Inverse Trigonometric Function: Find the Exact Value of $\sin^{-1}\left(\sin\left(\frac{7\pi}{3}\right)\right)$ $$\arcsin\left(\sin\left(\frac{7\pi}{3}\right)\right)$$
I cannot use this formula, correct? $f(f^{-1}(x))=x$
The answer in the book is $\frac{\pi}{3}$
How do I approach solving a problem such as this? 
The inverse sin function of $\sin\frac{7\pi}{3}$
Am I saying to myself there were 7 revolutions and $\frac{\pi}{3}$ a corresponding angle to $\frac{7\pi}{3}$

Let $\sin^{-1}\sin\dfrac{7\pi}3=x$ where $-\dfrac\pi2\le x\le\dfrac\pi2$
$$\implies\sin x=\sin\dfrac{7\pi}3$$
$\implies x=n\pi+(-1)^n\dfrac{7\pi}3$ where $n$ is any integer
If $n$ is even $=2m$(say), $x=2m\pi+\dfrac{7\pi}3=\dfrac{(6m+7)\pi}3\implies -\dfrac\pi2\le\dfrac{(6m+7)\pi}3\le\dfrac\pi2$
$\iff-3\le12m+14\le3\implies-2<-\dfrac{17}{12}\le m\le-\dfrac{11}{12}<0\implies m=-1$
Check for odd $n=2m+1$(say)
