^
+ — ——+... , that is, less than — : therefore e is incommensurable.
256 PLANE TRIGONOMETRY.
i88. Converging Series for the immediate calcu-
lation of Napierian Logarithms.
From (259) and (260) we have by subtraction
jgS jjgS /)p7
log, (1 +X)-I0ge (1 - X)=:2(X+ -^ + -g- + y + •••)
/l+x\ „. x'^ x^ x""
'°8'(tIv„-)=^(*+3+-6 + 7 + --''
. . . . 1+x . , , m— 1
in which let =wi, and therefore x= , thu8 we have
1-x m+1
which, however, converges too slowly to be of much utility in the
calculation of the logarithms of integral numbers.
In (202) let m = , and therefore
X m + l~2x + l' •
thus we obtain
1+x f 1^ 1 1 I
^^^* X -^|2x + l'^3(2^Tl)^+-[
or log,(l+x)==logeX+2{^ + |-„^+...}, (263)
which converges very rapidly, especially when x is large.
In the last equation, lot l+x=y'^, and therefore x=i/ — 1 and
2x+l=2i/2 — 1; then we have
log. ,/=log. (v.-l)+2(2^^ + 1^-^- + ...}
or logc (i/ + 1) =2 loge y - log, (y-1)
which also converges very rapidly when x is large.
By judicious substitutions, many other series for the calculation
of Napierian logarithms, may be deduced, but practically considered
CALCULATION OF NAPIERIAN LOGARITHMS. 257
they are now unnecessary, as tables have been already computed to
the highest attainable degree of accuracy. We shall therefore pursue
this subject no farther, but refer the student to the third example
at the end of this chapter for two other converging series, which may
be advantageously used for the same purpose.
Calculation of Napierian Logarithms.
189. By the last three formulee we are able to compute the
Napierian logarithms of all numbers ; but the properties established
in Articles 91, 92 and 93, render it necessary to apply them to prime
numbers only.
Thus, in (262) let m=2, then
• l°S'2=2(- + -(-) +-(-)+-(-) +....}
=.6931471....
In (268) let tK=2, then
f 1 1 /lv» 1 /lx» )
log. 3=log, 2+2 {-+-(-) +-(-) +.... I
=.6931471 +.4054650=1.098612.
log, 4=log, (2 X 2)=2 loge 2=1.386294.
In (264) let t/as4, then
log, 5=2 log 4-log.3-2{l+i(^^)V|(|^)V.... }
=2.772588 - 1.098612 - .064538
=1.609438. • '*•
loge 6=loge 3+log« 2=1.791759.
In (264) let i/=6, then
log. 7=2 log. 6-loge 5-2|- + -(-) + .... J
=1.945910.
Iog«8=log«2a=3 log, 2=2.079442. . 'i
loge 9=log« 32=2 log* 3=2. 197225.
log. 10=log, (5 X 2)=log, 5+log« 2=2. 302585.
&c. &c.
258 PLANE TRIGONOMETRY.
Hence the modulus of the common system is
1 1
logelO 2,302586
as was shewn in Art. 95.
=.4342944819..,,
Calculation of Common Logarithms.
190. Having computed the Napierian logarithms by the method
of the last Article, we may convert them into common logarithms
by (151), thus,
Log 2=. 43429448 log, 2=. 3010300
log 3=. 43429448 log, 3=. 4771213
&c. &c.
By means of M, the modulus, we may adapt the series of Art.
188 to the calculation of common logarithms ; thus, by (151)
log (l+a,)=Iog .+2M{^-i^ + |-jljj-.+ ....}. (265)
log (i/+l)=2 log y-\og iy-l)-2M
11 1
(266)
[2y^-l 3(2i/2-l)^
Having found the logarithms of prime numbers by the preceding
aeries, the logarithms of composite numbers are easily found by the
principle of Art. 91.
Thus, log 3360=log (2^ x 3 x 5 x 7)
=5 log 2+log 3+log 5+log 7
=3.5263393, &c.
Theory of Proportional Parts.
191. We shall now investigate how far the principle of propor-
tional parts can be depended on in finding the logarithm of a number
which is not found exactly in the tables. In the following investi-
gation we will assume that the logarithms are calculated to seven
decimal places, and that the tabie contains the logarithms of all
whole numbers from 1 to 100000.
THEORY OF PROPORTIONAL PARTS. 259
To shew that in general the Increment of the Loga-
rithm is approximately proportional to the In-
crement of the Number.
Let N and N+h be two numbers, the former containing five
digits and the latter six, the last {h) being after the decimal point.
Then we have
log {N+h) -log JV=log — ^ =log (l +-)
where M is the modulus .43429448 ....
Now, if N is not less than 10000, the second term of this series is
less than .000,000,002,2, which docs not affect the seventh decimal
place, and may therefore be neglected.
- Hence, as far as seven places of decimals at least, we have
log (N+h) -log N-^h,
which shews that the change of the logarithm is approximately pro-
portional to the change of the number.
We will now proceed to ascertain to what extent the principle of
proportional parts can be applied in the case of the logarithmic
trigonometrical functions.
ig2. To shew that in general the change of the
Tabular Logarithmic Function of an Angle is approx-
imately proportional to the change of the Angle.
Let 6 denote any angle and h any small increment such as 1' or
10", then we have
sin (B + li) sin ft cos /i + sin h cos ft
sin 6 sin '
=l+h cot ft, approximately J
since COS h=l and yin h=^h very nearly.
260 PLANE TRIGONOMETRY.
Therefore
Log sin (6+h) — Log sin 0=log (1+^ cot (?)
h^ cot* . ,
=M(h cot e + &c.) •
2
=Mh cot 6, approximately.
But when 6 is very small cot 6 is very large, and therefore the
h^ cot2 e
second term may be too large to be neglected.
Thus, if h=l' and 0=2', the value of the second term is .0000151,
which is far too large to be disregarded. If, however, h=10" and
0=2", the value of the second term is .0000004, which will affect the
seventh figure but not generally the sixth; therefore we conclude
that when is not very small,
Log sin (e+h) - Log sin d=:Mh cot d. (267)
that is, with the exception just stated, the change of the logarithmic
sine is approximately proportional to the change of the angle.
193. In a similar manner we find
Log cos (d+h) -Log cos e=-Mh tan 6 (268)
approximately.
When 6 is near 90° the second term, omitted in (268), is too large
to be neglected. Therefore, with this exception, the change of the
logarithmic cosine is approximately proportional to the change of
the angle.
194* In vhe case of the tangent we have " '' '
tan 04-tan h
tan {C+h):
i. - tan h taii
tan d+h
—tan d+hsec^ 6
1 — h tan 6
approximately,
., ^ tan (e+h) ^ „,
therefore =1+2^ cosec 26.
tan
and Log tan [O+h) ~ Log tan e=2Mh cosec 2fl (269)
approximately.
THEORY OF PROPORTIONAL PARTS. 261
When 6 is very small cosec 2d is very large, and the second term
is too large to be neglected; therefore, with this exception, the
change of the logarithmic tangent is approximately proportional to
the change in the angle.
195. A similar proof may be employed for each of the other
Logarithmic functions. The following are the results which may
be verified by the student :
Log cot {e+h) - Log cot e= - 2Mh cosec 2d. (270)
L 'y sec (d+h) - Log sec d=Mh tan 6. (271)
Log cosec {d-h)- Log cosec 6= - Mh cot 6. (272)
196. From the preceding Articles it is seen that the change of
the Logarithmic sine is equal to that of the Logarithmic cosecant,
but with the opposite sign. Hence a column of " Differences for 1' "
is printed in some tables between the columns of Logarithmic sines
and cosecants, serving to the former as a column of increments for 1',
and to the latter as a column of decrements for 1'.
In like manner the columns of cosines and secants have the same
differences for 1', and so also have the tangents and cotangents;
these columns serving as increments to the secants and tangents,
and decrements to the cosines and cotangents.
197. From the preceding investigations, and from an inspection
of the tables themselves, the student will see that the principle of
proportional parts is not applicable to angles which are very small
or nearly equal to a right angle. In the case of the Log sin and
Log cosec, the differences are irregular for small angles and insen-
sible for angles near 90° ; for the Log cos and Log sec the differences
are insensible for small angles and irregular for angles near 90°; for
the Iiog tan and Log cot the differences are irregular both for small
angles and for angles near 90°. For the methods of computing the
Logarithmic functions of angles near the limits of the quadrant, i he
student is referred to Art. 112.
.1.1 .
262 PLANE TRIGONOMETKY.
Examples.
U Prove that
1 1 I
log,a=7i{(l-a«)-fi(l-a«)2+i(l-a")3 + ....}
2. Prove that
^^"^ iif ■^1.21 M J ■^i.2;3l':atf~j ■*■••••
irhere N is any number and M the modulus.
3. J£ a, h, c be three consecutive numbers, prove that
log 6=i(log a+log cHm{^^H^^^ + ....}
and log c=2 log 6-log a-23f {-^^y-j+J^^^^-^ + ....|.
,11 1 1
4. Prove that — =— —4- ,^„^ -f , ,, „,^>. + . • ..
e 1.3 1.2.3.5 1.2 3.4 5.7
, „ ,^ e 1 1+2 1+2+3 1+2+3+4
6 Provethat-=— + ^^^ + -^^^+--^- + ....
6. Provethat2e=l+| + f^ + ^ + --A-- + ....
7. Find the modulus of the system whose base is -- .
Ans. -.91024.
8. Prove that log«(v/3i/^~'=l-i + i-l+A-A+....
9. Prove that loge 101 - loge 99=— very nearly.
50
1
10. To what base is - 5 the log of 32768 ? Ans. — .
o
11. The log of a number to one base is the same as that of its
reciprocal to another base; find the relation of the bases to each
other.
DE moivre's theorem, 263
CHAPTER XV.
DB moivre's theorem — EXPANSIONS OF CERTAIW TRIGONO-
METRICAL FUNCTIONS.
De Moivre's Theorem.
198. ' m he awy ratumal quantity j either integral or fractionclf
positive or negative, then
(cos B±\/ -1 sin 0)"*=cos md±\/ — Isin. md,
(1) Let the index be a positive whole number.
(cob d±\/-l sin 6/)2=(co8 e±V - 1 sin d) (cos B±V^-lBm 6)
=co82 e - sin2 6±V^ 2 sin 6 cos 6
=cos26±^-l sin2«,
(cos 0±n/ - 1 sin 0)='=(cos 2d±V - 1 sin 20) (cos e±\/-l rniO)
=co8 20 cos 6 - sin 2# sin 6
± y - 1 (sin 2ff 008 + cos 2fl sin 6)
s=co8 S^iv/- 1 sin 3^, and so on.
Suppose this law to hold for m factors, so that 3,.
(cos e±V^ sin 6)"'=coa md±>/'- 1 sin w^, -
then
(cos ^±v/-l sin ^)*"+^=(coa m8±\/ -1 sin wifl) (cos 6±\/~^lBm 6)
=:co8 m^ cos 6 - sin m^ sin ti
i v/ - 1 (sin m^ cos #+cos m^ sin ^) . ,
. srscos (mH- 1)W±n/-1 sin (m+1) 0.
If, then, the law holds for m factors, it also holds for m+1
factors ; but we have just shewn that it holds when m=:=3, therefore
264 PLANE TRIGONOMLTRY.
it holds when m=4, and by sucoeasive inductions w© conclude that
the formula is true for any positive integer.
Therefore we have
(cos d±\^-l sin (?)"'=cos md±\/ -iBinmB, (273)
when m is any positive integer.
(2) Let the index be a negative whole number.
(cos 6/ ±>/ - 1 sin 0)-'»= --
(co8 0±v'-i sin ey*
(cos« g + B in«6>) "* _
(cos ±v/^nr sill ^)"»
=(cos eT v/TT sin &)•,
by actual division
s=cos mdT\/ -1 sin m^, by (273)
=co8 ( - m#) ± v/ - 1 sin ( - m6>),
which proves the theorem when m is a negative integer.
m
(3) Let the index be a fraction — , either positive or negative.
n
(co8 d±>/-l sin 0)"*=co8 imd± v/-l sin md
'-e)±N/^sinrv(^
=008 » (— ©j ± V - 1 sm rv ( — 61
=(cos - 0± v^-l Bin - e)» , by (273)
therefor©
.wi ,. I / — T . w
(cos e±y/ -\ sin ^)'^ =corf- (^±s/ - 1 sin - 6, (274)
which proves the theorem for fractional indices.
199- A^9 long as m is an integer, both members of (273) can
have only one value ; but in the case of (274), in which the index is
a fraction, the first member has n difierent values in consequence of
the vH'^ root, while the second member has only one of these. Th6
second member, however, may be transformed so as to exhibit the
same number of values as the first.
DE moivre's theorem. 265
Thus, since cos 6=cob (2rn+ff)
and sin (^=sin (2r7r+^), we have
(cos e±>/-Tsin ey*=:{coa (2nr + d)±^/^Bm {2rn-\-e)\»
=cos - (2r7r+0)±vAri sin — (2r7r+&), (275)
n n
in which the second member has n different values corresponding to
the values 0, 1, 2, — n-1, of r, according to Art. 171. There-
fore the theorem is entirely general under the form (275).
200. To express an imaginary quantity of the
form (a-^hj^if by means of Trigonometrical
Functions.
Assume h cos d=a and k sin d=6,
whence tan 0=— and h^s/a'^ + h',
a
Then, by De Moivre's Theorem,
{a+h\^~^y=1^ (cos 0+\/~l sin e)"»
=(a2 +6^)2 (cos w^+v^^ sin mSy.
In a similar manner we find
(a + 6v/^)-=(a^+6=^)2-(co8 ^J^^W^ sin Hl!^^) •
J^sc.— Find the three values of
(y5" + 2v/"^)i.
Hero, tan^=-y5 or •0=41" 48' 37*
5
and A;=3 ; then we have, by giving r the valur'
0, 1, 2 in succession,
riK3(co8| + v/Tr8in |),
^3 (cos -j-+y^sin -J-),
.^ / 47r + .—- . 47r + ^.
f 3 (cos — — — + s/ - 1 sin — ^— )
266 PLANE TRiaONOMETRY.
201. To express the sine and cosine of an Angle
and of its multiple as Algebraic Binomials.
