Q. 4 B5.0( 6 Votes )

# <span lang="EN-US

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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