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Answer 1 · 2017-04-26T16:39:51

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

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

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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#