Evaluating $\arccos(\cos\frac{15\pi }{4})$ I have a problem with understanding of this exercise:
$\arccos(\cos\frac{15\pi }{4})= ?$
$\cos(\frac{15\pi}{4}-2\pi)=\cos(\frac{7\pi}{4})$
$\cos(\frac{7\pi}{4}+\pi)= -\cos(\frac{3\pi}{4})$
then $\arccos(-\cos\frac{3\pi}{4})$
All above I understand pretty clearly. We did it at school, too and this is a solution. I just don`t understand where $-x$ comes from.  It can possibly be $x=\cos(\frac{3\pi}{4})$ but what about the $-\frac{\pi}{2}$ then ??
$\arccos(-x)-\frac{\pi}{2}  = -(\arccos x- \frac{\pi}{2})$
$\arccos(-x)=\pi-\arccos x$
$\arccos(\cos\frac{15\pi}{4})=\arccos(-\cos\frac{3\pi}{4})=\pi-\arccos(\cos\frac{3\pi}{4})=\pi-\frac{3\pi}{4}=\frac{\pi}{4}$
or do you know any other method how to solve this ??
Thank you for your time.
 A: Here's one way to do it
$$\arccos\left(\cos\left(\frac{15}{4}\pi\right)\right)$$
$$=\arccos\left(\cos\left(\pi+2\pi+\frac{3}{4}\pi\right)\right)$$
$$=\arccos\left(\cos\left(\pi+\frac{3}{4}\pi\right)\right)$$
$$=\arccos\left(-\cos\left(\frac{3}{4}\pi\right)\right)$$
$$=\pi-\arccos\left(\cos\left(\frac{3}{4}\pi\right)\right)$$
$$=\pi-\frac{3}{4}\pi=\frac{\pi}{4}$$
A: Since $\arccos x: [-1, 1] \to [0, \pi]$, we need to find the angle in $[0, \pi]$ that has the same cosine as $15\pi/4$.  Since coterminal angles have the same cosine and 
$$\frac{15\pi}{4} = -\frac{\pi}{4} + 4\pi = -\frac{\pi}{4} + 2 \cdot 2\pi$$
we have
$$\cos\left(\frac{15\pi}{4}\right) = \cos\left(-\frac{\pi}{4}\right)$$ 
Since $\cos(-x) = \cos x$, we have 
$$\cos\left(-\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right)$$
Thus,
\begin{align}
\arccos\left[\cos\left(\frac{15\pi}{4}\right)\right] & = \arccos\left[\cos\left(-\frac{\pi}{4}\right)\right]\\ 
& = \arccos\left[\cos\left(\frac{\pi}{4}\right)\right]\\
& = \frac{\pi}{4}
\end{align}
where the final equality follows from the observation that if $\theta \in [0, \pi]$, then $\arccos(\cos\theta) = \theta$.
