Why does it seem more natural to think of $r$ as a function of $\theta$, rather than the other way around? When teaching functions in polar coordinates, the nearly universal practice is to consider functions of the form $r = f(\theta)$.  I think I have never seen any examples in which $\theta$ is expressed as a function of $r$.  My intuition tells me that the usual setup is more ``natural'' in some sense than the latter, but I am not sure whether that is just because it's what I'm used to, or if I am responding to some intrinsic difference between the two cases.  Is there any good reason why, for instance, it seems more natural to write $r = \sqrt{\theta}$ than to write $\theta = r^2$, despite the fact that they are (obviously) equivalent?
Edited to add: Here is an example of what I mean.  Consider the function $\theta = \cos (r)$.  It has a very cool graph:

Intuitively, if we think of $r$ as the independent variable, and ask how $\theta$ depends on $r$, the description is quite simple: as we move outwards from the origin, the angle oscillates back and forth between a minimum angle of $-1$ radian and a maximum angle of $1$ radian.  This is a perfectly reasonable function to think about, graph, and study.  And yet functions like this are (as far as I can tell) completely absent both from textbooks and from usual instruction.  I find this strange, and am wondering if anybody has any thoughts on why.
 A: Addendum-2 added to respond to the OP's editing of their original posting, re the function $\theta = \cos(r)$.

Addendum added to respond to a comment question.

Because a function must map each element in its domain to only a single element in its range.
For example, consider the function $r = \cos(\theta)$.
Then, for each angle $\theta$, there is a specific value $r$.  However, the function does not have an inverse.  For example if you try to express $\theta = g(r)$, then what is $g(1/2)$?  Even if you restrict the range to $0 \leq \theta < 2\pi$, for example, would $g(1/2)$ equal $\pi/3$ or $5\pi/3$?
One approach, is to force the inverse function $g(r)$ to exist by requiring that the range of $\theta$ be limited to $0 \leq \theta \leq \pi.$  In fact, this is exactly what is done with the Arccos function, and for this very reason.
Normally, when you have a function like $r = f(\theta)$, you want $\theta$ to at least take on the values in some half-open interval of width $(2\pi)$, and perhaps allow $\theta$ to take on the value of any real number.
By specifying $r = f(\theta)$, instead of attempting to specify $\theta = g(r)$, you don't have to worry whether a specific value of $r$ is associated with more than one value of $\theta$.

Edit
Certainly, you can contrive functions, like $r = 2\theta + 4$, where the considerations of the previous section are a non-issue.  However, normally, a function like $r = f(\theta)$ will be used when the function $f$ is performing some trigonometric action, such as some interaction between the sine and cosine functions.

Addendum
Responding to the comment question of Akiva Weinberger.

Don't we have the opposite sort of problem normally? $r=\sqrt{θ}$ and $r=\sqrt{θ + 2\pi}$ give the same curve.

First of all, $~\displaystyle r = \sqrt{\theta} ~$ is only defined on $~\theta \geq 0~$, while
$\displaystyle ~ r = \sqrt{\theta + 2\pi} ~$ is only defined on $~\theta \geq -2\pi.~$
Then, $\displaystyle ~\sqrt{\theta} ~$ is not equal to $\displaystyle ~\sqrt{\theta + 2\pi}. ~$
Second of all, the $~r = f(\theta)~$ approach is normally only used when $~f~$
relates to trigonometry
functions.
So, consider the alternate of function of
$~r = f(\theta) = \sqrt{|\cos \theta|}~$.
This function does not present any difficulty.  It is merely a function that is not injective.  The situation analogizes to setting $y = f(x) = |x| ~: x \in \Bbb{R}.$  You wouldn't be able to construct the inverse function $x = g(y)$, because (for example) $g(5)$ would ambiguously refer to $+5,$ or $-5$.
In the same way, when $~r = f(\theta) = \sqrt{|\cos \theta|}~$,
$f(\theta) = f(\theta + 2\pi)$. 
Notice what happens if you try to construct the alternative corresponding mapping of $\theta = g(r).$  What would $g(1)$ map to, $\theta = 0, \theta = \pi, \theta = 2\pi,$ or what?
You could not use the corresponding mapping to construct an inverse function because the original mapping was not injective.  This is typical for $r = f(\theta)$ functions that involve trigonometric operations.

In a similar vein, consider the polar function $r = \cos^2(\theta) + \sin^2(\theta).$

Addendum-2 
Reacting to the OP's editing of their original posting, re $\theta = \cos(r)$.
This is a function that I never thought of, because of my blind spot that the magnitude should be a function of the angle.
