Is this trig step correct? $\sin^{-1}(-\sin(x))$ = $-\sin^{-1}(\sin(x))$
Can the minus be taken out like this? 
 A: Generally, it is not true that
$$f^{-1}(-f(x)) = - f^{-1}(f(x))$$
Try to cook up a counter-example!
However, in this case, since $\sin$ is odd we get that
$$\sin^{-1}(-\sin(x)) = \sin^{-1}(\sin(-x)) = -x = - \sin^{-1}(\sin(x))$$
A: Yes, that is valid, since the $\arcsin$, like $\sin$ is an odd function.
Here, that means that $\sin^{-1}(-f(x)) = -\sin^{-1}(f(x))$, and in this case, $f(x) = \sin x$. So ultimately, we have that $$\sin^{-1}(-\sin(x)) = -x$$
A: It is only valid if you take the domain of $\sin^{-1}$ to be $[-1,1]$ and the range $[-\pi/2,\pi/2]$. There are other definitions for which this doesn't hold. For example, you could take the range to be $[\pi/2,3\pi/2]$. 
Any restriction to a symmetric subinterval (open or closed) will work as well.
Comment: From my comment above on the OP's question:
"It can be done only if you are using a specific domain/range in defining arcsine. See my answer below. Many of the other answers make this assumption, but it is patently false except in a very specific circumstance. When you work with inverses of functions that are not 1-1, you have to be very very careful or you can gloss over something and arrive at a wrong answer"
A: Using the definition of principal value of inverse sine function,
we can always find an integer $n$ such that $\displaystyle x=n\pi+(-1)^n y$ with  $\displaystyle -\frac\pi2\le y\le\frac\pi2$
$\displaystyle\implies-\frac\pi2\le-y\le\frac\pi2 $ 
$\displaystyle\implies\sin x=\sin(n\pi+(-1)^n y)=\sin y $
$\displaystyle\implies-\sin x=-\sin y=\sin(-y)$
$\displaystyle\implies\sin^{-1}(\sin(-y))=-y$ as $\displaystyle-\frac\pi2\le-y\le\frac\pi2 $
Similarly, $\displaystyle\sin^{-1}(\sin x)=\sin^{-1}(\sin y)=y$ as $\displaystyle-\frac\pi2\le y\le\frac\pi2 $
A: Yes you are correct. It can be taken out since the value of $sin(x)$ is taken as an angle.
