How can I show that $$ \sigma_\phi \circ \sigma_\psi=\rho_{2(\phi-\psi)}, $$ where $\sigma_\phi$ is a reflection about a line making an angle $\phi$ with the x-axis, and $\rho_\psi$ is a rotation about the origin with angle $\psi$. Is it possible to show this is true geometrically? Or is the only way to do this multiply the matrices that correspond with the rotation and reflection?
EDIT
The way I see it now is:
Consider a vector on the x-axis (wlog), the vector first moves $2\psi$, and then $2(\phi-2\psi)$. So we get: $2\psi+2(\phi-2\psi)=2(\phi-\psi).$
You said in the chat that you wanted a geometric interpretation of this problem, so here it is! Consider the point $X$ we're applying these two reflections on. Consider $X' = \sigma_\psi(X)$ and $X'' = \sigma_\phi(X') = (\sigma_\phi \circ \sigma_\psi)(X)$.
Let's draw a figure. Now, notice that line $X'X$ is perpendicular to the first line of reflection. Also, $X'X''$ is perpendicular to the second line of reflection.
Therefore, we conclude that $OX = OX' = OX''$; Therefore, there is a circle centered at $O$ through these three points:
Now what is $\angle XOX''$? Well, since rotation preserves angles, let's WLOG assume $X$ lies on the $x$-axis.
Now, $\angle X''OX = \angle X''X' + \angle X'X = 2\psi + 2\phi$. Therefore, the angle counterclockwise is $-2\psi + 2\phi = 2(\phi-\psi)$.