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.


1 Answer 1


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.


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 $\pi$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .