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