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Question
In ΔABC, A + B + C = π show that
cos2A +cos2B – cos2C = 1 – 2 sin A sin B cos C
Solution
We know that, cos2θ = `(1 + cos2theta)/2`
L.H.S. = cos2A + cos2B – cos2C
= `(1 + cos2"A")/2 + (1 + cos2"B")/2 - cos^2"C"`
= `1/2[2 + (cos2"A" + cos 2"B")] - cos^2"C"`
= `1/2[2 + 2*cos((2"A" + 2"B")/2)*cos((2"A" - 2"B")/2)] - cos^2"C"`
= 1 + cos (A + B) . cos(A – B) – cos2C
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. = 1 – cos C . cos(A – B) – cos2C ...[From (i)]
= 1 – cos C . [cos(A – B) + cos C]
= 1 – cos C . [cos(A – B) – cos(A + B)] ......[From (i)]
= 1 – cos C . (2 sin A sin B)
= 1 – 2 sin A sin B cos C
= R.H.S.
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