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$$\frac{\cos^3{x}-\sin^3{x}}{\cos{x}-\sin{x}} =1+\frac{1}{2} \sin{2x}$$

How do I start :( which identity do I use?

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Use the fact $a^3-b^3=(a-b)(a^2+ab+b^2)$. – emiliocba Jan 25 '12 at 13:29
up vote 8 down vote accepted

Hint: use the identity:


Solution follows:

We have $\cos^3 x -\sin^3 x =(\cos x-\sin x)(\cos^2 x +\cos x\sin x +\sin^2 x )$.


$$\eqalign{{ \cos^3 x -\sin^3 x\over \cos x-\sin x}&= {(\cos x-\sin x)(\cos^2 x +\cos x\sin x +\sin^2 x )\over (\cos x-\sin x)}\cr &=1+\cos x\sin x\cr &=1+\textstyle{1\over2}\sin 2x.} $$

(The last equality used the trigonometric identity $\sin(2x)=2\sin x\cos x$.)

As @Dilip Sarwate points out in the comment below, the above does not hold when $\cos x=\sin x$. In this case, ${ \cos^3 x -\sin^3 x\over \cos x-\sin x}$ is not defined. Of course, whenever ${ \cos^3 x -\sin^3 x\over \cos x-\sin x}$ is defined, it is equal to $1+{1\over2}\sin 2x$.

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Its 1+\cos(x)\sin(x) – user9413 Jan 25 '12 at 13:33
+1 But perhaps a note could be added as to what happens when $x = \pi/4$, say, and we cannot cancel $\cos x - \sin x$ from numerator and denominator? – Dilip Sarwate Jan 26 '12 at 22:36
$f(x)={\cos^3 x-\sin^3 x\over\cos x-\sin x}$ is not defined at $x=n\pi+\pi/4,n\in\mathbb Z$ and thus is not continuous at all $x\in\mathbb R$, while a related function $$g(x)=\begin{cases}{ \cos^3 x -\sin^3 x\over \cos x-\sin x},&x\neq n\pi+\pi/4,\\\frac{3}{2},&x=n\pi+\pi/4,\end{cases}$$ is continuous at all $x\in\mathbb R$ and equals $1 + \frac{1}{2}\sin 2x$ for all $x\in\mathbb R$. – Dilip Sarwate Jan 27 '12 at 3:52

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