Q4 of 23 Page 1

Solve Question:

Prove that sin6θ + cos6 θ = 1-3 sin2 θ. Cos2 θ

sin6θ + cos6 θ


= (sin2θ)3 + (cos2 θ)3


Using a3 + b3 = (a + b)3 – 3ab (a + b), we write,


= (sin2θ + cos2 θ)3 - 3sin2θcos2θ (sin2θ + cos2 θ)


Using (sin2θ + cos2 θ = 1), we get,


= 1 -3sin2θcos2θ


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