If A, B, C are angles of a triangle, then sin 2A + sin 2B - sin 2C is equal to

1) 4 sin A cos B cos C

2) 4 cos A

3) 4 sin A cos A

4) 4 cos A cos B sin C

Answer: (4) 4 cos A cos B sin C

Solution:

Given that A, B, C are angles of a triangle.

⇒ A + B + C = π

⇒ 2A + 2B + 2C = 2π….(i)

Now,

sin 2A + sin 2B – sin 2C

Using the formula sin x + sin y = 2 sin(x + y)/2 cos(x – y)/2,

= 2 sin(2A + 2B)/2 cos(2A – 2B)/2 – sin [2π – 2(A + B)] {from (i)}

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

= 2 sin(A + B) cos(A – B) + 2 sin(A + B) cos(A + B) {since sin 2x = 2 sin x cos x}

= 2 sin(A + B) [cos(A – B) + cos(A + B)]

= 2 sin(π – C) (2 cos A cos B)

= 4 cos A cos B sin C