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Question
In ΔABC, A + B + C = π show that
sin2A + sin2B − sin2C = 2 sin A sin B cos C
Solution
We know that, sin2θ = `(1 - cos2theta)/2`
L.H.S. = sin2 A + sin2 B + sin2 C
= `(1 - cos2"A")/2 + (1 - cos2"B")/2 - sin^2"C"`
= `1/2[2 - (cos2"A" + cos2"B")] - sin^2"C"`
= `1/2 [2 - 2*cos((2"A" + 2"B")/2)*cos((2"A" - 2"B")/2)] - sin^2"C"`
= 1 – cos(A + B) cos(A – B) – sin2C
= (1 – sin2C) – cos(A + B). cos(A – B)
= cos2 C – cos(A + B). cos(A – B)
In ΔABC, A + B + C = π
∴ A + B = π – C
∴ cos(A + B) = cos(π – C)
∴ cos(A + B) = – cos C .....(i)
∴ L.H.S. = cos2 C + cos C. cos(A – B) ...[From (i)]
∴ = cos C[cos C + cos(A – B)]
= cos C[– cos(A + B) + cos(A – B)] ...[From (i)]
= 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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