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CHAPTER I

ELEMENTARY TRIGONOMETRY

A knowledge of trigonometrical functions is essential for the proper understanding of various formulae of the differential and integral calculus. The present chapter is therefore devoted to the development of the elementary functions and their properties. The account is short, and for the purpose of studying the functions generally recourse should be had to a recognized textbook. The chapter has been included only with the object of enabling those who have not studied trigonometry to obtain sufficient knowledge to follow the remainder of the book.

Definitions.

Consider a straight line

₁

of indefinite length fixed in a plane. At a point

in the straight line is hinged another straight

line

OA,

also of indefinite length, capable of being revolved about

the hinge at

but only in an anti-clockwise direction. Then, as

revolves, it sweeps out an angle

XOA.

Amp

P_-A

X,U

N X

X, N

A'P

Fig.

Take any point P on the moving line

and drop a perpen

on to the fixed line

₁

OX.

Then, by the properties of similar triangles, the ratios between the sides of the right-angled triangle

PON

will be the same for all positions of P, for any one

2 ELEMENTARY TRIGONOMETRY

position of the line OA. These constant ratios are the trigonoratios of the angle XOA, and are defined thus, where a stands for the angle XOA :

P is the sine of the angle XOA and is written sin a, ^ON
O „ cosine XOA cos a,
ON tangent XOA „ 9)

tan a.
These are the principal ratios, and most trigonometrical problems can be solved by the use of these three ratios only. It is often convenient, however, to use the reciprocals of the ratios: the respective reciprocals are

OP the cosecant of the angle XOA written as cosec a,
r, the secant „ „ XOA, „ sec a,

NN, the cotangent XOA, „ „ cot a.

3. It is important to note that, even if the two triangles PON in the first two diagrams of Fig. I are geometrically equal, it does not follow that the trigonometrical ratios of the two angles given by the positions of OP are the same. In elementary plane geometry, the straight line joining any two points L and M may be indifferently denoted by LM or ML. On the other hand, the straight lines which enter into the definitions of the trigonometrical ratios have sign as well as magnitude, and the direction of the straight line determines the sign. To ascertain the correct sign to be given to a straight line, we proceed in the following manner. Imagine the plane in which the fixed line has been drawn to be divided into four sections by the straight line X₁OX and a straight line YO Y_l through 0 perpento X₁OX.
If OP be any position of the revolving line, we can arrive at the point P from 0 either by proceeding along OP or by the double journey ON, NP. In order to develop a logical system we must adopt a convention based on the direction to be taken to arrive at

DEFINITIONS

P from 0, and on the particular quadrant in which the point P lies. The convention is that lines drawn from 0 in the directions OX or 0 Y are positive and that those drawn from 0 in the directions

p A
Fig. 2.

OX₁ and OY₁ are negative. The line OA is called the radius and is always to be considered as positive. The perpendicular line NP is to be regarded as drawn in the direction N— P, i.e. from the line OX to the radius OA, and not from P to N.

In Fig. 2, therefore, we have

1.Quadrant XOY: ON positive and NP positive;
2.X₁OY: ON negative and NP positive;

₁

O Y

₁

negative and

negative ;

4.„XO Y₁: ON positive and NP negative.

These quadrants are called the first, second, third and fourth quadrants respectively.

It is evident that the trigonometrical ratios, being derived from

the ratios between ON,

NP, OP,

will have sign as well as magni

X(i)(ii)
Fig. 3.

tude. For example, for the angle a in Fig. 3 (i) all the sides of the triangle ONP are positive in direction and as a result all the trigoratios of the angle will be positive.

4 ELEMENTARY TRIGONOMETRY
On the other hand, in Fig. 3 (ii) we shall have
(positive)/(positive) (negative)/(positive) (positive)/(negative)

Negative angles.

If the revolving line be constrained to move in

clockwise

direction, it is said to trace out a negative angle. For example, let

the straight line OA, take up the position indicated, not by a

Fig. 4.

revolution passing first through the position

OA,

but by passing in

the opposite direction direct to OA, ; then the angle

XOA,

is a negative angle.

In the figure the angle

XOA

and the angle

XOA

₁

is —

Relations between the ratios.

