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Question
If A + B + C = 180°, prove that sin2A + sin2B − sin2C = 2 sin A sin B cos C
Solution
LH.S = `(1 - cos2"A")/2 + (1 - cos2"B")/2 - (1 - cos2"C")/2`
Hint: `[sin^2"A" = (1 - cos2"A")/2]`
= `[1/2 + 1/2 - 1/2] - 1/2 [cos2"A" + cos2"B" - cos 2"C"]`
= `1/2 - 1/2 [2cos("A" + "B") cos("A" - "B") - (2cos^2"C" - 1)]`
= `1/2 - cos("A" + "B") cos("A" - "B") + cos^2"C" - 1/2`
= cos C cos(A – B) + cos2C
= cos C [cos(A – B) – cos(A + B)]
= cos C [2 sin A sin B]
= 2 sin A sin B cos C
= R.H.S
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