Consider [tex]\arccos(-x)+\arccos(x)[/tex], where [tex]x \in [-1,1][/tex].
This means there exists a [tex]\theta \in [0, \pi][/tex] such that [tex]\cos \theta=x[/tex] and [tex]\cos (\pi -\theta)=-\cos \theta=-x[/tex].
Thus, taking the inverse cosine of both of these equations, we get that:
[tex]\arccos(x)+\arccos(-x)=\theta+(\pi-\theta)=\pi[/tex].
Subtracting arccos(x) from both sides, we get the required result.
