Why is the range of arctan $[ -\frac{\pi}{2} , \frac{\pi}{2}]$? I've been taught in school and it says on Wikipedia that the range of arctan is $[ -\frac{\pi}{2} , \frac{\pi}{2} ]$.
Why isn't it $[0,\pi]$ ?
 A: As the graph of the function $\,\tan x\,$ show, this is a very not $\,1-1\,$ function onto the reals, so just as it's done with $\,\sin x\,,\,\cos x\,$ we limit its range in order to get a $\,1-1\,$ onto, and thus invertible, function. As the period of $\,\tan x\,$ is $\,\pi\,$ , we can take any interval of length $\,\pi\,$ to do this...but...if the interval contains a point of the form $\,\displaystyle{x=\frac{(2n+1)\pi}{2}}\,n\in\mathbb{Z}$ we're going to have an ugly vertical asympote there, so we choose an interval of the form $$\left(\frac{(2n-1)\pi}{2}\,,\,\frac{(2n+1)\pi}{2}\right)\,\,,\,n\in\mathbb{Z}$$withy $\,n=0\,$ being customary.
A: It's simpler because it makes the function continuous.  In a sense, arctan is a "multiple-valued function" (but the prevailing modern definitions of function consider such things to be something other than functions).  I.e., there is more than one number whose tangent is $x$.  So which one do you call $\arctan x$?  The answer is, in a sense: the simplest one.
Draw a right triangle showing sides suitably labeled "adj", "opp", and "hyp".  Recall that tan = opp/adj.  Let adj = 1, so that tan = opp.  As opp goes up to $\infty$, look at what happens to the angle: you see it going up to a right angle, i.e. to $\pi/2$.  Now look at what happens as "opp" becomes negative, and then goes down to $-\infty$, and watch the angle going down to $-\pi/2$.
Draw the pictures carefully and you'll see what I mean.