From the definitions of the ratios we have at once

tan

= NP/ON = OP

_sin

cos a

sin

= NP/OP

cos

ON/OP

tan

= NP/ON

and similarly for the reciprocal ratios.

i.e.

positive,

i.e.

negative,

i.e.

negative,

(i).

Similarly,

cos a cot a = sin a

(ii).

Again, from any of the diagrams in Fig. r,

+ ON' = OP'.

(NP/OP)

(ON/OP)' --

(sin a)

(cos a)

IDENTITIES 5

A more convenient method of writing (sin a)², (cos a)², etc. is by omitting the brackets and denoting the squares of the ratios by sin^e a, cos' a, etc. The above relation is therefore

sin^e a + cos' a = I (iii).
Similarly, by dividing both sides of the identity
ON²+NP²=OP² by ON² we shall have

i + (NP/ON)² = (OP/ON)²

or I + tan^g a = sec^t a (iv).
Again, dividing by NP², we shall obtain
I + cot' a = cosec² a (v).

7. Identities.

Just as algebraic identities can be proved by the application of various fundamental rules, so the relations between the trigonoratios can be applied to the proof of trigonometrical identities.

Example 1.
Prove that tan a + cot a = sec a cosec a.
tan a + COt a = sin a/C0S a + COS a/Sin a
= (sin^g a + cos' a)/C0S a sin a
I/COS a sin a
= (I/C0S a) (I/sin a)
= sec a cosec a.
Example 2.
Prove that sec^t a — cosec² a = tan^g a — cot^e a.
Example 3.

Prove that tan a — tan + tan a tan /3 = o.
cot a—cot/3

Multiply through by cot a — cot /3 and the expression becomes tana—tan/3 + cot a tan a tan /3 — tanatan/3 cot /3 or    tan a — tan /3 + tan /3 — tan a, which is zero.

6        ELEMENTARY TRIGONOMETRY Alternatively :
I    I    tan a — tan
cot a — cots=    —    -
tan a tans    tan a tan s
tan a — tan
    = — tanatans.
' ' cot a — cot s
8. Magnitude of angles. Degrees.

The unit angle in elementary geometry is the degree. An angle of x degrees is denoted by x°. The degree is defined as the angle subtended at the centre of a circle by an arc equal in length to I/36o of the circumference. For arithmetical calculation the degree is a convenient unit, and we can obtain the values of the trigonoratios of many angles by reference to simple geometrical figures.

Example 4.
Find the sine, cosine and tangent of (i) 45°, (ii) 3o°, (iii) 6o°.

(i) Let ONP be an isosceles triangle, right-A

angled at

so that

ON = NP.

Then, if

be of unit length,

=ON

+NP

=I+I=2,

so that

= 1/2.

Therefore, easily,

sin 45

⁰

1/'

cos 45

⁰

I/1/z,

tan 45°

(ii)Take the angle XOA to be 30^°. Then the angle NPO is 6o° and

the figure

ONP

is one-half of an equitriangle of side equal in length to

OP.

If, therefore,

be of unit length,

and

ON =

1/3, so that

sin 30°

= Z,

cos 30° =

^1/

tan

⁰

I/1/3.

(iii) From a consideration of the above figure it is evident that sin 6o°

= 1/3/2,

cos 6o° = -, tan 6o° =

1/3.

9. Magnitude of angles. Radians.

A more convenient unit for analytical purposes is the angle subtended at the centre of a circle by an arc equal in length to the radius : this angle is called a radian. Since the ratio between the

PERIODICITY OF THE RATIOS 7

angle at the centre and the arc on which it stands is constant for all circles, it follows that the radian is the same whatever the radius of the circle : the radian may therefore be taken as a unit of measurement.

To obtain the number of radians corresponding to the number of degrees in an angle, all that is necessary is to multiply the number
of degrees by _180.This is easily seen to be so, for if x be the number of degrees corresponding to a radian, we have angle subtended by the arc equal in length to the radius _ radian

	angle subtended by half the circumference	18o°
i.e. so that	r/7rr = x/18o, x = 18o/7r, or IT radians = 18o°.

