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Prove that sin^6θ + cos^6θ = 1 – 3 sin^2θ. cos^2θ. - Geometry

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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

Concept: Trigonometric Identities
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