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Question
Prove that 2(sin6A + cos6A) – 3(sin4A + cos4A) + 1 = 0.
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Solution
sin6A + cos6A = (sin2A)3 + (cos2A)3
= (1 – cos2A)3 + (cos2A)3 ...`[(∵ sin^2A + cos^2A = 1),(∴ 1 - cos^2A = sin^2A)]`
= 1 – 3 cos2A + 3(cos2A)2 – (cos2A)3 + cos6A ...[∵ (a – b)3 = a3 – 3a2b + 3ab2 – b3]
= 1 – 3 cos2A(1 – cos2A) – cos6A + cos6A
= 1 – 3 cos2A sin2A
sin4A + cos4A = (sin2A)2 + (cos2A)2
= (1 – cos2A)2 + (cos2A)2
= 1 – 2 cos2A + (cos2A)2 + (cos2A)2 ...[∵ (a – b)2 = a2 – 2ab + b2]
= 1 – 2 cos2A + 2 cos4A
= 1 – 2 cos2A(1 – cos2A)
= 1 – 2 cos2A sin2A
L.H.S. = 2(sin6A + cos6A) – 3(sin4A + cos4A) + 1
= 2(1 – 3 cos2A sin2A) – 3(1 – 2 cos2A sin2A) + 1
= 2 – 6 cos2A sin2A – 3 + 6 cos2A sin2A + 1
= 0
= R.H.S.
∴ 2(sin6A + cos6A) – 3(sin4A + cos4A) + 1 = 0
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