How do I prove that $\arccos(x) + \arccos(-x)=\pi$ when $x \in [-1,1]$? Prove that $\arccos x + \arccos(-x) = \pi$ when $x \in [-1,1]$.  
How do I prove this? Where should I begin and what should I consider?
 A: An idea:
$$x\in [-1,1]\;\implies\;\exists\,!\,\,\theta\in [0,\pi]\;\;s.t.\;\;\begin{cases}\cos\theta=x\\{}\\\cos(\pi-\theta)=-x\end{cases}\implies$$
$$\arccos x+\arccos(-x)=\arccos(\cos\theta)+\arccos(\cos(\pi-\theta))=\theta+\pi-\theta$$
A: Well you can do this. add $-π/2$ to both sides two times.
$(\arccos(x)-(π/2) )+(\arccos(-x)-(π/2) )=π - 2(π/2)=0$
then I define my function $f$ in this way:
$f(x)=\arccos(x)-(π/2)$
now if I prove that my function $f$ is an odd function then I can have:
$-f(x)=f(-x)$
then I'll have :
$(\arccos(x)-(π/2) )-(\arccos(x)-(π/2) )=0$
and it is proven. and to prove that the $f$ is and odd I use it inversed function that I call $G(x)$ ($G$ is inverse of $f$) :
$\cos(x+π/2)=G(x)$
so : $G(x)=-\sin x$
because $-\sin x$ is an odd function then it's inverse function $f$ is also an odd function.
A: If $\arccos(x)=y,0\le y\le\pi$
The principal values of $\arcsin$ lies $\in\left[-\dfrac\pi2,\dfrac\pi2\right]$
Method $\#1$
$\cos[\arccos(x)+\arccos(-x)]$
$=\cos[\arccos(x)]\cdot\cos[\arccos(-x)]-\sin[\arccos(x)]\cdot\sin[\arccos(-x)]$
$=x(-x)-\sqrt{1-x^2}\cdot\sqrt{1-x^2}$
$\implies\cos[\arccos(x)+\arccos(-x)]=-1$
$\implies\arccos(x)+\arccos(-x)=\arccos(-1)=\pi$ as $0\le\arccos(x),\arccos(-x)\le\pi$
Method $\#2$
If $\arccos(x)\ge0,0\le y\le\dfrac\pi2, \arccos(x)=\arcsin\sqrt{1-x^2}$
Else if $\arccos(x)<0,\dfrac\pi2<y\le\pi, \arccos(x)=\pi-\arcsin\sqrt{1-x^2}$
A: Let $f(x)$ be your function on left-hand side, now compute $f'(x)$. 
