Prove that $\displaystyle \cos^4x- \sin^4x\equiv\cos^2x-sin^2x $ I have the trig rules but don't know what to do with them. Any help would be much appreciated
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Originally Posted by greghunter Prove that $\displaystyle \cos^4x- \sin^4x\equiv\cos^2x-sin^2x $ $\displaystyle a^4 - b^4=(a^2+b^2)(a^2-b^2)$
Originally Posted by Plato $\displaystyle a^4 - b^4=(a^2+b^2)(a^2-b^2)$ Thanks, that's really helpful
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