What's the intuition behind the 2D rotation matrix? Can anyone offer an intuitive proof of why the 2D rotation matrix works?
http://en.wikipedia.org/wiki/Rotation_matrix
I've tried to derive it using polar coordinates to no avail.

There's two ways to think of it. At this stage, I think seeing both might be helpful.
Hard way: Let $(x,y) \in \Bbb R^2$ be represented by polar coordinates $(r, \varphi)$. I mean the relations $x = r \cos \varphi, y = r \sin \varphi$. So, let $(x_\theta, y_\theta)$ be the point after a rotation of $\theta$. Clearly, we have: $$\begin{align} (x_\theta, y_\theta) &= (r \cos(\varphi + \theta), r \sin (\varphi + \theta)) \\ &= (r(\cos \varphi \cos \theta - \sin \varphi \sin \theta), r(\sin \varphi \cos \theta + \cos \varphi \sin \theta)) \\ &= (r \cos \varphi \cos \theta - r \sin \varphi \sin \theta, r \sin \varphi \cos \theta + r \cos \varphi \sin \theta) \\ &= (x \cos \theta - y \sin \theta, x \sin \theta + y \cos \theta) \end{align}$$
To find the matrix, in the standard basis, see that $(1,0)$ goes to $(\cos \theta, \sin \theta)$, and $(0,1)$ goes to $(- \sin \theta, \cos \theta)$. So, the matrix is: $$\begin{pmatrix} \cos \theta & - \sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} $$
Easy way: Let $R_\theta$ be the rotation. See that rotations are linear mappings, hence, it suffices to know the effect of the transformation on a basis, let's say... $((1,0), (0,1))$. Drawing, it is easy to see that $R_\theta (1,0) = (\cos \theta, \sin \theta)$ and $R_\theta (0, 1) = (- \sin \theta, \cos \theta)$. This way, the matrix is: $$\begin{pmatrix} \cos \theta & - \sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}$$
Finally, we can give the transformation: $$(R_\theta (x,y)) = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \cos \theta - y \sin \theta \\ x \sin \theta + y \cos \theta \end{pmatrix}$$
Maybe you don't agree with me on what point of view is easier, but if you get a clear idea of both, I'm happy.
