and

cos (A + B) = cos A cos B – sin A sin B............................(3).

In the figure of Art. 120, draw CA' at right angles to CA.

Then

4 NCA' comp. of ▲ NCA

= L NAC = A,

int. BCA' comp. of ext. 4 C = comp. of (A + B).

=

cos (A – B) = cos A cos B + sin A sin B...............................(4).

In the figure of Art. 121, draw MCA' at right angles to CA. Then LNCA': =comp. of NCA = 4 NAC = A,

and ext. BCM = comp. of int. C

sin A + sin B = 2 sin 1 (A + B) cos § (A – B)................... (5),

sin A - sin B = 2 cos

.......

(A + B) sin 1⁄2 (A – B)................... (6).

Let

R

POC=A; and QOC= B. Take OP=OQ. Join PQ,

cutting OC in C.

Let OR bisect angle POQ, triangle POQ, and base PQ.

Then (Art. 116, I.)

▲ POC+

COQ = 2 ▲ POR and ▲ POC - ▲ COQ = 2 ▲ ROC.

Δ

Δ

L COQ = 24 ROC.

cos B + cos A = 2 cos 1⁄2 (A + B) cos 1⁄2 (A – B)................. (7),

......

OC'R = comp. of

=

In the figure of last article, draw FOC' at right angles to OC.

Then and COQ comp. of QOF,

Also (Art. 116, II.),

▲ QOC' + ▲ POC' = 2 ▲ ROC′ and ▲ QOC′ – À POC'

126. To find the relations between the ratios of an angle and those of its half or double.

A

2A

A

M

R

and

Describe a semicircle QPR, centre O, on diameter QR.
Make RQP = A : and draw PM perpendicular to QR.
Then ▲ ROP at the centre = 2. ▲ RQP at circumference = 2A,
4 QPR in semi-circle a right-angle,

=

QPM = L RQP = A.

QR QP

QM- MR
QM-MR QM

2 sin A. cos A.

QP QR

QR

QR

QM + MR

cot A+ tan A

2

cot A-tan A

=

MP 2MP

OM 20Ꮇ

=

MP 2MP

=

2MP

QM-MR

=

=

2MP

2

1 - cos 24

=

2

QO + OM 1 + cos 2 A
20P

OP OM

MP

OP
OМ

(16) cot A

=

MP

MP

MP

MP

OM

MP

2

=cosec 24-cot 24.

+ =cosec 24+cot 24.

127. To find the tangent and cotangent of the sum or difference

In order to express this fraction in terms of tan A and tan B, we must divide its numerator and denominator by cos A. cos B.

128. another.

The preceding 24 formulæ are not independent of one
It will be useful to show their algebraical inter-

dependence. This is done in Arts. 129–132.

3

to prove

129. Given sin (A + B) = sin A cos B + cos A sin B
cos (A-B) = cos A cos B + sin A sin B

.(1),

.(4).

and

and

= cos (90° - C) = sin C,

sin (A+B) sin (90° C - B) = cos (CB).

.. (1) becomes cos (C – B) = cos C cos B + sin C sin B.

Writing A for C we have (4). Similarly

Given

sin (4

B) = sin A cos B – cos A sin B we can prove cos (A + B) = cos A cos B- sin A sin B

.....(2), .(3).