In applying the calculus to trigonometrical functions it is essential that angles should be expressed in terms of an absolute unit of measurement. Consequently, in all the work that follows, unless otherwise stated, angles must be taken to be measured in radians.
10. Periodicity of the trigonometrical ratios.
If we consider the definitions of the ratios, taking into account the signs as well as the magnitudes, it can easily be shown that there will be more than one angle having the same particular ratio. To take a simple example : in the following figure, let the radius

A    4
    X,    1N₁    0    N Fig. g.
take up the positions OA and OA₁, where the angle XOA is the angle a and the angle XOA₁ is the supplement of XOA, i.e. 7r — a.
Then, attending to the directions of the lines involved, we shall have
sin a = NP/OP = N₁P₁/OP₁ = + sin (7r — a),
cos a = ON/OP = — ON₁/OP₁ = — cos (TT — a),
    and    tan a = NP/ON = N₁P₁/— ON₁= — tan (7T — a).

8    ELEMENTARY TRIGONOMETRY
Again, from the figure it can be shown similarly that sin a = — sin (7r + a) = — sin (27r — a), cos a = — cos (7r + a) = + cos (27r — a), tan a = + tan (7r + a) = — tan (27T — a).
Fig. 6.

If now the radius make a complete revolution, so that, starting from the position OX it takes up the position OA after first tracing out the angle 27r, then it is evident that sin a = sin (27r + a) ; cos a = cos (27r + a) ; tan a = tan (27r + a).

We have, therefore, that
sin a = sin (7r — a) = sin (27r + a) = sin (37r — a) =
cos a = cos (27r — a) = cos (27r + a) = cos (47r — a) =
tan a = tan (7r + a) = tan (27r + a) = tan (37r + a) =      These relations may be generalised in the forms :
all angles having the same sine as a are the values of nrr + (— 1)ⁿ a,
cosine    „ 2n7r ± a,
tangent    „    n7r + a,
where n is a positive integer.

For example, it has been proved above that sin 3o° = In absolute measure this is sin 7r/6 = 2_i so that all angles whose sine is ,. are the successive values of {n7r + (— I)ⁿ 7r/6}, i.e. 7r/6, 57r/6, 137r/6, 177r/6, and so on.
It will be seen that if we replace n by 2m, so that only even values of the positive integers are taken into account, the general angle for sin a is zm7r + a; this brings the property of the sine into line with those of the other ratios. For all the trigonometrical functions, therefore, we may say that

f (x + 2m7r) = f (x).

GRAPHICAL REPRESENTATION OF THE RATIOS 9
If a function have this property it is said to be a periodic function with period arr.
A graphical representation (Fig. 7) shows quite clearly the periodic property of the sine, cosine and tangent.
4
3

MUMMA
1IMIXIMEMIMMMMMMISMOEWAINIPMWAMI
6-3
-1

-3

(i) Curve of sines

4(ii) » +~ cosines

(iii) ++ ~+ tangents

Fig. 7.Notes: (i) The tangent and cotangent are periodic with period 7r.

(ii) It can be shown quite easily by the consideration of a diagram similar to Fig. 6 that the generalised forms hold equally for negative integral values of n.11. Ratios of (Pr ± a).(i) If in Fig. 8 the angles XOP and XOP₁ are a and 17 — a respectively, and we make OP₁= OP, then by considering the geometry

of the two triangles ONP and ON₁P₁

it is easily seen thatON/OP = N₁P₁/OP₁, NP/OP = ON₁/OP₁, NP/ON = ON₁/N₁P₁;so that cos a = sin (n1T — a), sin a = cos (27r — a), tan a = cot (7r — a).

-2

I Kli,f    ~~1111=~II
I. Ii        1    --

10    ELEMENTARY TRIGONOMETRY
Similarly, cot a = tan (27r — a) and cosec a = sec (17r — a).

The angles a and 27r — a are called complementary angles, the " co " in cosine, cotangent and cosecant corresponding to the comangle.

(ii)If the angle XOP, = 17r + a as in Fig. 9

N, 0N X Fig. 9.

we shall have

ON/OP = N

₁

P,/OP,,

.•, cos

= sin

('17r

a) ;

also

NP/OP = — ON

₁

/OP, ,

.•. sin a = — cos

(27r +

a) ;

andtan

= — cot

(27r +

a).

