How do you find the \arcsin(\sin(\frac{7\pi}{6})) ?

The answer is:
${\displaystyle \arcsin \left(\sin \left(7\frac{\pi }{6}\right)\right)=-\frac{\pi }{6}}$
The range of a function ${\displaystyle \arcsin \left(x\right)}$ is, by definition.
${\displaystyle -\frac{\pi }{2}\le \arcsin \left(x\right)\le \frac{\pi }{2}}$
It means that we have to find an angle ${\displaystyle \alpha }$ that lies between ${\displaystyle -\frac{\pi }{2}}$ and ${\displaystyle \frac{\pi }{2}}$ and whose ${\displaystyle \sin \left(\alpha \right)}$ equals to a ${\displaystyle \sin \left(7\frac{\pi }{6}\right)}$.
From trigonometry we know that
${\displaystyle \sin (\varphi +\pi )=-\sin \left(\varphi \right)}$
for any angle ${\displaystyle \varphi }$.
This is easy to see if use the definition of a sine as an ordinate of the end of a radius in the unit circle that forms an angle ${\displaystyle \varphi }$ with the X-axis (counterclockwise from the X-axis to a radius).
We also know that since is an odd function, that is ${\displaystyle \sin (-\varphi )=-\sin \left(\varphi \right)}$
We will use both properties as follows:
${\displaystyle \sin \left(7\frac{\pi }{6}\right)=\sin (\frac{\pi }{6}+\pi )=-\sin \left(\frac{\pi }{6}\right)=\sin (-\frac{\pi }{6})}$
As we see, the angle ${\displaystyle \alpha =-\frac{\pi }{6}}$ first our conditions. It is in the range from ${\displaystyle -\frac{\pi }{2}}$ to ${\displaystyle \frac{\pi }{2}}$ and its sine equals to ${\displaystyle \sin \left(7\frac{\pi }{6}\right)}$. Therefore, it's a correct answer to a problem.