Question

sin^4x-cos^4x= ?

Answer 1

Maybe do it as a diff. of two squares first, a^2-b^2 = (a+b)(a-b)
\[\Large \sin^4x-\cos^4x = (\sin^2x +\cos^2x)(\sin^2x - \cos^2x)\]
then use the identity sin^2x+cos^2x =1, so the left set of brackets is just 1... so we have \[\Large \sin^2x - \cos^2x\]
I don't know where to go from here, because you didn't post any other information on what to do...