(iii) When the angle

XOP

is so small that

and

coincide,

sin o = o, cos o

tan o

= O;

and from the above

sin 1

₂

:77'

= cos o

cos

17r

= sin o = o, tan

27r

= oo.

12. Inverse functions.

From the identity sin 7r/6 = 2 we can obtain the inverse relation, namely, that 7r/6 is the angle whose sine is 1. The notation adopted for this is

sin-' = 7r/6.

This inverse notation is not to be confused with the algebraic notation for negative indices. Although a-¹ is equivalent to I/a, sin-' x is not 1/sin x, but the angle whose sine is x. We have generally from the above, that if sin a = x, then

sin^-' x = n7r + (— I)ⁿ a.

As a general rule it is convenient to take the inverse function as the numerically smallest angle (with the proper sign) giving the required value of the direct function.

PROJECTION II
Example 5.
Write down (i) the smallest positive angle, (ii) the general formula for the angle x, given that x = cos-' o + cos' ^-1/3/2 + cos' -I/1/2.

cos 17T =	0,	cos^-I o = ₂¹-7r;
cos 7r/6 =	1/3/2,	cos-¹ 1/3/2 =	7r/6;
cos 7r/4 = I/"V 2,		cos^-1 I/"V 2 =	7r/4.

Therefore x = 7r + 7r/6 + 7r/4 = I17r/I2, which is the smallest posiangle. The general angle is 2n7r + II702.
13. Projection.
If from the extremities of a straight line AB perpendiculars be dropped on to another straight line LM, produced if necessary,
B

Fig. io.

the part intercepted on LM by the feet of the perpendiculars is called the projection of AB on LM.

In Fig. Io A₁B₁ is the projection of AB and B₁A₁ is the proof BA. If AN be drawn through A parallel to LM, then A₁B₁ = AN. Call the angle NAB /3; then AN/AB = cos /3, so that

A₁B₁ = AN = AB cos P.

In other words, the projection of the line AB on the line LM is AB cos /3, where /3 is the angle between the lines AB and LM, both produced if necessary. As in para. 3, the lines are supposed to have signs according to the direction in which they are drawn: thus BA = — AB and so on.

The following proposition is important.
The sum of the projections of the sides of a triangle XYZ, taken in order, on any straight line in the same plane, is zero.

I2 ELEMENTARY TRIGONOMETRY

The projection of XY is LM,

YZ is MN,
ZXisNL,
so that the sum of the projections of XY, YZ, ZX is

LM+MN+NL=o.

L N M
Fig. ii.

As a corollary to this, we have at once that the sum of the projections of XY and YZ = the projection of XZ. For, denoting the projection of XY by (XY), etc.,

(XY)+(YZ)+(ZX)=o,
i.e. (X Y) + (YZ) = — (ZX) = (XZ).

It is easily seen that if ABCD...K be any closed figure, the sum of the projections of the sides AB, BC, CD, ... taken in order on any straight line in the same plane is zero.

14. The addition theorems.
Let the revolving line sweep out the angle a by taking up the position OA, and subsequently the angle _Pby the new position OB.
Y

(I)
(2)
Fig. I2.

Drop a perpendicular PN from any point P in OB on to OA. Project the sides of the triangle ONP on to OX.

THE ADDITION THEOREMS 13

(1)(OP) = (ON) + (NP)

= ON cos

a+NPcos(27r+a)

cos

a — NP

sin

OP cos (a +

OP cos /3 cos a — OP sin /3 sin a.

(2)

(OP)

(ON)

(NP)

=ONcos(7r+a)+NPcos(27r+a)

ON (—

cos

a) — NP

sin

OP cos (a + [3) = OP cos (7r — /3) (— cos a) — OP sin (7r — /3) sin a = OP cos /3 cos a — OP sin /3 sin a.

Therefore in both cases we have that

cos

+ /3) = cos

cos /3 — sin

sin [3

(vi).
By changing the sign of

[3,

cos (a — /3) = cos a cos [3 + sin a sin /3 (vii),
since it can be shown easily from the relations in para. to that cos (— /3) = cos [3, and that sin (— /3) = — sin [3.

