How to show that $\arcsin|\sin x|-\arccos|\cos x|=0$ for all $x\in \Bbb R$ How to show that $\arcsin|\sin x|-\arccos|\cos x|=0$ for all $x\in \Bbb R$
I tried to draw the graph and got that $\arcsin|\sin x|=\arccos|\cos x|$
But I have no idea how to prove it.
Thank you all!
 A: If $0\leqslant x\leqslant\frac\pi2$, then $|\sin x|=\sin x$, and $|\cos x|=\cos x$. Therefore$$\arcsin|\sin x|-\arccos|\cos x|=x-x=0.$$
If $\frac\pi2\leqslant x\leqslant\pi$, then $|\sin x|=\sin x$ and $|\cos x|=-\cos x$. So,$$\arcsin|\sin x|=\arcsin(\sin x)=\frac\pi2-x$$and\begin{align*}\arccos|\cos x|&=\arccos(-\cos x)\\&=\frac\pi2-x\end{align*}and therefore$$\arcsin|\sin x|-\arccos|\cos x|=\left(\frac\pi2-x\right)-\left(\frac\pi2-x\right)=0.$$Finally, use that fact that $x\mapsto\arcsin|\sin x|-\arccos|\cos x|$ is periodic with period $\pi$.
A: Compute the derivative of $f(x)=\arcsin\lvert\sin x\rvert-\arccos\lvert\cos x\rvert$ (where it exists):
$$
f'(x)=\dfrac{1}{\sqrt{1-\sin^2x}}\dfrac{\lvert\sin x\rvert}{\sin x}\cos x-\dfrac{-1}{\sqrt{1-\cos^2x}}\dfrac{\lvert\cos x\rvert}{\cos x}(-\sin x)
$$
This becomes
$$
f'(x)=\frac{\cos^2x\lvert\sin^2x\rvert-\sin^2x\lvert\cos^2x\rvert}{\sin x\cos x\lvert\sin x\cos x\rvert}=0
$$
due to $\sqrt{1-\sin^2x}=\lvert\sin x\rvert$ and $\sqrt{1-\cos^2x}=\lvert\cos x\rvert$.
Therefore the function is constant on every interval where it's differentiable. However, the function is everywhere continuous, so it's constant everywhere.
Since $f(0)=\arcsin0-\arccos1=0$, you're done.
A: Given the periodicity of $\pi$, it suffices to prove the equality over the domain $x\in (-\frac\pi2,\frac\pi2]$, where 
$|\sin x |= \sin|x|$ and $|\cos x |= \cos|x|$ and
$$\arcsin |\sin x|-\arccos |\cos x|\\
=\arcsin (\sin |x|)-\arccos (\cos |x|) \\
= |x|-|x|=0$$
A: Using Proof for the formula of sum of arcsine functions $ \arcsin x + \arcsin y $
$$f(x)=\arcsin|\sin x|+\arcsin|\cos x|=\arcsin(|\sin x|^2+|\cos x|^2)$$
For real $\cos x,\sin x;$
$$f(x)=\arcsin(1)=?$$
Now $$\arcsin(|\cos x|)+\arccos(|\cos x|)=?$$
