Reflections and Rotations I can see how this works in my head, but I can't prove it analitically.
Let $R_\theta$ be the matrix \begin{pmatrix}\cos\theta&\!\!-\sin\theta\\\sin\theta&\;\cos\theta\end{pmatrix} and $F_\theta$ = $R_\theta F$, where $F$ is \begin{pmatrix}1&\!\!0\\0&\;-1\end{pmatrix} Show that $F_\theta$ represents the reflection whose axis is the line that contains $(0,0)$ and makes an angle of $\theta/2$ with the x-axis.
That matrix turns an arbitrary point $(x,y)$ in $(x\cos(\theta)+y\sin(\theta), x\sin(\theta)-y\cos(\theta))$, but I don't know how to represent it.
 A: You are performing linear transformations in a vector space so, since linearity, you can find the the transformation  simply finding how the transformation operates on the vectors of a ortho-normal basis.
Let the basis be:
$$
\mathbf{i}=
\left[
\begin{array}{ccc}
1\\
0
\end{array}
\right]
\qquad
\mathbf{j}=
\left[
\begin{array}{ccc}
0\\
1
\end{array}
\right]
$$
In the figure you see how the transformations $F$ and $R_{\theta}$ operate.

The product $R_{\theta}F$ is an orthogonal symmetry on $x$ axis followed by a rotation of angle $\theta$. And from the figure you can easily see that this is the same as a orthogonal symmetry with respect the straight line passing through the origin and forming an angle $\theta/2$ with the $x$ axis.
The matrix representing this transformation has as columns the vectors $ R_{\theta}F (\mathbf{i})$ and $ R_{\theta}F (\mathbf{j})$, i.e.
$$
\left[
\begin{array}{ccc}
\cos \theta& \sin \theta \\
\sin \theta &-\cos \theta 
\end{array}
\right]
$$
as you have find.
A: Hint:$F$ will turn $(x,y)$ into $(x,-y)$, so $F$ is an reflection whose axis is $x$ axis. And thus $FR_{\theta}$ means first $R_{\theta}$, then $F$.
A: if you know that $\pmatrix{\cos \theta&-\sin \theta\\\sin \theta&\cos \theta}$ represents reflection, then we can figure where the mirror is just by looking at where $(1,0)^T,$ the $x$-axis is mapped to. in this case it is mapped to the line $y = \tan \theta x$ because the firat column of the matrix is $(\cos \theta, \sin \theta)^T$ and the mirror is halfway between the $x$-axis and $y = \tan \theta x.$ this is the line
$y = x\tan \frac{\theta}{2}.$ 