Since (cos 0+\/-l ain 0) (cosy-v/-l sin y)=coa» O+sin* 6>=1,
if we assume cos fl+v/ - 1 sin B=Xf
then COB 6 - V -1 sin 6= — ,
»
and by addition and subtraction we obtain
1 , I
2co8fl=x+— , and 2v- lain ^=05 ;
• '• X X '
also, cos m^+ \/ - 1 sin wi0=(co8 6+\/ -1 sin 0)"'=x**,
and cos md-V -1 sin m0=(cos 0- \/ - 1 sin 6)"*a= -- ,
»"*
by De Moivre's Theorem, whence as above
2 cos m^=x"*+ — , and 2\/'^sinm0=w'" . (27C)
ajm' y.m
If the index be fractional, we have by Art. 199,
ml ml
2 cos —(2r7r +<?)=»;" +— ,r , and 2\/ - 1 sin — {2rn + 0)=x" - -;r,
where r has the values 0, 1, 2, .... w - 1, so that oach member has
n different values.
This notation will be found very useful in subsequent investi-
gations.
202. To express the sine and cosine of the mul-
tiple of an Angle in terms of the sine and cosine
of the Simple Angle.
By De Moivre's Theorem we have
cos m#+\/ - 1 sin mB
=(co8 0+v^^-rsinO)"»
TRIGONOMETRICAL EXPANSIONS. 267
teoos^ 0+m co8"*-^ e y -T sin 6- ^^^ coa'^^ q ^{^-i q
X. z
rn(m-l)(m-2) .«_« ^ / — r • i ii ■
. —i — COB**"' d V ~1 sin' 0+ ....
1.2.3
by the Binomial Theorem.
Equating the real and imaginary parts of this equation, we have
cos w0=co8'» B - ^\" " cos"*-' B 8in« e
1. J
m(m-l)(m-2)(m-3) ^^^^^^,^^^^^^ (27V)
^ 1.2.3.4
to i(»»+l) *®"^8 if w is odd, and to iw+1 terniB i/ m is even.
m(m-l)(m-2) „^, ^ . ,.
Binm0=mco8«-i0Bin(9--^^-^— -^ cos'"-' sin'
.n(m-l)(m-2)(m-3)(m-4) ^^^^ e^m'd^..., (278)
^ 1.2.3.4.5
to i(m+l) terms if m is odd, and to — terms if w is even.
We may observe here that if m be even, the last term of the
expansion of (cos e+v/Tfain ef" is (-l)^" sin'« #, which is real;
Wl— 1
and the last term but one is m{-lY cos ^ si'^"*"^ ^' **'
m--2
v/^ m( - i)^ COS B sin"^i 0, which is imaginary. Thus, when m is
even, the last term of cos m6' is ( - l)^sin"' B, and the last term of
m— 2
sin «10 is w ( - 1) '' cos Bin"*-^ B.
Again, if w bo odd, the last term of the expansion of
(cos 0+v/^'l sin er is (-l)"^sin'" or v/-l (-i)^ sin"» 0, which
is imaginary ; and the last term but one is m( - 1)'' cos B sin'"-^ B,
which is real. Thus, when w is odd, the last term of cos imB is
m(— 1)' cos B sin"^! ^^ and the last term of sin m£> is ( - 1)^ sin"* B.
268 PLANE TRIGONOMETRY.
203. To ejfpress the tangent of the multiple of
an Angle in terms of the tangent of the Simple
Angle.
The quotient of (278) by (277) is
m(m - 1) (m - 2)
m cos"*-i sin - ^ ^^ cos"^ d sin" 6+..,,
, _ 1. J.o
tan md= ^ ^
m(m - 1) „ .
cos™ d cos'"-^ d 8in2 e+,,,,
m(m-l)(m-2)
m tang- tans 0+ . . . .
= r^T ■ m
m(m-l) .
1 tan3(9+....
1.2
by dividing numeratuv and denominator by cos"* 6.
204. To express the sine and cosine of an Angle
in terms of its Circular Measure.
a
In (278) and (277) assume m6=^ or m=— , thea we have
, sing a(a-e)(a-2d) ^^/sing^^
sin a=a 008*^1 g ^ ^ ^ cos»'^g ( — ) +....
6 1.2.3 ^ d
-6) „ / sin
1.2 '"'"-^"(t
a(a-6) „ /Singv'
cos 0=008"* - V - cos"»-2 g ( )+....
sm
JNow, when 4Ji=« , 6=0, cos g and all its powers = 1 and
6
and all its powers = 1. (Art. 74.)
Hence we have ultimately
(280)
.;,.r-. (281)
a3
1
1.2.3
' 1.2.3.4.5
a*
COS a=l - h
a<
1.2 1,2.3.4
By means of the last two series we may compute the sine and
cosine of any angle. Thus, suppose we require the sine of 12".
TRIGONOMETRICAL EXPANSIONS. 269
The circular measure of 12°=;^^=. 20943951, "
15
. .«o «««.o«.. (.20943951)' (.20943951)*
then sxn 12-20943951 - '-^^ ^^^^^^ -
=.20943951 - .00153117 +.00000335
=.20791169.... , .. : .
agreeing with the tables which give .2079117. ' '
205. To express the positive integral powers of
the cosine of an Angle in terms of the cosines of
its multiples.
Assume 2 cos ^=xH — ,
X
1 .=v"...
then 2 cos nd=x^ +— by (276). i
1 "
2''cos"^=(cc+— ) - . . .. /v, ,-• ;■, r.
, n(n-l) _ n(w-l) 1 1 1
1.2 1.2 X»^ «n-2 x" ,7
/ 1 \ / o 1 \ w(w - 1) / . 1 \
by placing together the first term and the last, the second and the
last but one, and so on ;
11 '
but x" +— =2 cos n0, x»*-2+ — -=2 cos (n - 2)6, &c. , ; .>
therefore •
2" cos" 0=2 cos n0+2n cos (n - 2)6+2. ^ cos (n - 4)6+
1 **
In the expansion of (x + — ) by the Binomial Theorem, there
are n+1 terms, therefore when n is even there will be a middle term,
/ n \th — 1
viz., the ( - + 1 j , which does not involve x, since x^ • =x*'=l.
This term is ;.■'• ' '-^^
n(n-l)(n-2). ...(hn+l)
1.2.3....^
270 PLANE TRIGONOMETRY.
When n is odd there will be two middle terms of the expanded
binomial, viz., the ^{n+lf^ and the ^(n+S)*^, whose coefficients,
however, will be equal, involving x and — respectively ; their sum is
X
n(n-l){n-2) ^(n + 3)
(-^)
1.2.3. ...^(n-1) ^ X
Therefore we have generally
n(n — V)
2^^ cos* 6— COB vJd+n cos {n - 2)g+ cos (w - 4)5+ . . . . ,
1. a
the last term being
i(n-2) (^ + 1) \
— — — :; ; n even.
(282)
^1. (^-l)(^- 2).. .-(i^ + l)
. n(n-l)(n-2 )....i (n + 3)
1.2.3 ^{n-1)
2o6. To express the positive integral powers of
the sine of an Angle in terms of the sines or
cosines of its multiples.
Assume 2>/ - 1 sin 5 ==x ,
X
then 2>/ - 1 sin iud=x^
x»
» 1 ••
and 2^(-l)2 sin" <^=(x )
- n(n-l)
=X»- «M5'^2+ - - x"-* - . . . .
1 . ^
^n(7i-l) 1^1 1
1.2 x""^ x«-2^x'*
the upper or lower sign being taken as n is even or odd.,
Arranging the terms as in the last Article we have
1, If n be even, it is of the form 4^ or 4w+2, since every even
TRIGONOMETRICAL EXPANSIONS. 271
number is either exactly divisible by 4, or divisible by 4 with a
remainder 2.
If 7i=4m, (V^l)<»* = 1=(-1)2'« =(-1)2.
n
If w=4m + 2, (^^)4«+2= - 1=( - iym+i—(^ _ 1 jT
— + 1) of the
preceding one, does not involve x, and is
n{n-l) (n-2) {^n+ 1)
1.2.3...T^ii '
n
according as — is even or odd; it has therefore the samo sign as
n
( - 1)2. Hence we have, n being even,
2n-i( _ i)Tsin» 0=co8 w9 - n cos (n - 2)0 + '^^^^^1 cos (rv - 4)5 - . . . .
1. a
,.(_1)t K^-1)(;-2).--.(^+1) . (283,
1 . J . o . . .. . ^
(2) If n be odd, it is of the form 4m+l or 4m 4- 3, since every
odd number is divisible by 4 with a remainder y^hich is either 1 or 3.
n—X
If»i=4m+1, (v/-l)4'»+i=(-l)a >/-!.
n— 1
Ifw=4m + 3, (\/^)*'"+3=( - 1)2 v/-l.
The sum of the two middle terms of the first of the above series is
n{n-l){n-2) ^(n4-3) . 1. "
1.2.3. ..4(n-l) ^*""x^
according as \{n-\-l) is odd or even; it has therefore the same sign^
n— I
as (-1)2. Hence we have, n being odd, , • .,
n— 1
(-1)2 2'»v/^sin''0=2v/^sinn(J-»i2\/"^sin(n-2)O r.f..
n{n-l) ^ ,
:. ^: >. ' + -j-^2v/-l sin(n-4)0.... n'.
n— 1
+' ^' i.2....ior-ir,.^^,""'*'
272 PLANE TRIGONOMETRY.
or dividing by 2\/ - 1,
( - iJ2-2«-i sin" e=sin nd - n sin {n - 2)0+ -.^?^^ sin (n - 4)5+ , . . .
n— 1
+ ( - 1) 2 'o ,, %\ sin 5. (284)
1.2.t}. . . .^(?i-l)
207. To express cos nB in a series of descend-
ing powers of cos 6, n being a positive integer.
Assume 2co8 0=a + — , then 2 cos n0=a** H — ,
a a"
Now, (l-a»)(l- — )=l-(a + — )a;+xa
=l-!B(6-a;),
\ih=a-\ — =2 cos 5.
a
Then log (l-ax) + log (l--)=log {l-a;(6-«)|.
Expanding both members of this equation by (260) we have
ax
I • • •
2 n I jp,
a 2a2 na"
x**~2 x"^^ x*
Equating the coeflQicients of «" in this identity, we find on th»
112
left-hand side the coefficient of x" to be — (a*» +— ) or — cos nd
n^ a*» ' n '
while the coefficient of x** on the right-hand side is found by taking
x"
out the coefficient of x" from the expansion of — (6 - x)" and of all
n
the terms that precede it. Thus, the coefficient of x*»
TRIGONOMETRICAL EXPANSIONS. 273
in — (6 - ae)" is -I- — ,
n n
a;**~^ 1
in — :(6-iK)"-i is -.(n-l)&«-',
w-1 n-1
w-2 n-2 1.2 *
n-r^ ' 1.2.3.... r
Therefore we have
2 6" 71-3
— cosn0= 6"-^+ -— r 6«^-.... r,,- r- T
n «i 1.2
(-ir(n~r- l )....(n-2r+l) _,, .
^ 1.2.3....r ^ '
or writing 2 cos B for 6 and multiplying by w, ,
2 cos »i^=(2 cos e)** - w(2 cos <9)*^H--7^ (2 cos e)"-* - . . . ,
1.2 •,•'',
(-l)'-n(n-r-l)....(w-2r+l) , ■" , _ '
+ 1.2.3.. ..r ^-(2cose)-'^ (285)
' r_.. • , •-• . - .,",...
Here we must observe that since none but positive integral powers
of (6 - x) appear, the index of b in the general term must also be a
positive integer, that is, r must not be greater than — ^
2o8. In (285) write — - ^ for 6, then we have when n is even,
( - 1)=^ 2 cos n^=(2 sin 0)« -n{2 sin 0)n-2+ -(*^^) (2 sin <l>)^ - ... .
. -» nin-r-V) (n-2r4-l)
+(-!)--. -^ rixT^T =^^2sin^)-2r. (286)
and when n is odd,
( - 1)2 2 sin n^=(2 sin 0)«- n(2 sin 0)«-2+!!^-ri (2 sin ^)'*-* - . . . .
: : .+(.1)^ . «('>-'- l)--(>»-«^+l) (2 ,i„ ,)^. (287)
274 PLANE TRIGONOMETRY.
209. To express cos nO in a series of ascend-
ing powers of cos d, n being a positive integer.
(1) Let n be even
In (285) r is limited to values not greater than — , therefore
writing for r in the general terra the values — , — -1, — -2, ....
2 2 2
n
..,,3, 2, 1, 0, in succession, the number of terms will be — +1,
and as r is diminished successively by 1, the terms are alternately
n
positive and negative, the first term having the same sign as ( - 1)^.
Therefore we have
2 cos n&=(- 1)-* { -^ '-^ —^ (2 cos ey
K 1.2.3 \n
n.^(^-l)....(n-2(^-l)+l)
, 1.2.3....(i«-l) (2cos«)«
n(in+l) ^{\n - 1) in -2{\)i- 2)+l)
+ 1.2.8....(^-2) (2 cos »)•
«(l«+2)(Jn+l)....(ri-2(i«-3)+l) \
.1.2.3....(^-3) ^(2«o.»)' + &<^|
^" ^ \ 1.2.3....^
«.in(|n-l)... .4.3 ,^ .. . ,,
?_A2 !. (2cos0V ' V
1.2.3.... (^-1) ^ ^ ■ "
n(in+l)in(^-l)....6.5 ^ , . :
1.2.3.... (iii- 2) "^ '
■• > f'l
.=(-l)T|2.!^.(2oos.)'+!iS^|M(2.o..).
_ nan+2)(i«+l)X^-l)(i,.-2) (g ^, ^y, ^ 4^ K
1.2.3.4.5.6 ^ )
WA
TRiaONOMETRICAL EXPANSIONS. 275
dividing by 2, and reducing coefficients we have
cos rK? =(-1)2 {1- -— cos«g+ /^ „ / cos^g
^ ' I 1.2 1.2.3.4
n» (n''-2') ( 7i'-4«) ) '•
(2) Let n be oc2dL
Since r cannot be greater than — , it may be i(n - 1), the integer
next less than — ; the terms are alternately positive and negative,
the first term having the same sign as ( - 1) *(**-^>, and the number of
terms will be^(M-l)+l or \{n+l). Writing forr, J(rt-1), i(n-3),
\{n - 5) ... .3, 2, 1, successively in the general term of (285), we
have as before, by reducing the coefficients and dividing by 2, ^,
cos n5=: - 1)2 {nco&e- \ ^ ^ cos^ g+ \ ^ »\ ^ cos*
' ( 1.2.3 1.2.3.4.6
,, 1:2.3.4.6.6.7 ^^^^'Y (289)
2Z0. To expand sin oo and cos x in series con-
taining the ascending powers of Xy independently
of De Moivre's Theorem. ^m!
