find inverse function in given domain Suppose that we have function $y=\sin(x)$ we need to find its inverse function, assuming that $D(f)=[-\pi/4.\pi/4]$
I know that inverse of $\sin(x)$ is $\arcsin(x)$, it would be answer of a given function too, but why do I need  $D(f)=[-\pi/4.\pi/4]$? I don't know, should I introduce some variable $c$, so that $y=\sin(x)$ will look like $y=\sin(x-c)$ or $y=\sin(x)+c$? Please give me a hint. In case I meet similar problem, like "find inverse of function $y=f(x)$  where $D(f)=[a,b]$" what should I do? As I know domain of given function and range of inverse function are the same, so it means that the range of $\arcsin(x)$ is $[-\pi/4.\pi/4]$, but how to proceed?
 A: Well if you look at the definition of a function, it states that one element in the domain CANNOT be mapped to more than one element of the range. 
So, when looking at inverses, you have to make sure that the function is injective(one to one) before you can claim the existence of an inverse.
Example: Take sin(x) from 0 to 2\pi. sin(0) = sin(2\pi). So an inverse of sin(x) over 0 to 2\pi is impossible since it would map 0 to 0 AND 2\pi.
This is why you have to pick your domain carefully. 
A: The map "sin" is strictly increasing on $[-\pi/4,\pi/4]$, and therefore it is injective. Its inverse, in this case, is simply the restriction $\arcsin_{|[-1/\sqrt{2},1/\sqrt{2}]}$ of the "global" inverse $\arcsin \colon [-1,1] \to [-\pi/2,\pi/2]$.
A: The typical range of $\arcsin(x)$ is $[-\pi/2,\pi/2]$, with domain $[-1,1]$. This is because we restrict $\sin(x)$ to the interval $[-\pi/2,\pi/2]$--a maximal interval on which $\sin(x)$ is one-to-one--and then take the inverse.
However, we didn't have to have a maximal interval, just some interval on which it's one-to-one, so the same principle will apply here, and you'll simply have $f^{-1}(x)$ as the restriction of $\arcsin(x)$ to the interval $[-1/\sqrt{2},1/\sqrt{2}]$.
