Show that for all $(x,y)$ there exists $(r,\theta)$... Given a problem wherein $(x,y) \in \mathbb{R}^2$, I often transform to polar coordinates by introducing the assumption that $(r,\theta)$ satisfy $x = r\cos \theta, y = r\sin \theta.$
Of course, it's only safe to introduce assumptions if you have an existence theorem. This motivates the following.
Theorem 1. For all $(x,y) \in \mathbb{R}^2$, there exists $r \in [0,\infty)$ and $\theta\in \mathbb{R}$ such that $x = r\cos \theta, y = r\sin \theta.$
It occurs to me that I have no idea how to prove this. Ideas, anyone? Note that the above result is equivalent to the following.
Theorem 1'. For the the unique function $f : [0,\infty) \times \mathbb{R} \rightarrow \mathbb{R}^2$ with defining property $f(r,\theta)=(r\cos \theta, r\sin \theta)$, it holds that $f$ is surjective.
Another interesting result adds uniqueness into the mix.
Theorem 2. For all real $\alpha$, we have that for all $(x,y) \in \mathbb{R}^2 \setminus \{(0,0)\}$, there exist unique $r \in (0,\infty)$ and $\theta \in [\alpha,\alpha+2\pi)$ such that $x = r\cos \theta, y = r\sin \theta.$
Again, this can be recast into the language of functions.
Theorem 2'. For all real $\alpha$, we have that for the unique function $g : (0,\infty) \times [\alpha,\alpha+2\pi) \rightarrow \mathbb{R}^2\setminus\{(0,0)\}$ with defining property $g(r,\theta)=(r\cos \theta, r\sin \theta),$ it holds that $g$ is bijective.
I wouldn't know where to start proving any of these.

Now there's also an issue here of how we're defining $\cos$ and $\sin$. I'm thinking it would best to use the following definitions.
Definition 1. There is a unique function $c$ satisfying the initial value problem $c''=-c,$ $c(0)=1,$ $c'(0)=0$. Call it $\cos$.
Definition 2. There is a unique function $s$ satisfying the initial value problem $s''=-s,$ $s(0)=0,$ $s'(0)=1$. Call it $\sin$.
If somebody knows how to prove the theorem(s) using another set of definitions, that is also fine.
 A: Start with the case that $(x,y)$ is in the first quadrant. Then there's a right triangle with vertices $O=(0,0)$, $A=(x,y)$, and $B=(x,0)$, and by high school trig $x=OB=r\cos\theta$ and $y=AB=r\sin\theta$, where $r=OA$ and $\theta$ is the angle $AOB$. 
Then extend the trig functions in the standard way to the other quadrants and prove the formulas work there, too. 
A: In Chapter 8 of Rudin's Principles of mathematical analysis there is a section labelled Trigonometric functions (beginning on page 167 of my edition!). There you will find an argument of the sort you are looking for. But even there the continuity of the argument function $(x,y)\mapsto \phi(x,y)$ is not addressed.
A: To prove something exists, you need to find it. Try $r$ such that $r^2 = x^2 + y^2$ and $\theta$ such that $\tan \theta = \frac{y}{x}$. Of course, this assumes that given a $x\in\mathbb{R}$, $x^2 \in \mathbb{R}$ (which can be proved using group theory) and given $x,y\in\mathbb{R}$ there is some $\theta \in \mathbb{R}$ such that $\tan \theta = \frac{y}{x}$. The latter needs a further restriction that $x \neq n\pi$ for any $n \in \mathbb{Z}$.
