How do you prove the identity #(sin^4)x-(cos^4)x#?

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Answer 1 · 2015-09-03T22:15:53

You probably mean this one:

#sin^4x-cos^4x = -cos 2x#

#sin^4x-cos^4x#

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

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

...using the difference of squares identity:

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

Now #sin^2x+cos^2x = 1#, so

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

#= sin^2x-cos^2x#

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

#= -cos 2x#

...using the double angle formula for #cos#