Prove the identity sin^2 x-cos^2 x=sin^4 x- cos^4 x?

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For the beginning, we'll manipulate the right side, only.

[(sin x)^2]^2 - [(cos x)^2]^2 = [(sin x)^2 - (cos x)^2][(sin x)^2 + (cos x)^2]

But, (sin x)^2 + (cos x)^2 = 1 (Pythagorean identity)

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

We notice that we've get the difference of squares from the left side:

**[(sin x)^2 - (cos x)]^2 = [(sin x)^2 - (cos x)]^2**

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