Question

If A + B + C = 180◦, prove that sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C

Answer 1

LHS = (sin 2A + sin 2B) + sin 2C

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

[sin(A + B) = sin(180° – C) = sin C]

= 2 sin C cos(A – B) + 2 sin C cos C

= 2 sin C [cos(A – B) + cos C]

{cos C = cos[180° – (A + B)] = – cos (A + B)}

= 2 sin C [cos(A – B) – cos(A + B)]

= `2sin"C"{2sin  (2"A")/2 sin  (2"B")/2}`

= 4 sin A sin B sin C

= R.H.S