Proving $\cos^{-1}(-x)=\pi-\cos^{-1}x$ without geometry. Let
$$
\cos^{-1}x=a
\implies x=\cos a
$$
and
$$
\cos^{-1}(-x)=b
\implies -x=\cos b
$$
Hence we have
$$
\cos a+\cos b=0
$$
Using
$$
\cos(A+B)+\cos(A-B)=2\cos A\cos B
$$
with
$$
A=\frac{a+b}{2} \\
\text{ and } \\
B=\frac{a-b}{2}
$$
we get
$$
\cos a+\cos b=2\cos\left(\frac{a+b}{2}\right) \cos\left(\frac{a-b}{2}\right)=0
$$
with which
$$
\cos\left(\frac{a+b}{2}\right)=0 \text{ or } \cos\left(\frac{a-b}{2}\right)=0\\
\implies a=\pi-b \text{ or } a=\pi+b
$$
Now, to choose between the two, I'm making the below argument:
To find the inverse of a function, the function has to be one-to-one. Hence in our case, both $\cos a$ and $\cos b$ have to be one-to-one, which is possible only when
$$
n\pi\leq a,b \leq (n+1)\pi \text{, }n\in\mathbb{Z}
$$
From the above, we get,
$$
-n\pi \leq \pi-b \leq (-n+1)\pi
$$
and
$$
(n+1)\pi \leq \pi+b \leq (n+2)\pi
$$
The ranges can be reconciled for $a$ and $\pi-b$ by taking $n=0$ and we get
$$
0 \leq a \leq \pi \\
0 \leq \pi-b \leq \pi
$$
But no value of $n\in\mathbb{Z}$ can simultaneously reconcile the ranges of $a$ and $\pi+b$.
Therefore, we conclude that
$$
\cos^{-1}(-x)=\pi-\cos^{-1}(x)
$$
From this I also learnt that the inverse function is meaningful when the angle is discussed in the range $[0,\pi]$.
Is this line of argument mathematically fool-proof?
 A: That looks correct, but it's much simpler to say that, if $x\in[-1,1]$,\begin{align}\cos\bigl(\pi-\arccos(x)\bigr)&=-\cos\bigl(\arccos(x)\bigr)\\&=-x\\&=\cos\bigl(\arccos(-x)\bigr)\end{align}and that therefore, since $\pi-\arccos(x),\arccos(-x)\in[0,\pi]$ and since the restriction of $\cos$ to that interval is injective, $\pi-\arccos(x)=\arccos(-x)$.
A: If $0\leq{x}\leq\pi\implies$ we can use the next formula relatively of the both parts of the provided expression by you:
$$
\bbox[lightgreen]
{
\cos(\arccos(x))=x.
}
\qquad\qquad\qquad\qquad{(1)}
$$
Therefore
$$
\cos(\arccos(-x))=\cos(\pi-\arccos(x)),
$$
But another formula also exists:
$$
\bbox[lightgreen]
{
\arccos(-x)=\pi-\arccos(x)
}
\qquad\qquad\qquad\qquad{(2)}
$$
Applying $(2)$ to the previous expression we get:
$$
\cos(\arccos(-x))=\cos(\arccos(-x)).
$$
Good luck!
A: First of all let us prove that $\arccos(x)+\arccos(-x)$ is a constant. In fact,  differentiate it we  get
$$
(\arccos(x)+\arccos(-x))'=-{\frac {1}{\sqrt {-{x}^{2}+1}}}+{\frac {1}{\sqrt {-{x}^{2}+1}}}=0.
$$
Thus $\arccos(x)+\arccos(-x)=C$ for some $C$.
Now, to find $C$ just put $x=0.$ We obtain that
$\frac{\pi}{2}+\frac{\pi}{2}=C  \implies C=\pi,$ as required.
A: Let $f(x) = \cos^{-1}(-x) + \cos^{-1}x$. Note that
$$
f^{\prime}(x) = -\dfrac{1}{\sqrt{1 - (-x)^2}}\cdot (-1) - \dfrac{1}{\sqrt{1 - x^2}} = 0 \quad \Rightarrow \quad f(x) = k, \quad k \in \mathbb{R}
$$
For $x = 0$, we have $k = \cos^{-1}(-0) + \cos^{-1}0 = \pi/2 + \pi/2 = \pi$. Thus,
$\cos^{-1}(-x) = \pi - \cos^{-1} x$.
A: We have
$$
\cos(\pi-\cos^{-1}x)=\cos\pi\cos\cos^{-1}+\sin\pi\sin\cos^{-1}x=-x
$$
Since $0\le \pi-\cos^{-1}x\le \pi$,
applying $\cos^{-1}$ on both sides of the identity $\cos(\pi-\cos^{-1}x)=-x,$ we get
$$
\pi-\cos^{-1}x=\cos^{-1}(-x)
$$
A: Another way:
Use Proof for the formula of sum of arcsine functions $ \arcsin x + \arcsin y $  replacing $y$ with $-x$  so that $xy=-x^2<0$
$$\arcsin(x)+\arcsin(-x)=\arcsin(0)=0$$
Now use Why it's true? $\arcsin(x) +\arccos(x) = \frac{\pi}{2}$