The series for sin x must vanish when a5=0, therefore it can
contain no term independent of as, nor can the even powers of x
enter into the series ; for suppose
sin a=^x+5a;2+(7x3+Da;*+&o.;
substitute — a for sc and this becomes
sin (-«)=- -4a+J9x2- 0*3+ J[)x< - &c. ;
but ;/ . sin (- 05)= - sin a, (Art. 41)
'f::" "^'' ,^^ =-Ax-Bqc^'Cx^-Dx*-&o.,
therefore '-, ^ B=-B, D=-D. &c.. which is absurd
unlesg^uv iJ=0, D=0, &c. ,
"■•'ih
276
PLANE TRIGONOMETRY.
therefore
and
but if
iJierefore
Bin x=Ax+Cx^+Ex'^+&o.,
sin » . _ _
■=A+Cx*+Ex*+&o.i
X
y.=o, -^^^=1, (Art. 74)
X
A=l, and we have
ein oc=x+Cx^+Ex'^+&c.
(a)
Again, the series for cos x must =1 when x=0, therefore its first
term is 1, and it can contain no odd powers of x ; for suppose
" ■ ' ' cob«=1+^x+Bx»+Cx'+jDxH&o., ; ' ■ -•
then rj ooa(~x)=l~Ax+Bx^-Cx^+Dx*-&o., : a • *
but (i cob(-»)=cosx (Art. 41) ' r' ..^ ;
"'• f 'v =:l+Ax+Bx^+Gx^+Dx*+&o.f
therefore ' "' ' A=-A, C=-G, &o., which is absurd
unless -4=0, C=0, &o.,
therefore cos x—l+Bx^+Dx*+&o.
Adding and subtracting (a) and (h) we get
cos X + sin x=l + x+ J5x« + Cx'^+Dx* + ^x*+&a
cos X - sin x=:l - x+jBx' - Cx^+Dx* - Ex^+&o.
In (c) write x+h for x, then it becomes ^' "^ - "" -' * • ' "
cos (x+/i)+sin {x+/i)=l+(x+^)+5(x+^)«+C(x+^)'+ .\ ' .
(&)
(c)
(d)
' f '.'■
but
cos (x+A.)+sin (x+?i)==co8 x cos fc - sin x sin ^+sin x cos h+cosxamh
=«os h (cos x+sin x)+sin h (cos x - sin x)
=(1+B;i2+DA<+ . . . . ) (l+x+BxH0x3+ . . . .)
+(fe+C7i3+jE7iH. . . .) (1-x+j5x2-Ox3+. . . .).
Equating the second members of the last two equations and
expanding we have
1+X+ Bx^+ 0x9 + Dx* +
h'{-2Bhx+ZChx^+U)}ix^ +
Ch? +4Z)/i3x +
Dh* +
otf = --
/ l+x+JBx'+Ox" +Dx< +
h-hx +Bhx^ -Chx^ +
Bh^+Bh^x +B*h^x'>+
Ch? 'Ch'x +
SINES, ETC., OF SMALL ANGLES. 277
Cancelling the terms common to both members of this equation
and equating coefficients of hx^ hx^f &c. , we have
8(7 = 5
-B = -
1.2
C = -
1
1.2.3
D =
H
1.2.3.4
B =
1
1.2.3.4.5'
&0,
Hence we have by substituting in (a) and (h)
■ .1 : , ' ■ ■ ■ ■'.-.'.'•;. 1. J)
a* X*
— ■ + ■ ■ ~ • •••
1.2 1.2.3.4
ax I. Sines and tangents of small Angles.
The last two formulae furnish us with the means of finding the
sine and tangent of a small angle, and conversely, of finding a small
angle from its sine or tangent J , , ....
Thus, when x is small
-' . ging;=x~ , very nearly, . ■> ,>'
1.2.3
' Let a be an angle containing n*, i. f . ui;«i ; ui: ; . «(;) t\
then n,.,.^,. '*="7rT; or «=nsinl', ,; ,, ^: m-I i^i.i jd
sin X
since sin l"=3circular measure of 1" very nearfy;
therefore sin x=m> sin 1 cos* x
or Log sin n"=log n+Log sin l^+^Log cos «- 10) '•^'^.*' -^
=log n+Log sin 1"- 1(10 -Log cos «)
=logw+4,6855749-i(Logsec»-10). (290)
278 PLANE TBIGONOMETRY.
And also log n=Log sin n"-Log sin l"4-J(Log sec x- 10)
=Log sin n"+Log cosec l"+i(Log sec x- 10)
=.Log sin n" +6^ 3144251+i(Log sec x - XO). (291)
That is, to find the sine of a small angle we have the following
rule:
" To the logarithm of the angle reduced to seconds add 4. 6855749,
and from the sum subtract ^ of its logarithmic secant, the character-
istic of the latter logarithm being previously diminished by 10 ; the
remainder is the logarithmic sine."— (Chambers's Logarithmic Tables,
Art. 27.)
To find a small angle from Log sin, we have this rule :
'* To the given Log sin add 5.3144251 and ^ of the corresponding
Log sec, the characteristic of the latter logarithm being previously
diminished by 10, and the sum will be the logarithm of the number
of seconds in the angle."
In like manner a formula may be established for finding the
tangent of a small angle, and conversely.
nn. , . , A sin X X COS» X X
Thus tanx= = = , ''
- ...,..0 ; ... ...,.:. COSX COSX ^^^^ ^ ■ , .
and if the angle x contain n% we have as before x=n sin 1", and
Log tan n"=log n+Log sin 1" - f (Log cos x - 10),
=:log «+Log sin 1" + f (Log sec x - 10). (292)
and log n=Log tan n" - Log sin 1" - f (Log sec x - 10),
=:Log tan n" + Log cosec 1" - |(Log sec x - 10). (293)
This method of finding the sine and tangent of a small angle is
generally known as Maskelyne's method, and was first given by him
in his Introduction to Taylor's Logarithms. It is not as convenient
•as that of Art. 112, which is known as Delambre'a method.
«
Examples. _ ,
L Prove that
sin 5x=5 cos* x sin x - 10 cos» x sin' x+sin* x.
!«; . cos 5x=coB *x - 10 cos' x sin« »;+5 cos x sin* x.
EXAMPLES. 279
2. Prove that
2* sin* a=10 sin x - 5 sin 3ac+sin 5ac
2* cos^ ac=cos 5x + 6 cos 3ic+10 cos x.
3, If tan 6=— , shew that
a
m
(a+bv/ - 1)" +(a - 6v/ - 1 ) » = 2(aH6f " cos - 6.
4. Prove that
cos 4x=l - 8 COS* x+8 cos< as.
sin 5x=lC sin/ x - 20 sin^ x+5 sin x.
5. By means of (280) prove the following rule for finding the
length of a small arc: *' From eight times the chord of half the arc
subtract the chord of the whole arc : one-third of the remainder ia
equal to the arc very nearly."
6. Prove that when 6 is small
2 sin 6/+ tan 6=Sdf nearly.
7. Shew that
3 S^-l 3 3*-l S'"-! y
"^'^^=4 ..i:^'' 172.3:475^ ■^••••±1.2.... (2n+l)^ ^^r
8. Find the three values of (-1) by De Moivre's Theorem.
280 PLANE TBIGONOMETRY.
.',!■• ' .'
:■■'. I '
CHAPTER XVI.
EXPONENTIAL FORMULiE — COMPUTATION OP THE NUMERICAL
VALUE OF TT — TRIGONOMETRICAL SERIES.
Exponential values of the sine, cosine and tangent.
aia. From (258) we have ..
t 1.2 1.2.3 1.2.3.4 '
in which let ds/ - 1 and - 6v--l be successively substituted for x,
then we find
the sum and difference of which are
=^2cos^, by (281). (294)
^ 1.2.3 1.2.3.4.6 ^
=2\/^8in0, by (280). (295)
The quotient of (295) by (294) is
>/~ltan6=-—zz — =- —, (296)
e +e e +1
by multiplying numerator and denominator by e ~ ,
ft;.' ■ ; .
EXPONENTIAL FORMUL-^. ^81
These formulae, which are due to Euler, axe reckoned amongst
the moat useful in Modern Analysis.
By the addition and subtraction of (294) and (296) we get
,.,., ,: /^^=cos9+v/^sin0 X -^^ (297)
,^d :e-*^=:cos0-N/^sin0; •-: (298)
or introducing the notation of Art. 201, we have * J
x= /^-^=cos0+v/^8ind """^^ (299)
^ l=e-'^-'=ooBd-^^Bme. "^ (300)
Hence (294) and (295) may be written
1 \
Zcoa6=x + —
X
2v/^sin0=x- —
X '
(301)
and if we .ubatitute «rf for fl in (299) »nd (300) we find ty addition
and subtraction , ,;
1 .
2 cos m0=x"*+-- -i^ . '.:> 'J v
a"* I (302)
2N/^sinme=»'"- —
213. To express the Circular Measure of an Angle
m terms of its tangent. .^, - ^A'.<i^.i
From (297) we have h
I e*''^"'=cos<9+/^sind
4 =co8(9(l+v/^tan0). ^-■^::.X^
i Express in Napierian logarithms, thus
I ey/:Zl=.\oge COS 0+loge (1+n/^T tan 0)
f , tan^e ^-^^ tan^g w/okq'v
' , . =logecos0+v/-ltan0+-^ ^"^ "1 &«m l>y(259).
282 PLANE TEIGONOMETRY.
Equating the real and imaginary parts of this equation we have
\^ ■ ^ tan2(9 tan*^ tan' ^ . ' "
0=Joga cos 6+—- -— + -— &a, (303)
■ ^ tan8 tan* 5 tan' tf .
A. .} 03= tane--^+-— ^+&c., (304)
the last of which is known as Gregory's series, and is convergent for
all angles whose tangent is not greater than 1, . .
If tan d=x, so that ^=stan~^ x, the series may be written
x^ x^ x"" x^
taii-ix=w--+---+- -(&o. ; (305)
214. To find the numerical value of ir,
TT TT
In Gregory's series let 6=--, then since tan — =1 we have
4 4
TT ^ 1 1 1 1
■' '=^(n"+^+"^+*^^-)-
(306)
This series converges too slowly to be of much use. To obtain
a rapidly converging series, let
TT
— =tan~^ m+tan~* n ■ >
4
Of tan~U=tan-iJ^^, by (253)
^, , ^ m-\-n , 1 — n
thereforb 1=, and m=:- 1
1-mn li-n
If we make n=:^ , we find wi=^ t : \. ,
TT 1 1 •
therefore --=tan-i— +tan-^ — - ^
4^0
^ ^-' ;;.'<>: . ;■ > i I
■J
1 l/lx» l/lx" I/I7 ^
2 - 3(2) -^5 (2) -7(2)+*^-
1 1/1\' 1/lv" 1/lJ ^
(307).
SERIES FOR COMPUTING tt. 283
which are Euler's series for the computation of tt. They converge
much more rapidly than (306), , ^ , t ;,.
215. Machin's Series for computing w.
1 ./. .": '.
1 5 2
tan-* 1 - tan-^ — =taii-i ^ =tan-i —
6 ^1 o
. I i ' ,■ • J / u V
2 J^
3 ~ 5
^ - ««x g -v«- 2 17
2 1 3 5 7
tftn-i — -tan-* — =tan-* — — -=tan-i —
■^•"3:6
7 1
tap-* - - tan-* -=tan ^=:tan-* -
17.5
9 1
46~ 5 . , /-I
%J J'i;i
n^jt . n >.,.'.,U7 X -^'vv^
=-t«n-* — .
■' •'■l^^.i.a . ^ 239 _
Adding these equations and cancelling the terms common to both
sides, we have
■ '"f;\ , ,5..:v -■•■-. ^-v
tan- * 1 - 4 tan-* ^= - tan-*
1 . ' ", t : v^ ^■- ) -i -i '1--'
239
1.^-11=4 tan-* --tan-*—,
'V:\ «,i^
ih«.for. -= ^ ^ , ^ h (308)
284» PLANE TRIGONOMETBY.
These serien converge quite rapidly, especially the latter. If we
take eigh! terms of the first and three of the second we find
7r=3. 141592653589793 ....
For other series for computing tt, see the examples at the end of
this chapter.
Expansion of certain Trigonometrical Equations
into Series.
216. Cfiven sin p'=sin P sm (z+p), it is reqmred to express p in
a series of multiples of z. [See (a), Art. 151.]
Multiplying both members of the given equation by 2v^^, we
have
2v/ - 1 sin p=sin P . 2v/^ sin (z+p)
or e^^'^-e-^^'^sinPCe^'+^'^-e-"-^^'^), by (295)
whence 6^^=^"""^^"^^',
l-8inPe'^-*'
and taking the Napierian logarithms of both members we have
2pv/rr=log (1 - Bin Pe-^"^) - log (1 - sin Pe' ^)
- sin Pe-*^ - ^ 8in« Pe'^^^-i 8in3 Pe"
> + ainPe'^-'H-i8m«Pe''^ +i8in3Pe''
by (260)
=8in P(e' ^-e-*^^)+ J sin« P(e^^-e-^^)
^ +*8in3P(e'*^^-e-^^=^)4-....
=8inP.2v/^sin a+^sin* P.2\/^sin 2«
therefore / ^ . +J ain3P.2v/Tr8in3.+....
p=Rin P sin z+^ sin« P sin 2»+i sins P sin 3«+. . . . (309)
Here, p is expressed in circular measure ; to find p in seconds we
must divide both members by sin 1", according to (128), and since
e - .... I
TRIGONOMETRICAL SERIES, 285
2 Bin l''=8in 2", 3 sin l"=Bin 3", &c., approximately, the last equation
may be written thus,
„ sin P sin z sin^ P pin 2z , sin^ P sin 32 , .„- ..v
'^--^v- + -Iter- + -STs" +•••• ^^^'^
^ac. —Given P=58' 10" and a=40% to find p. (Same as Ex. 64,
Chapter X.)
Bin P=8. 228380 sin' P=6. 4567 sin^ P=4. 685
sin z=9. 808067 sin 2z=9. 9933 sin 32=9. 937
cosec 1"=5. 314425 cosec 2"=5.0134 co3ec3"=4837
2243". 23=3. 350872 29". 07=1. 4634 0". 29=^. 460
|)=2243".23+29".07+0".29=37' 52 .6. *
Here we omit the symbol Log for the sake of brevity.
