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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10HINT: Have you seen an identity involving $a^3 + b^3$? – 2011-12-07
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2Maybe you have seen a factorization of $x^3-y^3$? – 2011-12-07
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0No. Don't think so – 2011-12-07
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1Then the phrase you want to search for is "sum of cubes" (or "difference of cubes"). – 2011-12-07
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0And it's $x^3 + y^3$ not $x^3 - y^3$ – 2011-12-07
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4So you don't know $x^3+y^3=(x+y)(x^2-xy+y^2)$? – 2011-12-07
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2Alex Yan: I'm not sure exactly if this is what André intended, but $x^3+y^3$ also has the form $x^3-z^3$ with $z=-y$; if you know one factorization you get the other. – 2011-12-07
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0@J.M. No I didn't. We didn't go that deep into factoring cubic functions. Just finding roots until you get a quadratic and go from there. I imagine the concepts aren't that different but it takes time to figure out the patterns of a cubic function – 2011-12-07
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0Damn slow internet. Mixed up the order of the comments – 2011-12-07
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0Sure, you can take that route to derive the formula for the sum of two cubes. How would you factor $1+u^3$, for instance? – 2011-12-07
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0Alex, did you use synthetic division to find (confirm) roots of polynomials? If so, then try putting "-y" in the box, and use the coefficients of $x^3 + y^3$: ($1, 0, 0, y^3$). When you have a remainder of 0, this will confirm that $x = -y$ is a root, and therefore $(x + y)$ is a factor, with the resulting coefficients (1, -1, 1) corresponding to $x^2 - xy + y^2$ – 2011-12-07
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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 $\cos^2x+ \sin^2x=1$.