Question #62e00

1 Answer
Apr 26, 2017

sin^4x-cos^4x=(sin^2x-cos^2x)(sin^2x+cos^2x)

=sin^2x-cos^2x=sin^2x - (1-sin^2x)=2sin^2-1

Explanation:

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First you need to identify that this is the difference of two squares.

sin^4x-cos^4x=(sin^2x-cos^2x)(sin^2x+cos^2x)

Now we use the identity sin^2x+cos^2x=1 and in the rearranged form cos^2x=1-sin^2x

(sin^2x-cos^2x)(sin^2x+cos^2x)=(sin^2x-cos^2x)*1

sin^2x-(1-sin^2x)=2sin^2-1