'• "'■ '■'
217. CUven tan x=n tan y, to express xma series of multiples of y.
Multiplying both members ol the given equation by v/-l, wo
have by (296)
. • • e -1 e -1 • ... ;.. • ,
e
— — : =n -pz — »
*,V^ (l + ^)e''^ ^-^4-(l-n) . ..V':. A.-:
(l-n)e +(! + ») =- '
li.'
r'lt
3y^. 1-n ^■•'-- ^•.•'- "
e +-;;
1-W 2yV-l , t .....
\ 1 4-1::^ e'^^-'
■:\\-<-rVyK>'- 1- \ 14-— 6
1+W ^r. ,..„...-..;,,•;.• .-,{^-
'■ ' ayV=i/l +me'
2i/>/^ /
--^
ssse"' ' 'I — ^ 1 1 I V S^r - ■ >: \v Jiiio
,^...;; ,.•?;, %!; « • • u+me ' . =7:^ it;- ■;; >a V
286 PLANE TRIGONOMETRY.
1-n
by putting m=-
•^ *^ ® 1+n
or e
,i^,2,^^lj-me___ ^ ^^ .j^ logarithms
1+me
(a; - y) 2v/ - 1 =log (l+me"^"^) - log (l+we^^^)
-I
me -^m'e +-J m^e -....
» =s-m2\/-l BinJ2t/+im^2s/-l sin 4y
- J m« 2\/"^ sin 6y+....
therefore «=i/ - m sin 2y +^ m« sin 4i/ - J m^ sin 6j/ + . . . . (311)
^*_,/' wsin2y m«Bin4y m» sin 6y
sin 1 sin 2s sin o
o ^» ** "1^ y ^ ...
21o. GwcH rm «;=; , fo emress x vn, a senea of nrnti-
1-ncosy
pUs oj % * . -. . ,
The given equation may be easily reduced to
sin aj=n sin (w+y), ,. ... , , . (a)
which is of the same form as the equation of Art. 216. Hence
writing x for p, n for sin P, and y f or « in (309), we have at once
«,=n sin y+l ri« sin 2i/+J n' sin 3i/+.... (313)
Formulee (309)-(313) are very useful in Spherical Astronomy.
The left-hand members are limited to acute angles, but entire gener-
ality may be conferred upon them by observing that the equation
tan x=n tan y is true when we write n'n+x for x and wV+t/ for y,
f^d therefore for x-y we may write ac-y-(m'-n.')7r or x-y-pn
TKIGONOMETRICAL SERIES. 287
where p is, like nf and m', any integer or ze 'o. Therefore (311)
may be written thus :
x=pTr+y - m sin 2i/+^m« sin 4i/ - . . . .
In the same manner the other series of Articles 216-218 may be
generalized. „ ,
219. In a triangle ABG, given two sides a cmd b and the included
angle C, to express either of the other angles by a series of multiples ofC,
5 sinB sin (A+G) ..'-.1
Since — =-: — 7= : — z
a sin A sin A
, . •'. ... .'-..,,.'■ ..,.,
we have sin -4=— sin (^+0),
which, compared with (a) of the last Article, gives by (313)
t
a . ^ a^ sin 2(7 , a^ sin 3(7
,„ a sin . a' sin 2(7 a' sin 30 , -o,..
which will be convergent when — is a proper fraction. - ^ '
■ r' ■'■•' ■ '
220. In a triangle, given two sides a and b and the included a/ngle
C, to express cby a series of multiples of C, '
From Art. 122 we have • r. •''■ ' •;
c2=a2+62-2a6cos (7 .-V'tV- .=ss
b ^ b ^
=a2(l-2-cos (7+-) '
=a2(l --(«+-)+-), by (301) -_^ ^^.^.
==aMl--aO(l--)
^ a ^ oo"/ ..r'
h b
2 U>ge C=2 loge a+loge (1 - - a5)+loge (l - -)
a tttC
288
PLANE TRIGONOMETRY.
2a''
62
3a^
63
ax ^a^x"^ 2a^x3
-...., by (260)
=2iogea — («+-)-7r^pH-i)--rT(» +-i)-""
a ^ X' 2a^ ^ x^' Sa^ ^ x^'
b 62 63
=2 logs a 2 cos G-—~ 2 cos 2C- —-- 2 cos 3(7-
a 2a^ ^'»3
3a''
loge C=logg a - (— COS C
62
2a2
63
COS
20-
63
3a=
63
cos
3(7+....)
or logc=loga-iltf(— cosO+— — cos20+— — cos3(7+....), (315)
^a 2a2 3a3 /' ^
where ilf is the modulus.
, <o ecq>re88 r in a series of rmdtiples
1+e cos d
of 6; e being Uss than 1. (The Polar £Jquation to the Ellipse.)
26
Assume e=- — — • , which is always possible, since 1+6^ > 26.
then
1+6
I_e2^/Ll^f and 6= =
h + b^f l + v/f-
_ ^ 1
V Let 2co8^==»+— ,
then 1+e cos ^=1 +
1+6
-. (-i)
therefore
=1TP <'+^> (^4)'
B=a
+6=
(1-62)2
1+63
+ 6a
_1
1 +
(1 - 6x+62a;3 - b^x^+b*x* -...,)\
.^ b b^ b\b*
x X* x^ as'
TRIGONOMETRICAL SERIES. 289
^(l::^l!|(i +&2+6*+?>«+b" +....)
+ ( • ) •••• J*
1
But (1 +62+6*+6« + ....)= JTi?
b
-(6 +63+b^+?>' + ..-.)=-][r6i
- (63 H-b"^ +6^ +6" + . . . .)= - ^TftJ
+ ( )= •••• W:^^ ''"
Therefore
__« ^^^'^ / 1 - 26 cos 0+262 cos 2d - 26^ cos 36+....}
=a v/i^'^ (1 - 26 cos 0+262 cos 26 - 26^ cos 30+ .... )• (316)
222. To shew that iV i''^ . .Vrr/^
cos - cos^ COS -....ad W:=-^^. ^v V-
8inx=2 sin -cos- .^_ .,
=2« cos I sin I cos I , since sin |=2 sin | cos ^ ,
r 20 ■ ■
290 * ■ PLANE TRIGONOMETRY.
^, X X , X X
=2 cos — cos - sin —:: COS —
=2'» COS I cos ^ COS ^ . . . .COS ~ sin ^.
_, . XXX X mnx
Therefore cos — cos — cos r^ , . . .cos — =
2- sin --
. a;
Bin —
X 2**
but 2" sin — =x —x, when n=oc , by Art. 74.
TT « 05 jc , . , sin aj ^ ^
Hence cos "^ cos — cos — ad inf.= . (OJ 7
4 2^ 2" X
Whence we also get • • '
XXX
aj=sin (B sec — sec — sec — ... .ad inf. (318)
2 2^ 2"
Multiplying (317) by cos x and expressing in logarithms we haV j
1 .1 ^1 * -• • .. 1 /sin 2«x
log cos a+log cos — +log cos — + ad mf.=log (-— — )
. 2 , . }L ^ 2x '
Summation of Trigonometrical Series.
223. To find the mim of the sines <md cosines of a series of angh?-
in arithmetical progression. ., ,.; .^ ,; ,> .v-^„ ^ , >. -
(1) sin x+sin (a;+i/)+sin (a!+2|/)+&c., to n terms.
By (54) we have
cos (a5--jr) -cos («+— ) =2 sin — sm x,
/ 1/ \ 3 y
cos (a5+-T) - cos («+-^ l/)=2 sin ^^ sin (x+y),
3 5 « '''"* -~*^^?^
cos (»+— y) - COS (x+— y)=2 sin — sin (x+2y) ,
.. ►• ^ 4 Ji
TRIGONOMETRICAL SERIES. * 291
cos (x+?^y)-c5«(x+?^iy)=2 sin fain (x+(n-l)i/),
Lot -S denote the sum of the series, then we have by addition
cos (» - ^) - cos (x+ -y- y)
s=2 sin " (sin x+sin {x+y)+am (x+2y)+ . . . .)
s=2Bin|--5
cos («5 - "I) - COB (36+-^- y; , . ,.
whence S=
2'^ a
(319)
sin(x+— ^l/)8ln-
(2) cos X+C08 (x+y) +cos (x+2y)+&c. , to n terms.
By (52) we have
.^ rin(.+|) -.in(«-p =2.in|co,«. ^^^.^^^^
•fa (a+|i/)-Bin (x+l) =2 sin | cos (x+y),
*--^ Bln(x+|i/)-flin(x+2^)=2 8in|co8(x+2t/),
\ "■*»»
am (x+?^ y) - sin (x+?^ y)=2 sin ^ cos (x+(u - l)y).
Let 8 denote the sum of the proposed series, then we have by
addition
Bm(x+?^-v)'-«in (x-|)=2Bm | (co.»+cos(x+y)+&o....)
292 PLANE TRiaONOMETRY.
=2 sin I ' 8'
2
sin (x+— ^ y) - sin [% - -)
whence 5—
2si„|
W — 1 WW
COB («+— ^ 1/) sin —
-J (320)
. V
sin —
2
224. Tf t/=x, 2a5, &c. , in succession, we find from (319) and (320)
Sin — - — X sin — •
Bina+8in2ac+8in3a8+&c...+sin»Mi; — (321)
sin —
('!;*.) ,— '■•'•- -^ ■ 2
. >~ -V sin'** ■na y«^«v
sin x+sm 3a54 am 5a;+&c. . . +sin (2»i- l)a;= -; . (322)
sm a
&c., &a« &o.
. , cos — - — X sm --
' ' * 2 2
oos«+co8 2a5+co8 3a+&c...+costM5 — - (323)
. «
r . . ,; e'.n —
. . ^ ■ .w I',,;- :;:'■■ ( •.. t":, - •■ 2 ■
,- . /« ^v sin 2n» ^„^^^
cos jc+cos 3«+cos oas+ftc... +co8 (2n-l)a;=— — : . (324)
2 sin «
&C., &c., &c.
225. The formuliB of the last two Articles enable us to find the
sum of the squares of the sines or cosines of a series of angles in
arithmetical progression ; thus, let
sin^ ac+sin^ (x+i/)+8in2 (a;+2i/)+&c.,
to n terms be the proposed series. ' ; ^
TRiaoNOMETlUCAL SERIES.
293
Since 2 sin^ x=l - cos 2x, 2 sin« ix+y)=l - cos (2x+2i/) Ac. ,
we have by substitution
2S=l-co8 2x+l-co8(2x+2i/)+l-co8(2x+4i/)+l-cos(2a;+6i/)+&o.,
to n terms
=» - (cos 2X+C08 (2x+2i/)+cos (2x4-4i/)+&c. , to n terms)
cos(2x+(n-l)y)8inny ^^ ^^^Q)
""*''" siny *
and
n cos (2x-\-{n-l)y) sin ny
*^~2 2 amy
In a similar manner we may find the sum of the series
cos" oj+cos^ {x+y)+coa^ {x+2y)+&G.,
to n terms by using 2 cos* x=l+coa 2x, &o. • - ^ '
226. To find the sum of n term of a aeries of the form i y ^
sin X cos y+svn2x cos Zy+dkc .... +sin nx cos (2n - l)l/.
.By (61) we have : , ^ ^ r ^ ^ , . j^ -. ^ ^.^ '
Bin (x+y) +sin (x-y)=2sinx cosy
• ' ' sin (2a+32/)+sin (2x - 3y)=2 sin 2x cos 3y,
&c., &c- 1^^ .
Thus by addition the proposed series is resolved into two others
which can be summed by the method of Art. 223.
227. TofindthemmofnUrmofth^seriAS ,, ,,^ ^
tan x+2 torn 2x+4 tan 4x+<jfcc.
By (119) we have . ""'
tan X = cot X - 2 cot 2x -— - v:^^^
2 tan 2x=2 cot 2x - 4 cot 4x
4 tan 4x=4 cot 4x - 8 cot 8x
2"-i tan 2«-i x=2'*-i cot 2'*-^ a - 2»» cot 2" x ;
therefore by addition -S=cot x - 2» cot 2» x. (325)
294 PLANE TRIGONOMETUY.
228* To find the sum of n terms of the series
cosec x+cosec 2x+cosec Ajr+<S;e,
By (120) \7e have
X
cosec X =cot -- - cot x
cosec 2aj=cot x - cot 2x
cosec 4x=cot 2x - cot 4x
cosec 2"-^£c=cot 2^^x - cot 2*^^x,
therefore by addition <S=cot -^ - cot 2»»-ia;. (326)
29.9* To find the swm. of n terms of the series "^oj
a sin d+a^ sin 2e+a* sin dd+dhc, ■-•■- '-.•' ^^i
Putting S equal to the sum and multiplying both members by
2n/^ we have ^ ^^'^Y}^ ».,V ^-"■»-^- ''■ V' ■■ ■ •-"' '•' ■■■''■ ^ ■ '■ '"
2v/'^iSf=a..2\/^sin e+a».2V'~rsin 2e+a3. 2N/^sin 3<?+&c.
=a (x ) +a» (x« — )4 a^ (x3 ) 4-&c. , to n terms
. by (302)
ox + a«x' + a3x3 + &<;. , to n terms
^'ij>tti'i.> t»V»
(a a* a' %
— I 1 I-&C., to n terms.)
X x« x3 '
ax (a*» x" - 1) ax-* (a" x-** - 1)
- ax-1 ax-*-l »(Colenflo'sAlg.,Art.l57)
an+2 (x» - X-") - a»+* (x"+Hx-^*) -t- a (x - x-*)
"" a8-a(x+x-i)+l *
therefore a/;:.! o-* <^?K: ■■'.
a«+» sin nfl - a"+i sin (n+l)0+a sm <?
^ a«-2acosm *. . ^^^^^
. If a be a propar fiaction and ns=oc , we shall have
a sin 9
' ' ""a8-2aco8(94-l "^
«s the sum of the series continued to infinity. '^' "^ ' ' ' '
EXAMPLES.
295
a!cos0 . g
230. Tojind the sum of the series
cose sin 6-^ cos^ 6 dn 26+^ cos* 6 sin dd - d;c , ad inf.
