Evaluating an inverse function by sketching a unit circle Problem I'm working on: 
"Evaluate the inverse function by sketching a unit circle, locating the correct angle and evaluate the ordered pair on the circle."
The function I got was $\cos^{-1}(0)$.
So okay, the answer is $(\pi/2)$ which I agree with. $\cos(\pi/2)$ is 0, so it makes sense that the inverse function of 0 would give us $\pi/2$. But what about $3\pi/2$ for instance? Wouldn't that be an answer as well?
 A: You are correct, $x=\frac{3\pi}{2}$ is also a solution.

$\cos^{-1}(0) \Leftrightarrow \cos x=0$
Now, find the angles in the unit circle of which cosine is $0$.
Those are $x_1=\frac{\pi}{2}$ and $x_2=\frac{3\pi}{2}$.


A: I suspect that in the context of function $\cos$  this command ("Evaluate the inverse function...") was given unto you. It can only be obeyed if you are familiar with the inverse function of $\cos$. You must know then that in this context $\cos$ is not a function $\mathbb R\rightarrow\mathbb R$ but a function $\mathbb [0,\pi]\rightarrow [-1,1]$. 
If $f:X\rightarrow Y$ is a function then its inverse (if it has one) is a function $g:Y\rightarrow X$ such that the compositions $g\circ f:X\rightarrow X$ and $f\circ g:Y\rightarrow Y$ are the identity functions $x\mapsto x$ and $y\mapsto y$. An inverse will exist if and only if $f$ is a bijective function. The inverse $g$ is unique and bijective (hence has its own unique inverse wich is $f$). Often $g$ is denoted as $f^{-1}$.
As a function $\cos:\mathbb R\rightarrow\mathbb R$ it is not surjective. That can be repaired by making it a function $\cos:\mathbb R\rightarrow [-1,1]$ and here $[-1,1]$ is off course a very 'natural' choice. However in this condition it is not injective yet, and to repair that its domain must be restricted. A bit less 'natural' but generally accepted choice is:  $\cos:[0,\pi]\rightarrow [-1,1]$. Then $\cos^{-1}:[-1,1]\rightarrow[0,\pi]$ and clearly $\cos^{-1}(0)=\frac{\pi}{2}$. 
If you were not familiar with the fact that $\cos^{-1}:[-1,1]\rightarrow[0,\pi]$ then it was not okay to command you to evaluate $\cos^{-1}(0)$. I hope that from now on you are fully equipped.
