Question

Prove by only manipulating one sin(^4)x-cos (^4)x=2sin (^2)x-1

Answer 1

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.