By (302) we have >*
2s/^S=cos e (X-C^V'-^' (x^ -x-)+'-^' (x»-x-)-&0,
^o^ x^_co^^^^'^^ log (1+x cose)
2 3
x-^cos0+^^— ^ 3 + &c. = -log (1+xr-i cos d)
1+XCO8 ,, . 2SV:=I _1+X£OS0
-'^^ 1+^^ cose * ''''''''''' /=U^^e* .>..o
and by composition and division, .^Uiu: • ■ •-
e'^^^-l _co3_g_(g-ar-^) _ 2 cos sin v^^ -
■ " 2SV-1 ^""S+coseCx + x-O 2+2cos2 *
2 cos e sin vA-1 --- '
thert-ore by (296) v^-ltanfe=- a(i+cos''e) "
and .- - ^-tan-^ (^(I^). ' ^
231. The investigation of Trigonometrical Series cannot be
fuUy carried on without the aid of the Differential Calculus. The
• student must therefore consult the treatises on that branch of mathe-
matics for further information on this subject. (See Todhunter's
Diff. Cal., chaps, vi., viL ; Clark's CaL, chaps, v., vi; Williamson's
Diff: Cal., chap, iii.) i v' • ■
^^' "''''' Examples.
1. Prove by the aid of (294), (296) and (296) the foUowing
identities : .:. ..
/^ V X ^ 1-CO80 \ , .\ - ^ /2\ cos 2&=<os« - sin* ft
(1) tan— ■= — r ^ • ^'
^ ^ 2 sm0 . „^
e B f) Bin g+sin 36
(3) sin 0=2 sin ^ cos - . (*) tan ^^-^^f^Q^Q^a^d'
(5) tan(45-0)tan(4^-3.)4^^ (6) 2 cos.0=H-coa 20.
296
PLANE TRIGONOMETRY.
2. If 2 cos e=x+o(r-\ and 2 cos <^z=iy+y-^^ then will
V 2v/-l sin (m^+n0)=:x'"2/'* -a;-«*t/--». •
3. Shew that ^"^^=-1
*"^ ^'r^ + 1 oC. . ~&c., ad inf.=0.
4. Shew that 7r=--v/— 1. log (— 1)
ana (^>-u-^+J^(|;._|_(^;,^.,^^.
6. Prove that versin «= - Ke'* - « "a" "^"^^a,
6. Prove that
(l+cos e+v^TIsin 0)" +(l+cos 5- v/TTsin e)»=2«+icos»- cos-
2.2*
7. If aH6«=2, and a6=v/rrtan 20, shew that a=cos 0^2 sec 2^.
a If 2 cos a„ =a„ +- , and ai+aa+a,+ . . . . +a, =2»r,
\SC?S- ^n
then will sCjaCaXa Xn =1.
9. Prove that
2 sin (aj±y v/^)=(ev +erv) sin x±(ev. e-*) >/^cos aj.
10. If 5i=tan ^+tan J5+tan 0+&c.=sum of the tangents,
^2=sum of their products taken 2 and 2,
Sg=avim of their products taken 3 and 3,
&c., ^ &o.,
shewthat >. '.w-'.') <:,■-•:-; :;i-- iji'.
tan(J[+^+C+&c.)=^^=#^~. ■ " ■'
11. In any triangle shew that
log —==.¥{ (cos 2Jl - cos 2B)+^(cos 4A - cos 4B) -
+i(cos 6ui - cos 65)+&c. }
iV •UT
EXAMPLES. ; 297
12. Prove that
loge sec 6=^ tan* d-^ tan* d+i tan' - i tan^ 0+&c . . . .
13. Prove by Gregory's series that
3eo-i-(-+-)=-(--l)(- + l)-— -(--l)(-+l)+&c
14. Prove that log tan* ( j+ d) =tan e+\ tana e+i tan^ 0+&c....
15. Shewth»t|=i+-L+^^+....
16 Prove that — =4 tan-^ — - tan""^ — + tan"^ — : •" ' '
4 5 70 99
and — =4 tan~^ — - 2 tan-* — — -+tan-i
4 5 408 1393'
17. K tan 2e=8in 2^, shew that • ; • ■' .. ^ ■ - ^} .^-
6=cos 2^ tan ^ - i cos 60 tan^ 0+i cos 10^ tan* ^ -> • • • •
Sum to n terms the following four series :
-. ';' t '-
n Q a a a ..>..'. ^
18. tan - sec e+tan — sec — +tan — sec — +. ... » •
2 4 a o 4
B
' Ans. tan d- tan — .
2*
1 1 '• 1
19. — sec 0+— ace d sec 20+ rr. sec 6 sec 20 sec 22 0+
2 2* 2''
... : ' .4m. cos - sin cot 2" Q.
20. sin («-y)+sin (2a;-3i/)+sin (3ae-5y)+.... ' -
sin {\{n + l)x-ny\ 8in(|fix-ni/)
Am. —— •
sm i(x - 2i/)
21. — log tan 2x+— log tan 22x+— log tan 2»x+ ....
1 « • o log 2 sin 2«+ia5
ilna. log 2 sin 2x — •
Sum to infinity the following twelve series:
^^ cos 6 cos^ A cos' .. 1 / . ^ .\
22. — ~- + — r-^ H — r— + . . . . Aim. log (cot — coseo fj,
\ i O A
298 PLANE TRIGONOMETRY.
cos 20 cos Ad cos 6d
23. 1 +
1.2 ^1.2.3.4' 1.2.3.4.5.6
Ans. ^ cos (sin 6) (c*-"'* Ve"""*).
24. sin a sin a+^ »in^ a sin 3a+i sin^ a sin 5a+ .... i
•-" Ans. ^ tan-^ (2 tan* a).
or or
25. X cos (a+i3)+-— r cos (a+2/3) +r-r-^ cos (a+3i3) + ....,
.4rw. e* "^^ ^ cos (a+oc sin /?) - cos a.
- - Tj'ii-. a
26. sin a-^ sin 2a+^ sin 3a- J sin 4a+.... .4m. - - .
27. sin ac sin ac-^ sin 2x sin*«+| sin 3x sin' «— .. .. . , ,,
Ans. cot~^ (1+cot cc+cot* aj).
28. cos 2^+^ COS* 20+i cos^ 26+ .... -. .. Ana. log cot d.
1 + 23 1 + 2'
29. (1+2) log 2+-p^ (log 2)3 + j-^ (log 2)» + . . . . ^m. 4.
30. 2 {ain^ ac - ^ sin* 2»+J sin* 3a; - i sin* 4«;+ }.
Ana. log sec x.
31. K cos x+i cos* x+J cos* x+ . . . .=/S>,
and cos x+i cos 2«+^ cos 3x+ . . . .=», u ji-. u.v,
shew that "" " <S-2s=log2.
^^ sin sin 29 sin SO sin 49 ^ , sin 6
1.2 ' 2.22 ' 3 23 ' 4.2* 2-008
3 4 5 -
33. 2 cos 9+— cos* 6 +— cos3 6 +— cos* 0+ . . . .
2 *> 4
^' . cos 9 , ,^ „.
An>s. - log (1 - cos 6).
34. Prove by Art 222 that - - .*iu,
(l - tan" — ) (l - tan* -^ (l - tan* rj) . . . .ad inf.=x cot x.
35. Shew that
- n n(?i-l) „ „ n(n-l)(n-2) ,
1+— a cos wxH — r-,r- <J* cos 2wx+ o ~q~~ " *'°'' ^'"'*'
1 1»^ -'{'■■u'i ■■"■'■■■■:(■. 1'2.3
■ "'" +&c.=r** cos n9,
. ^ ' „ , . „ a sin mx
where r*=l+2a cos mx+a*. and tan 9=^ . ■'<
1+a cos WW!
RESOLUTION OF EQUATIONS. 291)
,ii_ '•. r if-' !' ■• .. . . •' f.
, ,..'■/ ' c" ' ^ ,'•-■, li V'
CHAPTER XVII.
RESOLUTION OP CERTAIN EQUATIONS INTO THEIE SIMPLE AND
QUADRATIC FACTORS.
232. To resolve the Equation t ■ ^
a2»-2aj"cos<f.+ l=0. (329)
into its Simple and Quadratic Factors.
'iUiU :^^i,:>ine
By completing the square and extracting the square root we
liave
a:» =co8 0± \/ - 1 sin 0=co8 (2r7r+(^) ± v' - 1 sin (2r7r+0)
whence aj=|cos (2r7r+^)±v -1 sm (2/'ir+0)| »
sxcos ^±v/-lsm ^, „ *'(330)
by De Moivre's Theorem.
1.* Aa.*.\ I'. '• J . ■•..xs\.t\*it% i..*\\
Therefore the 2n values of x will be found by assigning to r the
values 0, 1, 2, .... n - 1, in succession ; thus, using first the upper
sign, we have :■ ; ,'■. 'v^nn ,i.i^- ^^ '^'i- '^^ imrA ^^n tk'iu^
(1) if r=:0, 8C=co8 — + v/'^8in— , .i?b'x.K ■^%
n n
"it + d> / — — - . 27r +
(2) rx=l, a;=cos |-v-lsm ,
(3) rs=2, a5=cos hv-lsm ,
* 2(n-l)7rH-^ . /— -- . 2(n-l)^+0
(n) r=n - 1, «=co8 h v - 1 sm — ;
^ ' n n
300 PLANE TRIGONOMETRY,
and using the lower sign.
(1) if r=0, «=cos -^ - \/rr sin ^ ,
n n
(2) r=l, ai!=cos - n/ - 1 sin ^ ,
n n
(8) r=2, x=cos ^ - V" 1 sm ,
n n
(n) r==n- 1, a;=cos -^ — ~ - v -1 sin -^ L_Lr ^
n n
Now, from the first of each of these groups we have the first two
■imple lectors
{as-cos s/^BUi — ) and f jc-cos — + \/^sin — ) ,
the product of which is «2-2a; cos — +1, the first quadratic factor;
from the second of each we have the second two simple factors
(a-cos ^->/-lsm ^) and (a-cos ^+\/-lsin — ~^\,
the product of which is a;8-2« cos — — ^+1, the second quadratic
factor ; and so on, there being 2w simple factors in all.
Therefore we have for the w quadratic factors
a6*»-2»"cos^+l= (a;2_2xcos-^+l)
w
• ' • "J , 27r + <6
x(a52-2a;cos -^-1)
n
' r-n X (X^ - 2x COS -^^+1) • ':
n
/ , « 2(n-l)7r+(4
;•,' X (a;2 - 2x COS ^-^+1 . (331)
RESOLUTION OF EQUATIONS. 301
Since the simple factors of a;2'»-2oc'* cos ^4-1 are obtained from
(330) by making r=0, 1, 2, m- 1 in succession, it is evident
that they are all included in the form
/ 2/'7r + , — — . 2r7r + 0v .„„^-
% - (cos ~~- ± \/ - 1 sin -\ (332)
Ex. Besolvc the equation x*-x^+l=0 into its quadratic and
simple factors,
^ere n—2, 2 cos ^=1, therefore <}>=—, and by (331)
o
If ^TT
fl6*-a2+l= (x2-2acos — +1) (»2-2xcos — +1)
6 o
= (x - cos — - v/ - 1 sin — -)
X -COS — -H-v -1 sm — I ■■.'..,
.... •; :■ ..n
X (x +COS -+ v' - 1 sin — ) .. ^
^ p o ' ' - •■;■'•» :/
'■'"■'■'■''■' '' ' ■' - >
,, . X (x +C08 — -\/^sin — ), by (332)
X (x +iv/3"+K^)
X (X +^v/"3-^\/"^. -w.^ .,•..■ , ,,'7' ,;;,fc.
Hence the four roots of x<-x2+l=0, are " • r\-^^^^*Pi: ,
Kv/3"±v/^ and ^(-v/3"±v/^. . m!
233. In (331) let x=l, then we shall have
2 - 2 cos <6=(2 - 2 cos — ) (2 - 2 cos -^^—^) (2-2 cos -^^) , &o.,
Of . , . to n factors,
23 8in2 ^=2«» sin' /- sin^ V^ si^^ — ^ , &c.,
2 2ii 2n 2n ' '
302 PLANE TRIGONOMETRY.
md writing 2nB for (p, dividing by 2" and oxtrafl>ting the square ro(>( ,
we have
Q O
sin tie=2'»-i sin d sin (#+— ) sin (e+—) sin (d+—) , &c. , (333)
, • - > , ; to n factors.
Let n^=— or fl=— ; — , then we have
1=2'*"^ sm — — sin — — sin — — , &c., to n factors. (334)
2n 2n 2n ' * ^
Again, in (331) let jc=— 1, then we shall have .:
(n even) 2-2 cos <f) )
(wodd) 2 + 2 cos^ j
= (2+2 cos ^) (2+2 ooB?^^) (2+2 cos ^^^), &-.,
whence, putting 2nB for and proceeding as before, we have
{n even) ± sin nO
{n odd) ±co8 riB
to n factors.
Dividing (333) by (335) we get
i =2«-i cos e cos (0+—) cos (&+-^) , &c. , (335)
(ti even)+l ) ^ / ff\ / 2r. „
/ jjx^ „ >=tan(9tan (0+— ) tan (0+ ), &c., 3; C)
(rj. odd)itan w0 j ^ n^ ^ n ^
to 11 factors.
If
In (336) let n0=— , then we have, whether n be even or odd,
l=*an — tan — tan — , &c. , to n factors. (337 )
4n- 4>i- 4ri, , ,
234. To resolve the Equation as" -1=0 into ity
Quadratic Factors, n being odd. >, .t m
In (331) let ^=0 and it becomes
(»»- 1)2=(X - 1)2 X («2 _ 2X COS - +1)
n ■ ■ V
, ,*''••• . , X (i«2 _ 2x cos — +1) . ,,' ..,
"-. }i{9a9-2xco^^ -^+1), •.;. (338)
RESOLUTION OF EQUATIONS. SOIj
Now, as n is odd, and as there are in all n factors, the number
of (juadratic factors in (338), exclusive of (x-iy, is even; and since
2{n-l)ir , 2k. 277
COS =C08 I2vr )=cos — ;
n ^ n' n
2(?i-2)ir / 47rv
cos =C08
n ^ n' n
I 47rv 4n-
{2n ) =cos — , and so on :
the first and the last of these factors are equal ; the second and the
last but one are equal, and so on ; hence uniting these equal factors
and extracting the square root, we have, when n is odd,
27r
a^ - l=(x - 1) X (x2 - 2x cos - +1)
91-
, 47r
X (x* - 20* cos - +1)
n
'■•■'■ ""-" ■■•''• ^' (n-l)fl- ,, •-■
X (x« - 2k cos +1). . } (339)
n
235. To resolve the Equation aj»-l=0 into its
Quadratic Factors, n being even.
