True or false? $\sin^{-1}(\sin(2\pi))=0$ 
True or false?
  $$\sin^{-1}(\sin(2\pi))=0$$
  If true, give an explanation as to why. If false, give a counter example.

My teacher stated that we are not allowed to use calculators for summer homework, but I just plugged this into my TI-84 anyway. It turns out that this statement is true. However, I do not know how to explain or show why. 
Any help is greatly appreciated.
 A: First, the answer you got is correct as you simply evaluate $\sin(2\pi) = 0$, and take the sine inverse of $0$ which is $0$. Secondly, what you really want to ask is when is it true $\sin^{-1}(\sin \alpha) = \alpha$. It turns out that this is true if $-\frac{\pi}{2} \le \alpha \le \frac{\pi}{2}$. On the other hand, it is always true that: $\sin(\sin^{-1}(x)) = x, -1 \le x \le 1$.
A: The graph of $f(x) = \sin^{-1}(\sin(x)$ is shown below. It can be seen that 
\begin{align*}
f(x) &= x \qquad -\frac{\pi}{2} \leq x \leq \frac{\pi}{2} \\
&= \pi - x \qquad \frac{\pi}{2} \leq x \leq 3\frac{\pi}{2}  \\
&= -\pi + x \qquad 3\frac{\pi}{2} \leq x \leq 5\frac{\pi}{2} \\
&\cdots
\end{align*}

A: So, many different $x$ values can be mapped to a given value for $\sin$; $\sin(x)=\sin(y)$ if $|x-y|$ is a multiple of $2\pi$. Therefore, in order to create the inverse sine function, we have to "choose" what interval the function maps to. Given one choice for $\sin^{-1}$, we could always add some multiple of $2\pi$ to get another choice for $\sin^{-1}$. The standard choice for $\sin^{-1}$ maps the interval $[-1,1]$ to $[-\frac{\pi}{2},\frac{\pi}{2}]$; since $\sin(x)=0$ only if $x$ is a multiple of $\pi$, this means that $\sin^{-1}{(0)}=0$, as this is the only multiple of $\pi$ in the range. Therefore, $\sin^{-1}(\sin(2\pi)) = \sin^{-1}(0) = 0$. 
The idea that given $f$ we can define $f^{-1}$ so that $f^{-1}(f(x)) = x$ is really only valid if $f$ is invertible, i.e. if no two distinct $x$ values map to the same value for $f$. $\sin$ is invertible if restricted to the interval $[-\frac{\pi}{2},\frac{\pi}{2}]$, which is where we get the standard inverse sine. 
A: As stated in the comments, the unit circle isn't necessary, but I think it gives a nice "intuitive" graphical explanation.  
Consider that for an angle $\theta$, $\sin(\theta)$ is the $y$ coordinate of the point on the unit circle.
 (https://en.wikipedia.org/wiki/Unit_circle)
Then it is simply a question of:


*

*What is the $y$ component for the point at $2\pi$ radians?

*What angle $-\pi/2 \leq \theta \leq \pi/2$ gives you a $y$ component of $0$? 

A: I'm surprised at the number of answers that are overcomplicating this and/or just incorrect.  This is actually very simple.
The answer is true.  We can see this by direct evaluation:
$$\sin^{-1}(\sin 2\pi) = \sin^{-1} 0 = 0$$
Let's see why this is the case.
First, $\sin 2\pi = 0$.  This is basic unit circle trig.  Therefore we just need to evaluate $\sin^{-1} 0$.  That is, we want to evaluate the inverse sine function at $0$.
The domain of the inverse sine function is $[-1,1]$.  The range of the inverse sine function is $\left[-\dfrac\pi2, \dfrac\pi2\right]$.
When we say "evaluate $\sin^{-1}0$" we're really saying "give me the value of $x$ (in the range of inverse sine) such that $\sin x=0$."  So we want the value of $x$ between $-\pi/2$ and $\pi/2$ such that $\sin x = 0$.  The only value of $x$ satisfying this is $x=0$.  Therefore, $\sin^{-1} 0 =0$.
A: First note that $\sin ( 2 \pi ) = 0$, so the question is asking for $\sin^{-1} (0)$. Of course, $0$ satisfies this. However, so does $\pi$, $2\pi$, $3\pi$, $4\pi$ .... and on and on.
As a technical note, the calculator only considers values from $-\pi/2$ to $\pi/2$ when calculating the inverse function. The reason for this is that if you look at the graph of sine, it is actually bijective on this interval, and there is a well known theorem that a function is invertible if and only if it is bijective. Thus by restricting to this interval, the calculator guarantees that it can give you a unique answer.