Again, by changing a to

in (vi),

cos(;7r+a+/3)=cos(27r+

a) cos /3—

sin(j7r+a)sin/3,

i.e.sin

[3)

sin

cos [3 + cos

sin /3

(viii),
and, by writing — /3 for

_[3

in (viii),

sin

(a —

[3)

sin

cos /3 — cos

sin /3

(ix).
The corresponding formulae for the tangents of the compound angles may be obtained thus :

sin (a + _/3) sin a cos _/3 + cos a sin /3

tan

^(a+=

cos

(a+/3) cos a cos

3 — sin a sin

sin

cos

sin

_—

cosacos/3

⁺

cosacos/3

—

tana+tan

(

cos

sin

1 —

tan

tan f

cos

_f3

cos

cos /3

and similarly tan

(a —

tan

—

tan

(xi).

+ tan

atan~

Note. We have proved the addition theorems for the following ranges of angles: (t) a + /3 < 27r; (2) a < z7r and 17 < /3 < 7r The method of projection can be applied in a similar manner to prove the theorems for angles of any magnitude.

14 ELEMENTARY TRIGONOMETRY

15.Sum and Difference formulae.

We have

sin

(a+/3)=

sin

acos/3+cosasin[3,

sin

(a —

3) = sin

cos /3 — cos

sin /3;

therefore, by addition,

sin (a + /3) + sin (a — /3) = z sin a cos /3.

Let a + /3 = y and a — /3 =8.

Then sin y + sin 8 = z sin

(y +

8) cos (y — 8)

(xii).
By subtraction, we have, similarly,

sin y — sin 8 = 2 cos

₂

(y +

8) sin

₂

(y —

(xiii).

From formulae (vi) and (vii) it can be shown in the same manner

that

cos y + cos 8 =

COS

₂

(y +

8) cos (y — 8)

(xiv),

cosy—cos8=zsin2(y+8)sin2(8—y)

= — 2 sin (y + 8) sin

₂

(y —

(xv).

These formulae can be proved by projection on the same lines

as those adopted for the proofs of the addition formulae.

16.Double angles and half angles.

From formula (vi) we have, by putting /3 = a,

cos 2a = cos

a —

sin

⁼

cc,

or, since cos

+ sin

cos

= 2 cos

⁼

a —

I =

I —

2 sin

(xvi).
Again, from the formula for sin

+ /3), putting

/3,

sin 2a = 2 sin

cos

(xvii).
The tangent formula (x) gives

tan

= 2 tan

a/(1 —

tan

(xviii).

By replacing 2a by a convenient formulae in terms of half angles

can at once be obtained, thus :

sin a = 2 sin ~a cos 2a,

= cos

2a + sin

2a;

ILLUSTRATIVE EXAMPLES    15
2sin2acosla .'. sina=
cos² 2a + sin² 2 sin la cos la
Similarly    cos a =
and, by division,    tan a =
17. Examples.

Some examples illustrative of the use of the above formulae for the proving of identities and for the solution of trigonometrical equations are given below.

Example 6.
sin 5a + Sin a sin 3a — sin a
sin 5a + Sin a 2 sin ₂(5a + a) COS ₂(5a — a) Sin 3a — Sin a 2 COS ₂(3a + a) sin ₂(3a — a) 2 sin 3a COS 2a sin 3a
2 COS 2a sin a    sin a
But sin 3a = sin (2a + a) = sin 2a cos a + cos 2a sin a

2 sin a COS²a + (I — 2 sin²a) sin a

= 2 sin a (I    sin²a) + (I — 2 Sin²a) Sin a

3 Sill a — 4 sin³a.
sin3a/sina=3 -4sin²a=3+2(I -2Sin²a)—2

= I + 2 CoS 2a.
Example 7.
If    cos a+cos/3+cosy+cosacos/3cosy=o,

prove that    tan la tan _z/3tan 2y = f I.
_TI— tan² 2a
low    cos a =
I + tan^ga
Let    tan la = a, tan 2/3 = b, tan 2y = c.
Prove that
I + 2 COS 2a.

