Checkout JEE MAINS 2022 Question Paper Analysis : Checkout JEE MAINS 2022 Question Paper Analysis :

If A + B + C = pi, then sin 2A + sin 2B + sin 2C is equal to

If A + B + C = π, then sin 2A + sin 2B + sin 2C is equal to

1) 4 sin A sin B sin C

2) 4 cos A cos B cos C

3) 2 cos A cos B cos C

4) 2 sin A sin B sin C

Answer: (1) 4 sin A sin B sin C

Solution:

Given,

A + B + C = π

2A + 2B + 2C = 2π

sin 2A + sin 2B + sin 2C

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

= 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 sin A sin B)

= 4 sin C sin A sin B

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