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Question
If A + B + C = `pi/2`, prove the following sin 2A + sin 2B + sin 2C = 4 cos A cos B cos C
Solution
L.H.S = (sin 2A + sin 2B) + sin 2C
= 2 sin(A + B) cos(A – B) + 2 sin C cos C
= 2 sin(90° – C) cos(A – B) + 2 sin C cos C
= 2 cos C [cos (A – B) + sin C] + cos(A + B) .....(∴ A + B = `π/2` – C)
= 2 cos C [cos(A – B) + cos(A + B)]
= 2 cos C [2 cos A cos B]
= 4 cos A cos B cos C
= R.H.S
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