16    ELEMENTARY TRIGONOMETRY
Then the condition that
cosa+cos_N + cosy+ cosacos_NC0Sy=o is the same condition as
I — a² I — b² 1 — c² (I — a²) (I — b²) (1 — c²)
I + a² + I + b² + I + c2 + (^I+ ^a2) (r +—b2) (^I+ c²) — o.
By a simple algebraic transformation this becomes
    4 (1 — ^a2b2c2)
(I +^a2)(^I+ b²) (1 +      _
c²) — ^o_'i.e.    — a²b²c² = o,
a²b²c²= r
or    abc = ± r;
i.e.    tan la tan _2N8 tan ly = f I,
which proves the proposition.
Example 8.
Prove that, if a + + y = 7r, then
r —cosa+cosP+cosy_tanla I — cosP+cosy+cosa tan 1/3 '
Ifa+P+y= IT, then Za=17r- z 09+7), S0 that
sin 2a=cos1(fl+y) and cos Za=sin 2 (f3+y).

(I — COS a) + (COS N + COS y)	2 sin^g la + 2 COS ₂(P +7) C0S 1	(3 -7)
(I + cos a) + (cos y — cos P)	2 cos² la — 2 Sin ₂(P + y) sin	(y — _/3)

Substituting for sin la and cos a as above, and dividing through by son    +. y) , i.e. by cot #    + y) or tan 2a, we obtain
cot ( + Y) co    + y)    cos z (_ ^—_Y)
05' + ^y) + sin 1 (f — Y)
2 cos ly cos _IN tan la

= tan g 2 cos Iy si    =
n 1₁3 tan 1⁶Example 9.
Solve the equation sin 9X + sin 5X + 2 sin²x = r.

sin 9X + sin 5X = I — 2 Sin²x

= cos 2X, i.e.    2 sin 7X cos 2X = cos 2X;
therefore, either    cos 2x = o     (a),
or    sin 7x = i     (b).

    ILLUSTRATIVE EXAMPLES    17
From (a)    2X = 1¹7x,
or    x = 7r/4 and generally x = (2n ± 2) a,
and from (b)    7X = a/6,
or    x = 7r/42    „    x = 4 [n + (— I)ⁿ ] rr.
Example 10.
Express the function A cos y + B sin y in terms of a single function of a single angle.
Let A = r cos 8, and B = r sin 6.
Then, since cos²8 + sin²8 = I, we have A² + B²= r²,
and, by division,    tan 8 = B/A,
i.e.    8 = tan-¹ (B/A).
We may write, therefore,
Acosy+Bsiny=rcosycos6+rsinysin8=rcos(y—8), where    r = + V/A² + B².
Similarly    A cos y — B sin y = r cos (y + 8).
Example 11.
To expand cos nx in an ascending series of powers of cos x and sin x, where n is a positive integer.
Now, cos 2x = cos²x — sin²x,
Cos 3X = 4 cos³x — 3 COS X

= cos³x + 3 cos³x — 3 COS X = COs³x — 3 COS X (I — cos²x)	2
= COS³x — 3 COS X sin²x = cos³x —	cos x sin²x,
	2.

C0S 4X = cOs²2x — sin²2x
= cos⁴x — 2 cos²x sin²x + Sin⁴x — 4 sin²x cos²x = cos⁴x — 6 cos²x sin²x + sin⁴x

= cos⁴x — 423 cos²x sin²x + 4=3'2 Sin⁴x. 2!    4^!This suggests the general form
cos nx = cos^' x — n₂ COS^H-2x sin²x + n₄ cos^n-4x sin⁴x — ...
+ (— I)m n2m COS^H-2"nx sin^2mx + ... ,
where nr stands for ^I₍n (n — 1) ... (n — r + 1). r.
Similarly, by expressing sin zx, sin 3X and sin 4X in terms of powers of cos x and sin x, the general form for sin nx would appear to be sin nx = n₁ cos^n-1x sin x — n₃ cos^n-3x sin³x + ...
+ (— ^I)^m-1n2m-1 COS^H-(2m-1)x sin^2m-1x + .... 2

18    ELEMENTARY TRIGONOMETRY
Assume these two formulae true for the positive integral value n. Then
cos (n + 1) x = cos nx cos x — sin nx sin x
= cos x (cosⁿx — n₂ coS^n—2x sin'x + ...)
— sin x (91₁ cos^n—1x sin x — n₃ cos^n—3x sin³x + ...), where the coefficient of cos^n—2m+1x sin'^mx is (— 1)m (ⁿ2m + 912m_1), which
iS(—I)m(n    ¹)2m^.
Similarly for sin (n + 1) x.

