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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.

goblin GONE
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  • What is your definition of $\cos$ and $\sin\ $? – Christian Blatter Mar 25 '13 at 12:19
  • @ChristianBlatter The definitions are under the horizontal line! – goblin GONE Mar 25 '13 at 12:23
  • Use pythagorean theorem together with the fact that $\cos^2 \theta + \sin^2 \theta = 1$, and $\cos$ and $\sin$ are $2\pi$-periodic. Thus you can define $\sin$ and $\cos$ by their power series, using if you want your differential equations, then rediscover everything about them, including basic geometric properties. – Jean-Claude Arbaut Mar 25 '13 at 13:22

3 Answers3

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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.

Gerry Myerson
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  • To be honest, I don't really know enough geometry to be able to tell if this is rigorous or not. If you are able to replace the phrase "by high school trig" with "by Theorem ..., whose proof can be found at ...," your answer will be more useful to me. – goblin GONE Mar 26 '13 at 07:02
  • Facts from high school trig are rarely published as theorems. If you have a right triangle then the ratio of the side opposite angle C to the hypotenuse depends only on angle C and not on the size of the triangle, by similar triangles. The value of this ratio is then called the sine of C. Also, the ratio of the adjacent side to the hypotenuse is called the cosine of C. If we call the length of the hypotenuse $r$, it is now immediate that the length of the adjacent side is $r\cos C$ and the length of the side opposite is $r\sin C$. But maybe I am missing your point. – Gerry Myerson Mar 26 '13 at 12:00
  • It's a good answer, I guess I was just expecting something else. I should probably learn axiomatic geometry at some point, and when I've done that I'm sure I'll look back at your proof and find it completely satisfying. – goblin GONE Mar 26 '13 at 12:10
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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.

Christian Blatter
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  • Thanks, I'll check it out. I think continuity is immediate. Since $\sin$ and $\cos$ are continuous, it follows that the functions $(r,\theta) \mapsto r \cos \theta, r\sin\theta$ are continuous. So $(r,\theta) \mapsto (r \cos \theta, r\sin\theta)$ is continuous. – goblin GONE Mar 27 '13 at 00:50
  • @user18921: I was talking about the continuity of the inverse $(x,y)\mapsto \phi(x,y)$. – Christian Blatter Mar 27 '13 at 09:05
  • Oh i see. Sorry I misread that part of your answer. – goblin GONE Mar 27 '13 at 09:32
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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 $n \in \mathbb{Z}$.

Code-Guru
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