$f(x)=\sin^{-1}(\sin x)$; $x$ is real then prove that $f$ is continuous for all $x$ but not differentiable for all $x =(2k+1)π/2$?

Question

My attempt is: $$\lim_{h \to 0} \frac{f((2k+1)π/2+h) - f((2k+1)π/2)}{h} = \lim_{h \to 0} \frac{\sin^{-1}(\sin((2k+1)π/2+h)-\sin^{-1}(\sin((2k+1)π/2)}{h} = \lim_{h \to 0} \frac{kπ+π/2+h-(2k+1)π/2}{h} = 1$$ Am I right upto this ?

Answer 1

Hint: using that $f$ is periodic and even (ckeck!), looking at $\pi/2$ is enough.

For $x\in[-\pi/2,\pi/2]$, $f(x) = x$. For $x > \pi/2$ but near to $\pi/2$ you can check that $$\sin(x) = \sin(\pi/2 + (x - \pi/2)) = \cdots = \sin(\pi/2 - (x - \pi/2)) = \sin(\pi - x)$$ and using that $\pi - x\in[-\pi/2,\pi/2]$ $$\arcsin(\sin(x)) = \cdots$$ Can you continue?

Answer 2

Graph of$sin^-(sinx) $ will help in more understanding.

You can first prove that it is periodic with 2$pi$ and not differentiable at $pi$÷2 and 3$pi$÷2 .... Then generalise it to$( 2k+1)pi÷2$

Not the best method.....