Another way of solving identities with fourth powers of `sinx` and `cos x` is to use the identities `sin^2x = 1-cos^2 x` and `cos^2 x = 1-sin^2 x`.
You usually want to start a trig identity from the side that looks more complicated, in this case the LHS.
`LS=sin^4 x-cos^4 x`
`=sin^2 x( 1-cos^2 x) - cos^2 x(1-sin^2 x)`
`=sin^2x - sin^2x cos^2 x -cos^2 x + sin^2xcos^2x` some terms cancel out!
`=sin^2 x - cos^2x` now get rid of the cosine
`=sin^2x - (1-sin^2x)`
`=sin^2x-1+sin^2x` collect like terms
`=2sin^2x - 1`
The identity `sin^4 x - cos^4 x = 2*sin^2x - 1` has to be proved
`sin^4x - cos^4x`
=> `(sin^2x - cos^2x)(sin^2x + cos^2x)`
=> `(sin^2x - (1 - sin^2x))*1`
=> `2*sin^2x - 1`
This proves that `sin^4 x - cos^4 x = 2*sin^2x - 1`