49. It may be remarked, that the easiest method of deducing formulæ such as these, is first to express sin 2 A and cos 2A in terms of sin A and cos A. Thus, let it be required to prove that

.. by adding and subtracting, we have

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

sin (A + B) sin (4 - B) = 2 cos A. sin B......(2).

Similarly it may be shewn that,

cos (A + B) + cos (A – B) ·

= 2 cos A. cos B...

(3).

cos (A-B) - cos (A + B) = 2 sin A. sin B..............(4).

51. To find the values of sin A + sin B, and cos A + cos B, in terms of the sines and cosines of (A + B), and 1 (A – B).

... sin A + sin B = 2 sin 1 (A + B). cos ↓ (4 – B).............. (1).

......

sin A - sin B = 2 cos(4+ B). sin (4 – B)......(2).

Similarly,

cos A + cos B = 2 cos 1⁄2 (A + B). cos 1⁄2 (4 – B)..

.....

.(3).

cos B - cos A = 2 sin (A + B). sin (4 – B)......(4).

These four formulæ might have been deduced from the formulæ of the last Article, by making,

in which case,

A+ BS, and AB = D,

A = 1 (S + D), B = 1 (S – D).

52. The formulæ of the last two Articles are necessary to be remembered; the following are not of such frequent

=

=

= (sin 4)2. (cos B) – (cos A)2. (sin B)2

(sin 4). {1 - (sin B)} - {1 - (sin A)2}. (sin B)2

(sin A)2 - (sin B)2.

Similarly, sin (A + B) . sin (A – B) = (cos B)2 - (cos A)2.

In like manner it may be shewn that

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

or,

55.

To prove that

=

(cos B)2 - (sin A)2.

sin n A+ sin (n − 2) A = 2 sin (n - 1) A. cos A,

cos n A + cos (n

... sin n A + sin (n − 2) A = 2 sin (n - 1) A. cos A......

.(1).

Again; in the same manner,

cos n A = cos (n − 1) A . cos A − sin (n − 1) A . sin A,

And, cos (n − 2) A = cos (n − 1) A . cos A + sin (n − 1) A . sin A ;

.. cos n A + cos (n − 2) A = 2 cos (n − 1)A. cos A.................... (2).

and by successive substitutions, sin 4 A, sin 5 A... cos 4 A, cos 5 A... might be found.

56. In like manner,

sin n A − sin (n − 2) A = 2 cos (n − 1) A . sin A...............(1),

cos (n-2) A cos n A= 2 sin (n - 1) A. sin A......(2).

57. From the latter of these formulæ we may, by means of successive substitutions, find the cosine of nA in terms of the sines of A and its multiples.

SO

Suppose n to be even, and = 2m.

... cos 2 (m − 1)A - cos 2m A = 2 sin (2m - 1) A. sin A,

cos 2 (m-2) A- cos 2 (m1) A = 2 sin (2m - 3) A. sin A,

cos 2 (m3) A cos 2 (m2) A = 2 sin (2m - 5) A. sin A,

cos 2 (m −m) A – cos 2 {m − (m − 1)} A = 2 sin A . sin A,

.. by addition we have,

(since cos 2 (m— m) A, or cos 0o, = 1,)

If n is odd, and = 2m+1, we should have got,

sin2m A+ sin2 (m-1)A+

cos (2m+1) A= cos A-2 sin A..

(1).

cos 3 A

= cos A - 2 sin A. sin 2 A = cos A - 2 sin A. 2 sin A. cos A

= cos 4-4 cos 4. {1-(cos A)} = 4 (cos A)3 - 3 cos A.