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.