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Question
Prove the following:
sin6A + cos6A = 1 − 3sin2A + 3 sin4A
Solution
L.H.S. = sin6A + cos6A
= (sin2A)3 + (cos2A)3
= (sin2A+ cos2A)3 - 3 sin2A cos2A (sin2A + cos2A)
... [∵ a3 +b3 = (a+b)3 – 3ab(a+b)]
= (1)3 – 3sin2A cos2A (1)
= 1 – 3sin2A cos2A
= 1 – 3sin2A (1 – sin2A)
= 1 – 3sin2A (– 3sin2A)(– sin2A) ...[Multiply each term in the parentheses by – 3sin2A]
= 1 – 3sin2A (– 3sin2A)(– sin2A)
= 1 – 3sin2A + 3sin4A
= R.H.S.
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