How to show that $\arcsin|\sin x|-\arccos|\cos x|=0$ for all $x\in \Bbb R$

Question

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!

Answer 1

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 $\pi$.

Answer 2

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.

Answer 3

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

$$\arcsin |\sin x|-\arccos |\cos x|\\ =\arcsin (\sin |x|)-\arccos (\cos |x|) \\ = |x|-|x|=0$$

Answer 4

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 $$\arcsin(|\cos x|)+\arccos(|\cos x|)=?$$