Verify $\arccos{x}+\arcsin{x}=\frac{\pi}{2}$

Question

Following is the proof of the above proposition given in my textbook.

Let, $\arccos{x}={\theta}_2$ and $\arcsin{x}={\theta}_1$

$$\sin({\theta}_1 + {\theta}_2)=\sin{\theta_1}\cos{\theta_2}+\sin{\theta_2}\cos{\theta_1}$$

$$\implies \sin({\theta}_1 + {\theta}_2)=x^2+(\sqrt{1-x^2})(\sqrt{1-x^2})$$

$$\implies \arccos{x}+\arcsin{x}=\arcsin{(1)}$$

$$\implies \arccos{x}+\arcsin{x}=\frac{\pi}{2}$$

Problem:

When, $\cos{\theta_1}=-\sqrt{1-x^2}$

and $\sin{\theta_2}=\sqrt{1-x^2}$

Then, $${\theta}_1 + {\theta}_2=\arcsin{(-1)}$$

$$\implies \arccos{x}+\arcsin{x}=-\frac{\pi}{2}$$

Doesn't it disprove the given proposition?

Answer 1

No, because $\theta_1=\arcsin x\overset{\text{def}}{\iff}\sin\theta_1=x\enspace\textbf{and} -\frac\pi2\le\theta_1\le \frac\pi2$.

On this interval, $\cos\theta_1\ge 0$.

Answer 2

I think you should care about domain/range matters in your argument.

However the safest way to prove the given identity is to derive it: $$ (\arccos x+\arcsin x)'=\frac{-1}{\sqrt{1-x^2}}+\frac{1}{\sqrt{1-x^2}}=0 $$ thus $x\mapsto\arccos x+\arcsin x$ is a constant function in the connected components of its domain, which is $D:=[-1,1]$, which is connected, thus our function has to be constant in the whole $D$. Now just evaluate the function in a smart point (like $x_0=0$) to conclude.