# 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?

-

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 $[-1/\sqrt{2},1/\sqrt{2}]$.

-
so my inverse function is $arcsin(x)$ right yes?and what should be constraints?i mean domains,ranges – dato datuashvili Jul 4 '12 at 14:54
Well, as you, yourself noted, the range of the inverse function will be the domain of the original, so the range will be $[-\pi/4,\pi/4]$. If you note that $\sin(x)$ is increasing on that interval, it follows that its least and greatest range values will be achieved at $-\pi/4$ and $\pi/4$, respectively--specifically, they will be $-1/\sqrt{2}$ and $1/\sqrt{2}$, which you can get by plugging in. Since it's continuous, all values in between those will be obtained as well, so the range of $f$ (the domain of $f^{-1}$) is $[-1/\sqrt{2},1/\sqrt{2}]$. – Cameron Buie Jul 4 '12 at 15:06
thank you very much,the same method is used if given function on given range is decreasing yes? – dato datuashvili Jul 4 '12 at 15:13
Yes. ${}{}{}{}$ – Cameron Buie Jul 4 '12 at 15:15

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.

-

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 $\arcsin \colon [-1,1] \to [-\pi/2,\pi/2]$.

-
Your first $1/\sqrt{2}$ should be $-1/\sqrt{2}$. – Cameron Buie Jul 4 '12 at 14:48
Oops! Thank you! – Siminore Jul 4 '12 at 14:55
we have not to change anything?answer would be simple arcsin(x) with it's basic attributes? – dato datuashvili Jul 4 '12 at 15:03
@dato No! The answer is $f^{-1} \colon [-1/\sqrt{2},1/\sqrt{2}] \to [-\pi/4,\pi/4]$, $f^{-1}(y)=\arcsin y$. This is not arcsin. – Siminore Jul 4 '12 at 16:28