# Domain and range of composition of trigonometric functions.

Let's assume that $f(x)=\sin(\arcsin(x))$, then the domain is $[-1,1]$, and so is the range, and $f(x)=x$.

However, for $g(x)=\arcsin(\sin(x))$, the domain is $\mathbb{R}$, the range is $[-\pi/2,\pi/2]$, but $g(x)=x$ only for $x \in [-\pi/2,\pi/2]$. How could I rewrite $g(x)$ for the remaining values of its domain?

Also, if I were to do similar compositions with other trigonometric functions, would there be something interesting? (i.e., something very different from the methods we used above.)

Any help would be appreciated.

Note that $g(x) = \arcsin(\sin x)$ is periodic with period $2\pi$. $$g(x+2\pi) = g(x)$$

Also, $g(x)$ is an odd function, since all the functions in composition are odd.

$$g(-x)=-g(x)$$

On the principal value branch, i.e. $x \in \left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$, $$\arcsin(\sin x) = x$$

When $x \in \left[\frac{\pi}{2}, \frac{3\pi}{2}\right]$ then

$$(x-\pi)\in\left[\frac{\pi}{2},\frac{3\pi}{2}\right] \\ \arcsin(\sin (x-\pi)) = x-\pi \\ \arcsin(\sin x) = \pi-x$$

You now have the definition of $g(x)$ over an interval of length $2\pi$.

Note: For $\arctan(\tan x)$, the result would be trivial, as it is periodic with $\pi$ period. So the principal value branch would be repeated over periods of $\pi$