Prove the identity: $ \cos^3x+\sin^3x=(\cos x+\sin x)\cdot(1-\sin x\cdot\cos x) $ I was able to change both sides to $\cos x-\cos x\cdot\sin^{2}x+\sin x-\sin x\cdot\cos^{2}x$, which is kind of long. Is there a shorter way, such as factoring the left side? If so, how can I do it?
How would I approach this identity: $\cos^3x+\sin^3x=(\cos x+\sin x)\cdot(1-\sin x\cdot\cos x)$?
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algebra-precalculus
trigonometry
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0Oh yeah, we used synthetic too. We just didn't do a lot of factoring of polynomials like we did with quadratic equations. – 2011-12-07
1 Answers
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Observe that $ \begin{align*} \cos^3x+\sin^3x &= (\cos x+\sin x)(\cos^2x-\sin x \cos x +\sin^2x) \\ &= (\cos x+\sin x)(1-\sin x \cos x ) \end{align*}$ noting that $\cos^2x+ \sin^2x=1$.