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Question
Prove that sin6A + cos6A = 1 – 3sin2A . cos2A.
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Solution
L.H.S. = sin6A + cos6A
= (sin2A)3 + (cos2A)3
= (1 – cos2A)3 + (cos2A)3 ...`[(∵ sin^2A + cos^2A = 1),(∴ 1 - cos^2A = sin^2A)]`
= 1 – 3cos2A + 3(cos2A)2 – (cos2A)3 + cos6A ...[∵ (a – b)3 = a3 – 3a2b + 3ab2 – b3]
= 1 – 3 cos2A (1 – cos2A) – cos6A + cos6A
= 1 – 3 cos2A sin2A
= R.H.S.
∴ sin6A + cos6A = 1 – 3sin2A . cos2A
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