Factor a difference of squares:
\(\sin^4x-\cos^4x=(\sin^2x-\cos^2x)(\sin^2x+\cos^2x)\)
This reduces to
\(\sin^2x-\cos^2x\)
due to the Pythagorean identity. By the same identity, you have
\(\sin^2x-\cos^2x=\sin^2x-(1-\sin^2x)=2\sin^2x-1\)
and you're done.
