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Question

Sin6 A+cos6A=1-3Sin2Acos2A

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Solution

Consider LHS:
sin6A+cos6A = (sin2A)3 + (cos2A)3

= (sin2A + cos2A)3 − 3 sin2A cos2A(sin2A + cos2A) [Since, (a+b)3 = a3 + b3 +3ab(a+b)]

= (1)3 − 3 sin2A cos2A(1) [Since, sin2A + cos2A = 1]
= 1 − 3 sin2A cos2A
= RHS

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