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.