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Prove that sin6θ + cos6θ + 3sin2θ cos2θ = 1

Answer»

Solution : We know that sin2θ + cos2θ = 1

Therefore, (sin2θ + cos2θ)3 = 1

or, (sin2θ)3 + (cos2θ)3 + 3sin2θ cos2θ (sin2θ + cos2θ) = 1

or, sin6θ + cos6θ + 3sin2θ cos2θ = 1



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