<s>[INST] 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?
 [/INST] 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,
$\cos^{-1}(-x) = \pi - \cos^{-1} x$.
</s>