To rotate the point $(x, y)$ around the origin $(x_0, y_0)$ by the angle $\theta$, we use the formula (where $R$ is the standard Cartesian rotation matrix):
$$\begin{bmatrix} x' \\ y' \end{bmatrix} = \begin{bmatrix} x_0 \\ y_0 \end{bmatrix} + R\begin{bmatrix} x - x_0 \\ y - y_0 \end{bmatrix}$$
(re-center $(x, y)$ relative the the "origin" $(x_0, y_0)$, rotate as usual, then re-offset by $(x_0, y_0)$).
This can be rearranged as:
$$(I - R)\begin{bmatrix} x_0 \\ y_0 \end{bmatrix} + R \begin{bmatrix} x \\ y \end{bmatrix}$$
Is there a geometric interpretation for $I - R$?
I tried playing around with it a bit and was able to show that
$$I - R = 2\sin\left( \frac\theta2 \right) \begin{bmatrix} \sin\left( \frac\theta2 \right) & \cos\left( \frac\theta2 \right) \\ -\cos\left( \frac\theta2 \right) & \sin\left( \frac\theta2 \right) \end{bmatrix}$$
This looks like a new rotation matrix (with $\sin$ swapped for $\cos$, and at a half-angle), but I couldn't see a way to interpret this. I tried playing with the axis/direction by which we define polar coordinates, but couldn't get anything to match up with this, though I think that's the right track.
But even if that's nailed down, I don't know what it means to have the $2 \sin \left( \frac\theta2 \right)$ scaling this matrix.
Any ideas?