How do you factor the expression and use the fundamental identities to simplify #sin^4x-cos^4x#?

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Answer 1 · 2017-02-15T13:42:24

#=(sinx-cosx)(sinx+cosx)=-cos2x#

Since

#a^2-b^2=(a-b)(a+b)#,

it is:

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

#=(sinx-cosx)(sinx+cosx)(sin^2x+cos^2x)#

On the other hand, it's:

#(sin^2x+cos^2x)=1#,

then it's

#=(sinx-cosx)(sinx+cosx)=sin^2x-cos^2x#

or

#-cos(2x)#