Question

Prove that i) sin 2A = 2 sin A COSA
ii) cos 2A = cos²A-sin^2A​

Answer 1

Step-by-step explanation:

Example

If A, B, C are angles of a triangle, prove that

     sin 2A +sin 2C = 4 sin A sin B sin C

Solution

Since A, B, C are angles of a triangle, we have A +B +C =  = 180°.

L.H.S. = (sin 2A +sin 2B) +sin 2C

= 2 sin [(2A +2B)/2] cos [(2A -2B)/2] + sin 2C

= 2 sin (A +B) cos (A -B) + 2 sin C cos C

= 2 sin ( -C) cos (A -B) + 2 sin C cos ( -(A +B))

= 2 sin C cos (A -B) -2 sin C cos (A +B)

= 2 sin C [cos (A -B) -cos (A +B)]

= 2 sin C (2 sin A sin B) = 4 sin A sin B sin C = R.H.S.

Answer 2

Step-by-step explanation:

sin(a+a)=sinacosa+cosasina

sin(a+b)=sinacosb+cosasinb

sin2a=2sinacosa

2)cos(a+a)=cosacosa-sina sina

=cos^2a-sin^2a

cos(a+b)=cosacosb-sinasinb