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प्रश्न
Prove that: sin6θ + cos6θ = 1 - 3sin2θ cos2θ.
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उत्तर
sin6θ + cos6θ
= (sin2θ)3 + (cos2θ)3
= (sin2θ + cos2θ)(sin4θ + cos2θ - sin2θcos2θ)
= 1 x [(sin2θ)2 + (cos2θ)2 + 2sin2θ.cos2θ - 3sin2θ.cos2θ]
= (sin2θ)2 + (cos2θ)2 - 3sin2θ.cos2θ
= 1 - 3sin2θ cos2θ.
= RHS
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