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Question
Prove that sin6θ + cos6θ = 1 – 3 sin2θ. cos2θ.
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Solution
LHS = sin6θ + cos6θ
= (sin2θ)3 + (cos2θ)3
= (sin2θ + cos2θ) (sin4θ + cos4θ - sin2θ⋅cos2θ)
= (1)[(sin2θ + cos2θ)2 - 2sin2θ⋅cos2θ - sin2θ⋅cos2θ]
= (1)[(1)2 - 3sin2θ⋅cos2θ]
= 1 - 3sin2θ ⋅ cos2θ
= RHS
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