• I ■ .1. . ^1 1 (•••«•
When n is eoen the number of quadratic factors in (338), exclusive
of (x- 1)2, is odd, and there is a middle factor, the — , which will
not combine with any other. This factor ia
^ n
2
«• - 2x cos -«»— +I=x2+2x+l=fa;+l)«.
n '
'/ '^
Hence uniting the other factors and extracting the square root, we
have, when n is even,
27r
x" - l=(x - 1) (x+1) X (x2 - 2x cos — +1)
n
L.1 1 s) :> '] t '; - J •' 'tr X (x3 - 2x cos - +1) ' - '^^ -^~' '^'
47r
n
- • - .,/,'f^-'£i
'^ x(x2_2xcos^^ ^+1),
(340)
304 PLANE TRIGONOMETRY.
236. To resolve the Equation 05" + 1=0 into its
Quadratic Factors, n being odd.
In (331) let (j)=Tr and it becomes
(«" +1)»= (x2 - 2x cos -+1)
x(x* -2» cos — 1-1)
n
x(x2-2a;coB — +1)
71
^(^..2xcoB-^?^?^^^+l). (341)
Since there are n quadratic factors and n is odd, there is a
middle factor which will not combine with any other. This factor
is evidently
x« - 2x cos ^-fl=(x+l)«,
f... , n ■. r- , ,-::'
and it ia easily shewn, as in Art. 234, that the factors equally dis-
tant from the first and last, are equal; hence uniting the equal
factors and extracting the square root, we have, when n is odd,
flc" + l=(x + 1) X (x* - 2x cos —- f-1)
n ■ I
X (x' - 2x cos — hi)
n
x(x'-2x cos — hi)
n . .
{n-2W
X (x» - 2x cos ^ ^ 4-1). (342)
n
237. To resolve the Equation a;"+l*0 into its
Quadratic Factors, n being even.
When n is even the number of quadratic factors in (341) is even )
therefore we have ^^
RESOLUTION OP EQUATIONS. 805
IT
as* + 1= (x^ - 2x cos -+1)
n
x(«'-2x cos — +1)
11
x(x'-2»cos — +1)
n
x(«»-2xco8^^^^^7r+l). (343)
n
238. The simple factors in each case may be found by resolving
each of the quadratic factors ; or, they may be found from (332) by
putting 0=0 for the form x" - 1=0, and <p=n for the form x" +1=0.
Ex.1. — Find the quadratic factors of x^ +1=0,
Here n=5, then by (342) we have
aj»+l=(x+l) (x» - 2« cos 36''+l) (x' - 2x cos 108°+1) :, p
=:(x+l) (x» - ^^^^x+1) (x'+^^^lflx+l).
By putting each of the quadratic factors equal to and solving
bhe equation we obtain the five simple factors.
Ex. 2. — F <d the roots of X* +1=0. , > ,f . <.».
By (343) we have
aj*+l=(xa _ 2x cos 45''+l) (x^ - 2x cos 135'+1) ' '■"
. =(x2-\/2'x+l)(x»+v^x+l). .--r,
Bence x== -= — and -= .
v/2 v/2
"if'
239. In (339), divide both sides by x - 1, thus
27r
x'^i+af-»+ x+l= (x^- 2x008 — +1)
X
'.':'' , x(xa-2xcos-^^^^^7r+l)
n
. n « W-1 ^.
X (x^ - 2x cos o"-!-!),
n-1 '''
uiiere being — r — quadratic factors.
21 ,::. - ,• ' ,;;
'Vila *••^"l^-
T<>
m'H
■■■.. 't
.r
T
-i)
306 PLANE TRiaONOMETRY.
Put x=l, then we have
n— 1
n=2 » (l-ooB -)(l -COB -)...(! -cos iili7r)(l- cog -!LLl;r).
o«_i • « 27r 47r n - 3 vt - 1
3*2**"^* flin* -- sin* — ... sm* — r — tt sin' ir,
2n 2n 2n 2*1 '
or
T iTT" • 2ff . 47r . n - 3 . n - 1 ,, , ,
n^=s2 ^ Bin — sin — .... sin — - — ir sin ir. (344,
2n 2n 2n 2n ^
240. In (340), divide by ao-l and put x=l', then we have,
w — 2
there being —r — quadratic factors,
"~ 27r\ /- 47r\ ,, n-4 v,^ M-2 ,
n=s2.2 2 (X-cos —)(l-cos —)...(! -cos — ^^7r)(l-cos
rt
-^;.-
47r .«»i-4 ..n-2
^«-«,«27r.,47r .„»i-4 .,
=2.2*^' sin' — sin' -- .... sm' — r — tt sin'
■IT,
2w 2n 2n 2»i
or
1 fl-i
2^ . 47r . n-4 . m-2
n'=2 ' sin — sin — .... sin — - — «• sin —- — tt. ' (346)
2n 2w 2n ^r ^
241. In (342), divide by x+1 and put x=l; then we have,
there being quadratic factors,
nr-\
1=2 ' (l-cos— ) (l- COB — )...(! -cos — ^^7r)(l-C0S — - — tt),
,.„'r.o3^ .-n-4 .-n-2
=2^1 Sin* — sin* —....sm* — — tt sin* -- — tt,
2n 2n 2n 2» , .1 .p •
or
1=2 * Bin — sin — .... sin -— — re sin —- — tt, (346)
2n 2n ^ 2n
242* In (343), put x=l and we have
2=2^ (l-COS— ) (l-COB — )...(1-C0S — - — 7r)(l-C08-- tt)
\ n^ ^ ' n' n n
or
- r-j— . ir . Sir . n — 3 . n — 1 r . . /njt-,\
1=2 ' sm — Sin —....sin — - — ir sin — - — ir. ; ^ (347)
2n 2n 2n 2n c >'^
RESOLUTION OF EQUATIONS. 307
343. To resolve sin x and cos x into factors.
The series for ain x, Art. 210, may be written thus
.i„,=,(l._|_^_|_.4e.), (348)
from which we see that x is one factor, and the factors of the series
within the parenthesis must evidently be of the form
x«
a
where a is constant but has a different value in each factor.
For suppose the factors to b«)
1- — , 1-—, 1- — , 40.,
^he product of these will give a series of the same form as that
vithin the parenthesis. Now, the required factors must reduce
/he second member to zero when the first member is zero, that is,
nrhen x=0 qx^=^±w^\ therefore the general form of the factor is
1 ;
a
and since this must be equal to zero when x=±n7r, we have
a ,
whence a=n'7r«, ;« ^ , - o'
x«
therefore the general factor is 1 — --; ^ . '
=• n«7ra * -■ ■■' - '1 :,<.;
then making n=l, 2, 3, &c., in succession, (348) becomes
«nx=x(l-— )(1-— )(1-— ).... "(349)
Again, the factors of the series for cos x, Art. 210, must also be
of the form
-. ■' 1--; :, - , ' •
• a
IT
but the first member is zero for «=±(2n+l) — , where n is any
iteger including zero ; therefore we have '
308
PLANE TRIGONOMETRY.
l_<>:±}ll!=^
22a
whence
(2n+l)^7r«
2^
22x2
therefore the general factor is 1 - 72~TTTa^ '
and making n=0, 1, 2, &c. , in succession, we have
22x2
2^x2,
22x2,
COS x=(l- — ) (1- — ) (l-gi;;:,) ....
^50)
Logarithmic sines and cosines.
244. The last two series furnish us with the means of calcu-
lating the logarithmic sines and cosines without first computing the
natural sines and cosines. . -. . j. . .
In (349) and (350) put x=m—, then
m'x
•V C08m-=(l--)(l-^)(l-g2)....
mK
and expressing these by logarithms we have
log sm m^=log ^+log m+log (l- ^) +log (l- ^; +*c-i
logco»m-=log (1-^J +log (l-^)+log (l-g^)+*c-
Developing the second members of these equations by the loga-
rithmic series, and arranging according to the powers of m, we
obtain
WITT «■ , ^ / 1,1 1 t \ ^
logsin— «=log-+logm-m2. J-(^ + ^ + g-2 + *c.)
, M,l 1 1 ^ V
2 ^2* 4* 6* '
• Ac.
'»r:j; Fl ii <:/■■
."..■^■i '^
\'
a
(851)
■i * -"
Examples.
309
log cos -^=
Mfl 1 1 ^ V 'S
1 ua 32 S^* '
M,l 1 1 ^ V
-m*' — I— H — + — +&C. I
^/l 1.1 ^ \
&o.
(352)
By summing the constant numerical series, substituting the
value of the modulus M and giving m different values, the loga-
rithmic sine and cosine of any angle may be easily computed. Of
course 10 must be added to these expressions to give the tabular
logarithmic sine, &c.
1. Prove that
Examples. .
fe-*=2(l + -—){l +-—)....
and
e* -e-«=2«(l +— ) (1 +;;;:— )
2. Shew that sin 20" sin 40° sin 60° sin 80°=- .
lo
3. If 0=cos 0, shew that 0=42° 20' very nearly,
4. Eliminate d between the equations '^^"^
l-a2
a2 cos* 6= — - — , tan a=tan3 — .
3 . 2
jf.'.. -'r
'. i< ■: • . u
'fir.
Ans. sin' a+cos* o=(2a)S.
'H>i\-i)
II
310 PLANE TRIGONOMETRY.
s^>
MISCELLANEOUS EXAMPLES.
1. Shew that cos^ (45° - e)—\ (1 + cosec 2B) sin 2d,
2. If m tan {a~e) sec^ e=n tan H sec^ (a - «), find ft
^n». e=i(a-tan-^ — — tan a),
3. In any triangle prove that
coa--+co8--+cos — =4cos (45°-—) cos (45°-—) cos (45°- -).
a a u ' 4 * 4 * 4
4. From the figure of Art. 121 prove formulce (186) and (187)
geometrically. . , ,
5. In any triangle prove that tan — +tan - =cos — sec — sec — .
u a M A A
6. Solve the equation sin 5x=sin x+cos 3x.
TT
, . Am. «=(2n+l) - .
7. If a, 6, c, d be the sides of a quadrilateral circumscribed about
a circle, shew that the area of the quadrilateral is v'ahcd sin 6, where
26 is the sum of two opposite angles. ,. ,.. ,,,..:
A B C
8. In any triangle shew that sin — +sin 77 > cos — .
2 J 2
9. In the ambiguous case, o, 6 and A being given, if Cj , c^ are
the third sides of the two triangles, shew that the distance between
the centres of the circumscribing circles is ^(c^ - c,) cosec A.
10. Prove that
(1+tan — +sec — ) (1+tan — -sec ^)=2 tan — .
* 2 2* A <o «
11. Shewthatcos33°45'=i{2 + (2-v/2)*}*
MISCELLANEOUS EXAMPLES. 31]
12. If tan {d - a) tan (g+g)= "^ °"^ ° , shew that
' ^ ' 1+2 cos 2a*
e=i cos-i (2 cos2 2a).
13. If two circles whose radii are r, r^ , touch each other exter-
nally, and if d is the angle contained by the two common tangents tc
these circles, shew that sin 0=-illlIiA— - — i .
14. If i\ yT^yTg be the radii of the circles inscribed in the quadri-
laterals AO, BO, CO of the figure of Art. 152, prove that
n , ♦'a . »*8 «
r — Ti r — r^ T — r^ r
9
T
where r is the radius of the inscribed circle.
15. In any triangle prove that
tan2-+tan2--+tan2-
2 2 2 ■ V/ ..;■
=412=
16. Shew that sec 72° - sec 36°=sec 60°.
17. In the ambiguous case, prove that the circles which pass
respectively through the middle points of the sides of the two
t iangles, are equal, and that their common chord is equal to half
the side which is common to the triangles.
18. If the hypothenuse AB of a right-angled triangle be divided
In D by a line which bisects the right angle, prove that
ADiBD: : 1 -tan ^{A - B) : 1+tan ^{A - B).
19. If a+;9+>'=180°, prove that
« '-' 3 y , a+3 p+y a+y
cos — +C0S --+COS — =4 cos — — cos — .— CCS — — ,
2 2 2 4 4 4
20. Prove that sin 5° sin 15° sin 26°. . . .sin 85°=2 K
812 PLANE TRIGONOMETRl.
21. In any triangle prove that
J2 _ g2 g2 _ ^2 ^2 _ 52
sin 2A+——- sin 2B+ sin 2(7=0.
a2 6' c2
22. If ^1 , rj , rg denote the radii of the circles inscribed between
the inscribed circle and the sides containing the angles A, B, C
respectively, prove that
v/ r-ir, + v/ri'i-j + v/r^,=:r.
23. If sin-i e+sin-i -=^ , then <?=--(6 - 2/¥).
2 4 17
24. In any triangle if cos ^, cos B, cos (7 be in arithmetical pro-
gression, s-a, s-b and s-c are in harmonical progression.
a-1 6-1
25. Shew that tan-^ a - tan~* 6=tan~* : — tan~^
a+1 6+1
— + 6] sin - =co3^ -.
Ans. 6=^nr — a, where n is any odd intege*.
a
27. Given tan — =cosec d - sin 0, then cos'0=^(\/ 5 - 1).
28. If the sines of the angles A, B, C oi a, triangle are in arith-
metical progression, shew that
. A . C . A-B . B-G
Bin — - : sin — : : 8in — --— : sm — - — ••
^ A a 44
3/1
29. Given log 2 and log 3, find the logs of -J — and 135.
30. If D is the middle point of the base BC of a triangle ABCf
prove that
• . ^ 1-^ 6 sin -4
sin BAD= •
y(6Hc2H-26ccos^)
31. If the centres of the escribed circles of a triangle be joined,
prove that the distances of the centres of the escribed circles of this
new triangle from the centre of its inscribed circle are
8BBmi{B+G), 8B Bin i{A + C), SiJsin J(^+B),
where B is the radius of the circle circumscribing the ociginal
triangle. ' . •. ^^ •'•...■ ."'<:^ ;;'• .; i-:- 1 r .ft; .>».
MISCELLANEOUS EXAMPLES. 313
32. Through the centre of the circumscribing circle of » tri-
angle ABCi AOD is drawn to meet BG in D, prove that
OD'.BD: CD=coa A : sin 20 : sin 2B.
33. In any triangle prove that cot A= ; •
abc
2ww w' — ti' ff
34. Prove that sin-^ + sin-^ =— •
m^+M* m^ + n^ 2
35. If p, a, r be the perpendiculars from the angles of a triangle
ABC, upon the opposite sides a, h, c respectively, shew that
a sin A+h sin B+c sin 0=2(p cos A+q cos jB+r cob 0).