If, therefore, the series are true for n they are true for n + 1. But they are true for 2, 3, 4, ... ; therefore they are true for 5, 6, ... and for any positive integer.

EXAMPLES 1

1. Write down the sine, cosine and tangent of the following angles: 150°, ¹35⁰, 75⁰°, 210°.

2.Express the angles in the above question in radian measure.

3.Explain carefully why the following relations are impossible: (a) sin A = 1^.2 ;(b) sin^e A = 2 — cos'- A; (c) sin A = •8 and cos A = •7; (d) tan A = •8 and sec A = •q; (e) sin A = •5, cos A = .4 and tan A = •6.

4.Give in radians the smallest positive angle satisfying the equations :

(a)

sin

= ;

(b)

sin

Ix =

1/3/2;

I ;

(d)

cosec x =

1/2;

(e) cos

5.Determine cosec _/3: (a) sec 8 = 8;(b) cos /3 = •1o8;(c) tan _/3 = •501.

6.Prove the identities :

sin

⁴

a — sin

a = cos

⁴

a — cos' a;cos' a — sin

= 2

cos'

a — I = I — 2

sin

a — cos' a =

(I +

sin a cos

(sin a — cos

a);

sin

cos

= tan y /(

tan

y) ;

sine

_/3

tan

+ sin

= tan

/3 ;

(tan

A —

tan

cos

= sin

cos

B —

cos

sin

cos' /3 (3 —

tan

/3) =

3 —

4 sin

/3 ;

cos' /3

—

sin

y —

1 —

tan

sin

y —

tan

/3 tan

y '

cos

A —

sec

= (cos

A —

sin

tan

A);

(sec /3 + tan

/3)

(cosec /3 — cot (3) = (sec

3 —

(cosec /3 +

I).

EXAMPLES 19

7. Write down the complete solutions of the following equations:

(i) sin x = 1/V 2;    (ii) sec x = 2;
(iii) tan x = — I/1/3 ;    (iv) cos (¹~7r + B) = o;
(v) sin B = cos B.
8. Complete the identities :
(i) sin-1 2 + sin^-1 1^/3/2 =    (ii) cos^-1 o + cos^-1 I =
(iii) tan-¹ A/3 + tan-¹ I/A/3 =
(iv) sin^-1 (— 1) + cos^-1 o =    (v) cos^-1 + 2 sin-1 2 =

(vi)cot^-1 co + tan^—1 I =

cos

^—1

(— + 4 sec

^—1

(—

2/13) _

sin

(—

+ sin-

(—

z) _

sec

^—1

2 +

sec

^—s

(—

2) +

sec

^-1

(—

cosec

^-1

+ cosec

^-1

(—

+ cosec

^-1

(—

I) =

9. Solve the equations :

(i) cos 4X = sin

5x;

(ii) sin

= cos lox;(iii) tan

= cot

(l7r

x);

(iv) cosec

= sec

(37r — 2X);

(v) sin

(n7r — 3X) = cos (zn7r —

4X).

lo. Write down, in terms of ratios of the angle 8 alone,

sin

(37r/2 +

B); cos

(37r/2 —

0);

tan

(57r —

0);

cot

(57r/2 —

0);

cosec

(277'

0);

sec

(777'/2 +

0);

cot (—

0);

sin (—

37r —

0);

cos (—

27r + 0).

11. Show that(i) given A = ^sin—13/5,		cos A = 4/5;
(ii)„	B = cos' ^-12/13,	sin B = 5/13 ;
(iii)„	C = sin^—' 8/17,	cos C = 15/17;
(iv)„	D = tan^—' 2,	sin D = I/1/5.

