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Question
If A + B + C = `pi/2`, prove the following cos 2A + cos 2B + cos 2C = 1 + 4 sin A sin B sin C
Solution
L.H.S = (cos 2A + cos 2B) + cos 2C
= 2 cos(A + B) cos(A – B) + 1 – 2 sin2C
= 1 + 2 sin C(cos(A – B) – 2 sin2C)
∴ cos(A + B) = cos(90° – C) = sin C
= 1 + 2 sin C [cos(A – B) – sin C]
= 1 + 2 sin C [cos(A – B) – cos(A + B)]
= 1 + 2 sin C [2 sin A sin B]
= 1 + 4 sin A sin B sin C
= R.H.S
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