36. Shew that
cos (9 - i cos 20+^ cos Sd - &c. , ad inf. =loge {2 cos -- j •
37. If cos" j3 tan (a+d)=6m^ (3 coi (a - 6), find 0.
J.?i«. tan &=\/tan (a+/3) tan (a-fi),
38. If logo 6=m and logj, a=n, shew that ^" = I— •
logft w \n
39. Prove that
cos {^+x>/^)= — ;=-{(e* +e-*)-v/^(e« -e"*)} •
V4 ^ 2^2^ :<1<i
40. Shew that log {x+y\/^)~^ log (x'+i/O+tan-i — v/^.
41. If c be the hypothenuse of a right-angled triangle whose
sides are a and &, prove that
log»=iaog2+loga+log6)+M { (?^)%i(^)Vi(t^)";.. }
42. Solve the equation cos"^ — — 4- sec"^ «=^ •
^n». x=^^(v/3~9-v^).
36
43. In any triangle prove that
i „) , sin^ ^ + sin2 iJ + sin^ C— 2 COB ^ cos -B cos (7=2.
314 PLANE TRIGONOMETRY.
44 If the tangents of the angles of a triangle are in a geometrical
progression whose ratio is n, prove that
sin 2C=n sin 2A.
45. The shadows of two vertical walls at right angles to each
other, whose height are a and b feet, are observed at noon to be
c and d feet respectively in breadth; shew that if a be the sun's
altitude and j3 the inclination of the first wall to the meridian,
d^\ ad
i/c" a*\ , ^ aa
46. Sum to n tenns cos as-fcos 2x4- cob 3x-H ....
. nx
sin —
. 2 (n+l)x
Arts. J— ■ cos — —^— .
. X 2
sin —
2
47. Given the perimeter=2«^ the area= A > and the angle ^ of a
triangle, find a.
a* - A cot —
Aim, q — I
*-■
48. Given the area= A > the angle C and a+h=my find the sides
of the triangle.
Ans. a=|(m-f K m'"^— 8A cosec (7).
49. Given JB, r and p the perpendicular from G on c, to solve
the triangle.
. , . ^v »' sec fl , ^ . ^v (i> - >*) sec
^«.. cos 4(^+B)=;7= . 00. i(^-5)=^JL^^.
where sin 0=
-Jf-
60. Given the perimeter=25, C and jp the perpendicular from G
ou c, to solve the triangle.
V G
Ans. c=s cos^ 6, where tan* 0=_- cot — .
51. If a )3, y be the distances between the centres of the escribed
MISCELLANEOUS EXAMPLES. 315
circles, and a', (i', y' the distances between the centre of the inscribed
circle and the centres of the escribed circles, prove that
• =— , and that
a^Y s
, , , ri-r A r^-r B r.-r C
a : /T : y : : cos — : — ; — cos — : oos — .
' ' a 2 6 2 c 2
52. If z and »' be the meridional zenith distances of the moon or
a planet as Been from two observatories on the same meridian, and
whose latitudes are <t> and cp' respectively, one being N and the other
S, r the earth's radius, D the distance of the moon or planet and P
the horizontal parallax, shew that
sin «+ sin «*
(«+«') -(0 + 0')'
(2 + 2') -(0 + 0')
sin «+sin «/
63. If a quadrilateral can be inscribed in a circle and can also
have a circle described about it, the area of the quadrilateral is
equal to the square root of the product of the four sides.
54. Shew that, if Q and G be the centre of the circumscribed
circle and the intersection of the perpendiculars of a triangle,
QG'^=B^ {1-8 cos A ooa B ooa G). ^r
55. If a, Py y, d be the angles of a quadrilateral, prove that
cos o+cos ;9+cos y+cos rf=4 cos i(a+/3) cos K/^+y) cos ^(a+y).
56. Shew that
(1 +Bec 2x) (1 +sec 4x) (1 +sec Sas) .... (1+sec 2*» a5)=tan 2**« cot x,
57. If tan-i (»+ 1) v^ - tan-i -^^=cot-i 4 v^^ find x,
V 2
. . Ana. 6 or - 2.
58. In any triangle if a ^, 6 2, c^ be in arithmetical progression,
shew that
Bin3.B / aa-c' \a
aia B V 2oc /
59. Prove that 2» cos 6 cos 26 cos 2« cos 2"
=cos + cos 3d + cos 50 + + cos (2'»+» - 1)^.
316 PLANE TRIGONOMETRY.
/
60. If
tan A sec -4 f tan B sec 2?-ftan C sec (?-|-2 tan A tan B tan (^=0,
prove that
sec^ A-\-9fyfi^ B + sec* 0=1 ±2 sec A sec B sec 0.
61. Straight lines -40, £0, CO are drawn bisecting the sides
BGy AGf AB of a triangle in the points D, E, F respectively ; if
»*i> *'a> ^8 *^6 the radii of the circles circumscribed about the tri-
angles EOF, FDD, DOE, shew that
OA^.r]^OB^.j>^_OC^^Tl^l a2+5«+c2
o2 6^ c2 ' 3 ■ a3 6'-' c2 '
a ^ a "^ a
y ^ y
1 a 3
a
62. If cos 6= ■■■ ^ r-=i , shew that
v/a2 + 6a
f(a + 6v/-l)+ f (a-6v/ri)=2«/(a3+&a)oos|-.
A« « .-/... sin ^ sin 20 sin 3(9
63. Sum to infinity — - —
23 2*
Ans.
6 d '
9 tan — hcot —
2 2
64. If -4, 5, 0, D are the angles of a quadrilateral, shew that
tan -4-ftan B+tan O+tan D
cot ^-l-cot jB + cot + cot D
:tan A tan B tan G tan D.
65. If a and /? be the roots of the equation jb*-|>x+9=0, prove
that
tan-^ a+tan-^ )3=tan-^ - — .
1-q
66. Find 6 from the equation
"■ 8in(30+a)+cos (30-o)=cos(45°+0)
■ * ' Ans. 0=45° and sin 20=j(^ cosec (45'*-f-a) - 1) •
TT
67. If i>=rA } shew that .
COB fH-COS 3^i-COS 5^+ +COS 11^=— . 'f ' I'r
MISCELLANEOUS EXAMPLES. 317
68. Prove the formulfe,
^ , /TT , 0v . cos 3/9 8in*30 sin 4/?
l±8m 0=2 8m2 (— ± —) , sin2 6 +coa' —-—=—- — .
4 ^ o «> 4
69. If sin (x+y\/ ~ l)=/3(co8 a+sin a\/- 1), shew that
tana= oot X and B^=k(e''y+e-^ -2 cog 2x).
70. The area of a regular polygon circumscribed about a circle is
a harmonic mean between the area of an inscribed regular polygon
of the same number of sides and a circumscribed polygon of half
the number of sides.
71. If 3 sin 0=sin (a—B), shew that
•0=-- sin a - - • — sin 2a+- • -- sin 3a -&o.
72. Shew that
e-* cos - i e-^ cos 30+^ e"^ cos 5(9 - ad inf. =h tan"^ ,
e* — e-*
73. Prove that loga m=loga b . logs c . logo d logj m.
74. Shew that
/ e 27r + Att + Bs/ e 2n + B 4:Tr+B\ '
(tan — +tan — - — +tan — — W cot — +cot — - — +cot — — — j=9.
75. A common tangent is drawn to two circles which touch
each other and whose radii are r and 3r ; shew that the area of the
curvilinear triangle bounded by the common tangent and the two
circles is
n
6
(4v/3--7r)r^. :, s, m: ^,.
76. Shew that sin* {x+y)+cos^( «- i/)=l+sin 2x sin 2y.
77. Given tan (tt cot 6)=cot (tt tan 6), find B.
Ans. tan 0=^(2n+l± v/4n(n+ 1) - 15).
78. Shew tlMit when n is an odd integer,
coan (—-«?)=( -1) " sinwc. r;> e/r .
318 PLANE TRiaONOMETRY.
79. Shew that
(X * ^ / * * \*
sin - +008 — ) = (sin --- - coi - ) .
80. In a right-angled triangle, given a+b- c=m and the angle A^
shew that ^~~7^ ^^^ 2 ^^^^^ \^° ""2 /*
81. K straight lines be drawn from any point in the circum-
ference of a circle to the angular points of an inscribed regular
polygon of n sides, shew that the sum of the squares of these lines
=2nx (radius)*.
3 y/T + l
82. Solve the equation x^ — -7iX= — 7:^=—»
2 4^2
Ans. >/¥ cos 12% - y/2 cos 48% - v^ cos 72*.
83. A circle is inscribed in an equilateral triangle ; an equilateral
triangle in the circle ; a circle again in the latter triangle, and so on :
if r, fi, rj, r,, &c., be the radii of the circles, prove that
r=ri+r,+rg+&c., ad inf.
84. The radii of two circles which intersect are r, r^ , and a the
distance between their centres ; prove that the common chord
=— I (r+ri+a) (r+Vi-a) (r-ri-\-a) (-r+r.+a) } .
a ^ ' '
A C 4- & A
85. Prove that in any triangle tan (B+—) = — - tan — .
^ 2' c-o 2
86. Find the value of cos-^ J( - 1)*", m being any integer, and
express the different values by one formula.
Ans. {6n±i(3-(-l)-)}|.
87. If p be the perpendicular from the angle A oi & triangle
upon the opposite side, shew that
he a sin J. 4-6 sin ^+c sin (7
«= — . ^
2 6c cos A+ac cos B+ah cos C
88. The straight lines which bisect the angles A and ^ of a tri-
MISCELLANEOUS EXAMPLES. 319
angle ABC, meet the opposite aides in D and E respectively ; shew
that the area of the triangle CED is
A B
A sin — sin -- sec \{C-A) sec |(C-5),
where A is the area of the triangle ABC.
89. In any triangle prove that
g' cos JjjB-C) b^ cos i^{C-A) c" cos iU-B) _
COB ^{B+C) COS i{C+A) COB ^{A+B) ^ o+o -t-oc;.
90. If the straight lines which bisect the angles A, B, C of a,
triangle ABC meet the opposite sides in D,E, F respectively, shew
that the area of the triangle DEF is
„, . A , B . C B-C C-A A-B
2 A sm — - sm — sm — sec — - — sec — - — sec — - — ,
« ^ 2 '' 2 Jt 2
where A is the area of the triangle ABC.
91. If Pii Pii Pi denote the perpendiculars drawn from the centre
of the circumscribed circle of a triangle to the sides a, 6, c respec-
tively, shew that
. /a 6 c \ ahe
4(-+-+-)= ,
^Pl Pi Pt' PiPiPz
and A:(p^ ^-p^ +p\)=a^ cot* A+h"^ cot* B+<i^ cot* C. ' ' ' '
92. Shew by means of a trigonometrical formula that if
a+6+c=a6c, then
2a 2& 2c 2a 26 2c
■;. V!
l-a* l-62^1-c2 l-o* l-fca l-c**
93. Solve the equation
1+tan X tan --).
,. 2' ' /■ ♦
IT
Ans. af=(2n+l) — .
4
94. If X be the length of the line which bisects the angle ^ of a
triangle, and is terminated by the base, ^ the angle it makes with
the base, shew that
(a+6+c) (sin ^- sin — )
2 sm ^ cos —
320 PLANE TRIGONOMETRY.
95. In the Figure of Art. 153, shew that the area of the triangle
0,0,0, is
abc
~2
r .1 . iv c ,1 iv J? . ,1 iv A\
{ {-+t) tan -+(-4--) tan -4-(t + -) *»" 77 r
96. In the Figure of Art. 153, join 00^ and shew that the area
- , . _ . ahc G
of the triangle 00,0- is ; — cot — -.
97. Shew that -=- tan ^^ + - tan - + -, tan ^4 + ----
98. If a, 6, c, d denote the sides of a quadrilateral, and ^ the
sum of two opposite angles, shew that
(area)''=(3 - a) (« - h) (a - c) (a - d) - ahcd cos* — .
where 2s=a+b+' d.
99. If P denote the point of intersection of the perpendiculars
from the angles of a triangle on the opposite sides, prove that
P^2=42ja_a3,
100. Shew that the area of the triangle formed by joining the
points of contact of the inscribed circle, or an escribed circle of a
P A
triangle is -~^r- , where Re is the radius of the circle.
101. If a, 6, c, d be the sides of a trapezium taken in order, and
a the acute angle between the diagonals, shew that the area of the
trapezium is :^{(a'^+c'^)~(6''+d^)} tan a, and examine this result
when a=— .
2
102. If p, q, r be the perpendiculars drawn from the angles
A, B, C to the opposite sides, shew that
1 1
— +—
q r
= 6.
1 1
— + —
p r
1 1
— + —
p q
1 1
6^
1 1
— + -
a c
-' 1 1
a
103. If a and h be the opposite and parallel sides of a trapezium,
a being the greater, and 6 and the acute angles at the extremity
of a, shew that the area is ^(a* - 6^) sin 6 sin ^ cosec {6+<t>)'
MISCELLANEOUS EXAMPLES. 321
1111 7r»
104. Shew that p+^+^+^a+^c., ad inf.=-.
105. In any triangle prove that
A =sin — sm — sm — ( -;; — - 4- -; — r- + -; — t: I *
2 2 2Vsin^ sin 2i sin /
106. If a and /? be two different values of B which satisfy the
cos ^ sin ^ 1 , ,
equation 1 ; — = — , shew that
a c
a cos ^ (a+(3)=h sin ^ (a+/9)=:c cos ^ (a - /3).
107. The angles .4, 5, (7 of a triangle are in descending order of
magnitude; if another triangle be constructed having two of its
angles (A - B), {B - C) and sides m, w, p, then will an-\-cm=bp.
108. In any right-angled triangle, G being the right angle, prove
that a' cos A-\-h^ cos B=abc.
109. Three circles are so inscribed in a triangle that each touches
the other two and two sides of the triangle ; prove that the radius
of that lide^ which touches the sides ^£, J. (7, is <r'*^«^«
// B. , G x\
[(l+tan -)(l+tan-)
X
r
1+tan — •
where r is the radius of the inscribed circle of the triangle.
110. If all the angular points of a regular polygon be joined, and
r be the radius of the circumscribed circle, prove that the sum of
TT
all the lines including the perimeter of the polygon is nr cot — .
2n
111. If jB, r be the radii of the circumscribed and inscribed
circles of a regular polygon of n sides, and R', r' the corresponding
radii for a regular polygon of 2n sides, and having the same peri-
meter as the former, shew that iJ+r=2r' and Rr'=R'^.
112. Solve the equation cos 7^+7 cos 6/=0.
, Ans. 0=(2n+l)^.
