Verify the identity sin^4x-cos^4x=2sin^2x-1

We'll shift (sin x)^4 to the
right:

- (cos x)^4 = -(sin x)^4 + 2(sin x)^2 -
1

We'll multiply by -1:

(cos
x)^4 = (sin x)^4 - 2(sin x)^2 + 1

We recognize in the
expression from the right side a perfect square:

(cos x)^4
= [(sin x)^2 - 1]^2 = {-[1 - (sin x)^2]}^2 = [1 - (sin
x)^2]^2

But, from Pythagorean identity, we'll
get:

(sin x)^2 + (cos x)^2 = 1 => (cos x)^2 = 1 -
(sin x)^2

We'll raise to square both
sides:

(cos x)^4 = [1 - (sin
x)^2]^2

We notice that we,ve get LHS = RHS,
therefore the given identity (sin x)^4- (cos x)^4 = 2(sin x)^2 - 1 is
true.