12. Find the values of
sin (A + B) where A = sin—1 3/5 and cos (A + B) „ A = cos^-1 3/5 sin (A — B) A = sin-¹ 12/13 cos (A — B) ,, A = sin^—s 5/¹3
sin 2A „ A = sin^-1 15/17;

cos 2A ,, A = cot^-1 2 ; tan (A + B) „ A = sec^—1 5/4 tan (A — B) ,, A = sin^—s 3/5

tan 2A „ A = sin^-1 •83 ; cot (A — B) ,, A = cos-¹ •7
B = cos^—1 12/13 ; B = sin ¹8/17; B = sin^—s I/1/5; B = tan^-1 ;

B = cot^-1 12/5 ; B = cosec^-1 A/5^-;

20 ELEMENTARY TRIGONOMETRY
Prove the following identities :

13.sin (A — B) cos B + cos (A — B) sin B = sin A; tan (A + B) — tan A
14.1+tan(A+B)tanA=tan B;

sin3a=3sina—4sin

a;tan 3a

(I —

3 tan

a) = 3 tan a — tan

a;sin 7r/6 + sin 7r/3 = 2 sin 477

COS

77/12 ;sin 3/3 + sin 5/3 = 2 sin

_4/3

cos [3;tan 20 =

(1 —

cos 0)/sin 0;sin(0+477)—cos(0—17r)=o;

21. COS(47r+0)+COS(17T— 6)=

"V2 COS

22.Cos (a + /3) + sin (a — /j) = 2 sin (47r + a) COS (47r + /3);

+ cos 3A + cos 5A + cos 7A = 4 cos

cos

cos 4A;(cos

+ cos 3A) cos 4A = (cos 3A + cos 5A) cos

2A;

25.tan A + tan B = sin (A + B)/cos A cos B ; z6. cos 2P cos N — sin 4P sin /3 = cos 213 cos ₃P;
27.CoS 48 = cos⁴ 8 — 6 cos^t 8 sin^e 8 + sin⁴ 8 ;

(a+b)+(a—b)tan

0=(a+bcos20)sec

cos4A=3+4sin2A—2(cos

A+sinA)

⁴

;

tans(A+B)—tanz(A—B)=2sinB/(cosA+cosB);

(tan

+ tan

sin

(A — B) =

(tan

A —

tan

sin

B) ;

32. (co

A —

cos'

= cos

+ B) cos (A — B).

Solve the equations :

2 sin x + 5 cos

x= 2;

cos

x) —

sin (a + x) = 'V2

^—

;Cos

^—1

x = 477 — Cot

^—1

2 ;

4cosx=2tanx+3secx;

sin

sin 3X = sin

_5x

sin

7x;

tan^—1 ^x + I +tan—1x — ¹ = tan^—1 (— 7);
x—1 x

39.COS X + COS 2X + C0S 3X + COS 4X = 0.

X = tan^—1 + tan^—14 + tan^—1 1, . Express x in the form tan^—1 k.

Prove by projection that

cos

+ cos (27r/3 +

+ cos (z7r/3 —

4z. Plot the curve sin x/cos

from o to

and hence solve approxi

the equation

cos 2X = sin

Show that

2 cos IA = (± y

I +

sin

A +

'V 1 —

sin

A).

36.

38.

EXAMPLES 21

How would you find the cosine of the half-angle, given the sine of the whole angle, from this formula? Explain your answer by obtaining sin 7r/I2.

44.IfA+B+C=rr,prove that E tan A = tan A tan B tan C. Prove also that E sin 2A = 4 sin A sin B sin C.

45.Prove that ztan^-1 {(Q+ b~# tan 2x~ =cos^-1 {a + b cos x

46.Solve the equation 8 sin x + 1/3 sec x = cosec x.

If sin (B + C — A), sin (C + A — B), sin (A + B — C) are in arithmetic progression, prove that tan A, tan B, tan C are also in arithmetic progression.Solve the equation sin-¹ (I — x²)ⁱ+ tan-¹2x = 7r.

49. Obtain a and /3 from the equations sin(a+/3)cos(a—)9)_ , cos (a +

/3)

sin

(a—/3)=I.

50. By multiplying all the way through by 2 sin I8, sum the series

sin

a+sin(a+/3)+sin(a+2P)+...tonterms.

Show that

cos a + cos (a + 27r/n) + cos (a + 47r/n) + ... + cos (a + 2n — In/n) = o.