92
322 PLANE TRIGONOMETRY.
113. In any triangle ABC, if the lines which bisect the angles
A, B, G and terminate in the opposite sides be denoted by a, /?, y
respectively, prove that
aj3y ^a 6'^a c'^6 e'
and -^^A
o&c^--- ' (a+6)(6+c)(a+c)
114. Sum to infinity
1 1 1 i _1 1 £. _* ■
(33-l)32--(32-l)32+-(3'-l)3'»-&a
> 3 5
115. In any triangle prove thalf
1 2 «i
116. In any triangle, the square of the distance between the
centre of the inscribed circle and the intersection of the perpen-
dicularsis 4Ba-2Ur .
a+b+e
117. If h and h be the diagonals of a quadrilateral and ^ their
angle of intersection, shew that the area is ^hk sin <p.
, V^+1 )-^^: ,1 IT .
• 118. Shew that tan-i —=- tan-^— ==-r-
v/2-1 >/2 4
119. Eliminate 6 from the equations < .
, ,. r 3a cos + a cos 35=4m
3a sin ^ - a sin 30=4*i ^'^
• i Result, a^ - m^ - n'^=3a^ m^ n*
120. Solve the equation tan-^ ( — -) - tan-^ ( — r)=-:r . ' " •
V3-I'
121. Prove that e+tan-» (cot 2d)=tm-^ (cot ff). '
MISCELLANEOUS EXAMPLES. 323
122. In any triangle prove that
A B C ^ r
C082 — +C0S8 — +cos« - -1 *+-r-»
2 2 2 OB
123. Sum to w terms
t»n-i «H-tan-i , , ^ +tan-^ - — T-r-+&c.
l+1.2aja 1+2.3x2
Ans. tan~* rue.
124. Shew that sin^ 6 cos^ 0=- sin 0- - sin 50+- sin 3d.
8 16 16
126. Shew that sin 9°=| {Vz + y/b -Vb^ y/~d] - .
126. Given sin 2(a+0)+sin 2a=2 sin 26, find d.
Ans. 0=tan~^ (3 tan a) - a.
127. In any triangle prove that
A ^ B 0,11.
acoss — +6 cos* — +c cos' — =A It:"' — I*
2 2 2 ^i( r'
128. In the figure of jLrt. 152, jgijove that - V
OA.OB.OC^.^F+BD4^)=^4.}^mT.BD.CE.
129. In the figure of Art. 153, prove that if JBj, Uj, iZ, denote
the radii of the circles described about the triangles OiBC, O^ACy
O^AB respectively, BiB^B^—2R^r,
130. A circle whose radius is r is inscribed in a quadrilateral ;
if tift^fta, ft denote the tangents into which the sides of the quadri-
lateral are divided at the points of contact, prove that .
ya j—1 . ^—1 . J— 1 . ^—1
131. The line joining the tops of two towers of unequal height
makes an angle a with the horizontal plane on which they stand,
and the distance between the extremities of their shadows when the
sun is in the same vertical plane as the towers is h ; if /3 be the sun's
altitude, shew that the distance between them is
h cos a sin fi coscc (a - /?).
324 PLANE TRIGONOMETRY.
132. If a, p, y denote the perpendiculars drawn from the centre
of the circumscribing circle of a triangle to its sides, shew that the
radius of this circle is the positive value of li in the equation
123 _ („2 4.^2 +y2) jB _ 20/37=0.
Shew also that ^+f^+^=- , where a, 6, c are the sides of
ao oc ac 4
the triangle.
133. In any triangle shew that •
tan2 —+tan2 — +tan'» "^ > ^»
(a+6) cos (7+(a+c) coa P+(&+c) cos A > 2xany side.
134. Find d from the equation
6 , e
— = (cos -
a
AvLit. fan /45''4— W± „
V2
5cos2 ~ = (cos -- v/2 sin -) - (v/2-1),
2 ^ is '•>
d 3
Ans. tan (45°+-) =±
135. In any triangle prove that
& - o cos a* - 6^*
JR=
2 cos A sin 2c sin {A-B)
136. If a, &, c be in arithmetical precession, the arithmetical
mean between the logs of a and c is
137. I'ind a and /3 from the equations
2tani(a4-/8)=-J-t— ^,
cos a+cos j8
l-v/2 » ,
2tan*(a-i8)— -. "'
,_ -^V cos a+cos /3
.' ^ , . . Ans. 0=80", i8=:45°.
1 . , 1 T
138. Shew that 2 sin-* — =+ am"^ -7==,=T •
V^IO v50 *
MISCELLANEOUS EXAMPLES, 825
139. Sum the series
1 o 1 « 1 . , . .
l+coe x+— cos 2x+ cos 3ae+— — -r— cos 4x+ . . . .ad inf.
2 1.2.3 1.2.3.4
Ans. e*^* cos (sin x).
140. If a, /?, y be the distances of the centre of the inscribed
circle of a triangle from the angles, and if a~^+Y~^=2^^, then
**!> *'a> ^1 3^® "^ arithmetical progression.
TT 11111
141. Shew that -z^=l+-^ --=---=- + -77 + 7::- &o.
2\/2 3 5 7 9 11
i42. If the bisectors of the angles of a triangle ABC be produced
to meet the circumference of the circumscribed circle in the points
D, E, F, and if p, q, t denote the sides of the triangle DEFj prove
par
that R'^= — , where It is the radius of the circumscribed circle.
a+h + c
143. li a, Pf-y be the three values of x (unequal) which satisfy
the condition
a b - I
+ _ +c=_o ;: ; > ■■ . vi
cos X sm X ;,
tana tan^(/3 + y) -"i "■■■
shew that = ;, — rr* ,, . ,.
tan 7 tan^(a + jS) m' '
144. If a circle be inscribed in a triangle and the points of con-
tact A', B', G' joined, and if JSi, iJj, i?, be the radii of the circum-
scribed circles of the triangles AB'G'y BA'G'f GA'B' respectively,
prove that
4 B ■
jBj : JB, : jB,=cosec — : cosec — : cosec — .
a ^ S
145. If ABG be a triangle, and the equation
x^+'i/+z'^+2yz cos 2A+2xz cos 2B+2xy cos 20=0
be satisfied by real values of a, y, z, then
V y z
' I
ain 2A sin 2B sin 2(7
326 PLANE TRIGONOMETRY.
146. In any triangle prove that
sin 2 A {h cob G-c cos By +anal. +
- • . =2 cos A cos B cos G {a^ tan A +anal. + ).
a
147. Adapt cot — =a(l+cos o+cos /?+cos y) to logarithmic com-
putation by the use of an auxiliary angle.
Ans. cot — =4a cos — cos — (a+<^) cos —{a - <p)
cos — (/34-y)cos-(^-y)
6 a £
where cos — = — — — ^— ^— •
2 a
148. If Pi, Pa, jPs be the perpendiculars from any point within a
triangle upon the sides, Pj, P^, P^ the perpendiculars from the
angles upon the same sides respectively, shew that
149. Prove that the sum of '1;h««QjJ|res of the ten lines joining
the centres of the five circles, four of which touch the sides of the
triangle, and the remaining one is the circumscribed circle, is equal
to 15 times the square of the diameter of the circumscribed circle.
150. If perpendiculars be drawn from the bisections of the sides
of a triangle to the circumference of the circumscribed circle, shew
that the sum of these perpendiculars is equal to 272 - r.
151. In problem 132, if a=3i, /3=4i, >'==4i, find B. Ans. 8^.
152. In any triangle shew that
, . ^ ab ' ^ he ' "" ac ' 2
153. Three circles whose radii are r^, r,, r, touch ©ne another,
Oj, Oa, 0, being their centres and A the point of intersection of
their common tangents at tlie points of contact ; if a^ , a,, a, denote
MISCELLANEOUS EXAMPLES. 327
the distances AOi, AO^, AO^ respectively, and R the radius of the
circle circumscribing the triangle O^O^O^, prove that
154. A circle is inscribed in a triangle, and any three triangles
are cut off by tangents to the circle ; if r^ , rj , r^ be the radii of the
circles inscribed in these three triangles, then the sum of the areas
of the triangles is
■^ ( - ^i+'Ti+r^) +-^ (»*i - rj+rj +- {r^+r^ - r,), . :
where a, b, c are the sides of the original triangle.
155. Sum the series ad infinitum •
• cos d cos^ 6 cos' 6
cos 0+ — - — cos 20+— -— cos 80+— —r cos 40+....
Ans. e cos (0+| sin 2/9).
156. If a, /?, Y be the lengths of the lines which join the feet of
th« perpendiculars from the angles of a triangle on the opposite
sides, shew that *^:
a P y _ a'^-]-h'^+c^ ''
^"'"i^'^c^" 2ahc * V?
where a, h, c are the lengths of the sides opposite to the lines a, /3, y
respectively. '''^
167. In any triangle if a, h, c, be in arithmetical progression,
shew that r^, r^, r^, the radii of the three escribed circles which
touch the sides a, b, c respectively, are in harmpnical progression.
f".
■\
328
TLAJfE TRIGONOMETRY.
APPENDIX.
Geometrical Demonstrations of (45) and (46).
I. We will here add a few geometrical demonstrations of the
very important formulae of Arts. 45-5^. >
I. — When (x+y) is an aiHgle in the second quadrant
In Fig. 1 let the angle BAG=x and the angle CAD=y. Con-
struct the Figure as in Art. 45. ; ,
sin {x+y)=
PM
.•>'k '■•;;,('
AP
JQN PK
~AP'^AP
ii'"'
QN AQ PK PQ^
^AQ AP PQ AP
=sin X cos y+Gos x sin y.
coa(x+i/)=— 2p
AN KQ
■/■i 'j'l^'i , {I'/fi,.--,!-
'AP AP
AN AQ KQ, PQ
AQ AP PQ AP
r=:cos X cos y - sin x sin y.
APPENDIX.
329
II, — When (x+y)is an angle in the third quadraaU,
In Fig. 2 let the angle BAC=x
and the angle CAD=y.. Construct
the Figure as in the preceding dia-
gram, then we have
sin (x+i/)=-
PM
AF
QN PK
~" AP AP
~ AQ ' AP PQ ' AP'
=: - sin QAN cos PAQ - cos QPK sin PAQ
= - sin (180" - x) cos (180° -y)- cos (180" - x) sin (180° - y)
= sin X cos y+coa x sin y. (Art. 39.)
cos (ic+j/)=-
AM
AP
AN KQ
AP AP
AN AQ^_^ PQ_ ' ' ^ !:
~AQ ' AP PQ' AP
=cos QAN cos PAQ- sin QPK sin PAQ
=co8 (180°- x) cos (180°- y) -sin (180°- x) sin (180°- y)
=co8 X cos y - sin x sin y. (Art. 39. )
The student will observe that in the quadrilateral AMPQ, the
opposite angles QAM, QPM are together equal to two right angles,
and since the angle QAM=x, the angle QPK=1S0° - x.
, f^i ■;■;■;■»'
330
PLANE TRIGONOMETRY.
III. — When (x+y) is an angle in the fourth quadrant.
In Fig. 3 let the angle BAC=*:x
and the angle CAD=y. Construct
the Figure as in the two preceding
diagrams, then* we have
sin (x+y)= -
PM
AP
QN PK
-n' • *
COS {x+y)=
AP AP
QN AQ^ PK PQ^
AQ ' AP PQ' AP
sin QAN cos PAQ +cos QPK sin PAQ
sin (360° -x) cos y+cos (360° -x) sin y
sin X cos y+coB x sin y.
AM
AP
AN NM
'AP AP
AN AQ KQ PQ
AQ AP PQ AP
=cos QAN cos PAQ+&m QPK sin PAQ
=cos (360° - x) cos y+sin (360° - x) sin y
=co8 X cos y - sin x sin y.
Here the opposite angles QAM, QPM of the quadrilateral
AMPQ, are together equal to two right angles ; therefore the
exterior angle QPK is equal to the angle QAN, but the angle
g^i^=360° - X, hence the angle QPE:=360° - x.
APPENDIX.
331
2. The geometrical proof of sin {x - y) and cos (x-y) in each of
the quadrants is similar to those of sin (x+y) and cos (x+y) which
we have just given. We will, however, give the proof of one case,
viz. , when x is an angle in the fourth quadrant and y an angle in
the first quadrant, x-y being, in th Figure, an angle in the third
quadrant.
In Fig. 4 let the angle BAC=x
and the angle DAC=y. Construct
the Figure as in Art. 48, then wo
have
PM
6m(x-y)=- —
NQ KQ
AP
NQ
AP
AQ_
AP
KQ PQ
AQ AP PQ AP
= - sin QAN cos PAQ - cos QAN sin PAQ
= - sin (360° - x) cos y - cos (360° - ac) sin y
= sin a cos y - cos x sin y.
coa{x-y)=~
AM
AP
,
PK-AN
! ■
AP
AN PK
AP AP
*
AN AQ
PK PQ
AQ AP
PQ AP
= cos QAN cos PAQ - sin PQK sin PA Q
= cos (360° -- x) cos y - sin (360° -x) siay
= cos X cos y+sin x sin y,
since the angle PQK is equal to the angle QAN.
332
PLANE TRIGONOME'J'RY.
i<UMBERS OFTEN USED IN CALCULATIONS.
Ratio of the circumference of ) „ lAimnfsnf
a circle to its diameter. . . \ =^=3.1415926536
v/7 =1.7724538509
:r»=9. 8696044011
Napierian base =e=2. 7182818285
Modulus of common logs =Af=. 4342944819
log« TT =1.1447298858
Unit of circular measure =w'=57°. 2957795130
" **• " =3437'. 7467708
" " " =206264". 80624
sin 1" =.0000048481
sin 2" =.0000096963
sin 3" =.0000145444
Mean diameter of the earth =7912 miles "
Equatorial radius of the earth =20923599.98 feet
Polar radius of the earth =20853657. 16 feet
English mile =5280 feet
Geographical or nautical mile .... =6076 feet ,
v/l=1.414214 ^ #""2=1.259921
v^3=1.732051 #" 3"=1.442250
v/I=2.236068 f ¥=1.709976
Logarithms
0.4971499
0.2485749
0.9942997
0.4342945
9.6377843
0.0587030
1.7581226
3.5362739
5.3144251
4.6855749
4.9866049
5.1626961
3.8982863
7.3206364
7.3191823
3.7226339
3.7836163
THE END.
J
.2?
&
i^-hH
^#4^4 ^v-c
i^
"^"^-^^Aulfi ^ /
N ^ ^ \ A, -^ — ^=^ ^ ' ^^ ^ ^ ^
*«?■