Explain why it is necessary to restrict the range of inverse trig functions? This is very confusing. Please help and use lower level vocabulary that is easy to understand.
 A: \begin{align}
\sin30^\circ & = \frac 1 2 \\[8pt]
\sin150^\circ & = \frac 1 2 \\[8pt]
\sin(-210^\circ) & = \frac 1 2 \\[8pt]
\sin 390^\circ & = \frac 1 2 \\[8pt]
& \vdots
\end{align}
So $\sin^{-1}\dfrac 12 = 30^\circ\text{ or }150^\circ\text{ or } -210^\circ \text{ or } 390^\circ\text{ or }\cdots\cdots\cdots\text{ ?}$  The answer is that one picks the one between $-90^\circ$ and $+90^\circ$. That is the restriction of the range.
A: In order a function to have an inverse, there has to be a bijection (a function that covers all the range of the codomain and also there is a 1-1 relation between the elements of the domain and the codomain). So, take for example $\sin:\mathbb{R}\rightarrow\mathbb{R}$. The range of $\sin(x)$ is not $\mathbb{R}$, it's $[-1,1]$, also there isn't a 1-1 relation, because for example $\sin(0)$ and $\sin(2\pi)$ have the same value.
That's why, in order to get an inverse function, we need to restrict both the domain and codomain of the functions. That's why, for example, $\sin^{-1}(0) = 0$ and not $2\pi$, because the proper definiton of sin and arcsin in order for both to be inverse should be:
$$ \sin: [-\frac{\pi}{2},\frac{\pi}{2}]\rightarrow [-1,1]$$
$$ \sin^{-1}: [-1,1]\rightarrow [-\frac{\pi}{2},\frac{\pi}{2}]$$
There we have the bijection we need.
