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Prove the identity (sin^4)x - (cos^4)x = (sin^2)x - (cos^2)x
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\[\sin ^{4}x - \cos ^{4}x = \sin ^{2}x - \cos ^{2}x\]
apply difference of squares to the left once, and look for an identity
sin^4x-cos^4x=(sin^2x+cos^2x )(sin^2x-cos^2x ).....(2) (sin^2x+cos^2x )=1.................(1) hence from 1 and 2 1((sin^2x-cos^2x )
why use a double-angle formula? shivams' two-step answer is fine
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