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Question
If A + B + C = 180°, prove that sin2A + sin2B + sin2C = 2 + 2 cos A cos B cos C
Solution
L.H.S = `(1 - cos2"A")/2 + (1 - cos2"B")/2 + (1 - cos 2"C")/2`
Hint: `[sin^2"A" = (1 - cos2"A")/2]`
= `3/2 - 1/2[cos2"A" + cos2"B" + cos2"C"]`
= `3/2 - 1/2 [2cos("A" + "B") cos("A" - "B") + 2cos^2"C" - 1]`
= `3/2 - cos("A" + "B") cos("A" - "B") - cos^2"C" + 1/2`
= 2 + cos C cos(A – B) – cos2
= 2 + cosC[cos(A – B)(cos(A + B)]
[cos(180° – C) – cos C – cos C]
= 2 + cos C [cos(A – B) + cos(A + B)]
= 2+ cos C[2 cos A cos B]
= 2 + 2 cos A cos B cos C
= R.H.S
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