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

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agneslund | Student, Undergraduate | eNoter

Posted February 25, 2011 at 7:58 PM via web

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

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giorgiana1976 | College Teacher | Valedictorian

Posted February 25, 2011 at 8:02 PM (Answer #1)

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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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