The derivative of $\arccos(\cos(x))$

I was asked to show that $\frac{d}{dx}\arccos(\cos{x}), x \in R$ is equal to $\frac{\sin{x}}{|\sin{x}|}$.

What I was able to show is the following:

$\frac{d}{dx}\arccos(\cos(x)) = \frac{\sin(x)}{\sqrt{1 - \cos^2{x}}}$

What justifies equating $\sqrt{1 - \cos^2{x}}$ to $|\sin{x}|$?

I am aware of the identity $\sin{x} = \pm\sqrt{1 - \cos^2{x}}$, but I still do not see how that leads to that conclusion.

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Since you're aware of $\sin{x} = \pm\sqrt{1 - \cos^2{x}}$, simply take the absolute value of both sides. –  Kevin Oct 7 '12 at 17:35
Actually, for complex $x$ the correct result $\sin(x)/\sqrt{1-\cos^2(x)}$ may not be simplified to $\sin(x)/|\sin(x)|$. –  GEdgar Oct 7 '12 at 19:00

In general, $\sqrt{a^2} = |a|$, as $\sqrt y$ refers to the non-negative number whose square is $y$. Then since $1-\cos^2x = \sin^2x$, we have $\sqrt{1-\cos^2 x} = |\sin x|$.

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$\sqrt{1-\cos^{2}(x)}=\sqrt{\sin^{2}(x)}$, which is $|\sin(x)|$ by definition.

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The function $f:\ x\mapsto\arccos(\cos x)$ is even and $2\pi$-periodic, since the "inner" function $\cos$ has these properties. For $0\leq x\leq \pi$ by definition of $\arccos$ we have $f(x)=x$. Therefore $f$ is the $2\pi$-periodic continuation of the function $$f_0(x)\ :=\ |x|\qquad(-\pi\leq x\leq\pi)$$ to the full real line ${\mathbb R}$, and $f'$ (where defined) is the $2\pi$-periodic continuation of $$f_0'(x)={\rm sgn}(x)\qquad(-\pi<x<\pi)$$ to all of ${\mathbb R}$. One way to express this extension is indeed given by $$f'(x)\ =\ {\sin x\over|\sin x|}\qquad (x\ne \pi{\mathbb Z})\ ,$$ but we can do without computing sines: $$f'(x)=(-1)^{\lfloor x/\pi\rfloor}\qquad (x\ne \pi{\mathbb Z})\ .$$
If $y=\arccos(x)$ then $x=\cos(y)$ and so $$\frac{dy}{dx}=\frac{1}{\frac{dx}{dy}}=\frac{-1}{\sin(y)}$$ if $\sin(y)\neq0$. Here, $$0<y=\arccos(x)<\pi, \sin(y)>0$$ and $$\sin(y)=\sqrt{1-\cos^2(y)}=\sqrt{1-x^2}$$ It means that $$\frac{dy}{dx}=\frac{d}{dx}\arccos(x)=\frac{-1}{\sqrt{1-x^2}}$$ where in $x\in (-1,1)$.