how would you prove sin4x - cos4x = sin2x-cos2x as well as sin4x + 2sin2xcos2x +cos4x = 1
:confused:

In both cases, you'll want to factor the left side of the equation. Treat sin^4 x - cos^4 x as a difference of two squares and treat sin^4 x + 2sin^2x cos^2x + cos^4x as a perfect-square trinomial. Also the 1st. Pythagorean Identity will come into play ; sin^2x + cos^2x = 1.

I think you have written the question incorrectly.

These identities are not correct.

I think you should have written:

1. (Sinx)^4 - (Cos x)^4

= [ (sinx)^2 - (cos)^2 ] [ (sinx)^2 + (Cos x)^2) ]

= (sinx)^2 - (cos)^2

2. sin^4 x + 2 sin^2 co^2 x + cos^4 x

= (sin^2 x + cos^2 x)^2

= 1^2

= 1