I am trying to show that
$$ F = \left(\begin{array}{cc} \cos \theta & \sin \theta \\ \sin \theta & -\cos \theta \end{array} \right ) $$
is a flip about a line through the origin.
What I tried: Let $v \in \mathbb R^2$, $v = (r \cos \varphi, r \sin \varphi)$. Then
$$ vF = (r \cos (\varphi - \theta), -r \sin (\varphi - \theta) )$$
I notice that if $\theta = 2 \varphi$ then $v$ maps to itself. From this I conjecture that this is a flip about the line through the origin of angle $\theta / 2$.
But there I have two problems:
(1) I don't know how to prove this conjecture. Please can you help me?
(2) I computed $(1,0)F = (\cos \theta, \sin \theta)$. If it was a reflection around the line $\theta$ shouldn't $(1,0)F = (\cos 2\theta, \sin 2\theta)$?
To start with, a more convenient way to write $vF$ is by using $\cos(x)=\cos(-x)$ and $\sin(x)=-\sin(-x)$ to find $$vF = (r\cos(\theta-\varphi) , r \sin(\theta - \varphi))$$
Now, a line given by an angle $\beta$ reflected about a line with angle $\alpha$ gives a line with angle $\beta + 2(\alpha-\beta) = 2\alpha - \beta$ (draw a sketch). This gives the equation
$$2\alpha - \varphi = \theta - \varphi \to \alpha = \frac{\theta}{2}$$