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 $\frac{7\pi}{3}$

You have to consider the restricted ranges of the inverse trig functions. Since $\sin\left(\frac{7\pi}{3}\right)=\frac{\sqrt{3}}{2}$, you want $\sin^{-1}\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi}{3}$, since the range of inverse sine is $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right).$
