What's the angle of rotation of a product of two reflections? Let $F_1$ and $F_2$ be two arbitrary reflections about two lines in $\mathbb R^2$. 
I've been trying to work out the angle of rotation of $R_1R_2$. To this end I drew pictures in which I reflect one point first along $R_1$ then along $R_2$. Then my plan was to calculate the angle of rotation between the point and its image but the problem I ran into was that I don't have the center of rotation. 

Could someone help me and explain to me how to find the angle of $F_1
 F_2$ (the product)?

Edit If possible using a geometric argument. 
 A: 
The figure gives a simple geometric answer. 
Let $O$ the fixed point of the two reflections that is the fixed point of the rotation. Than:
$P'$ is the relfection of $P$ in the line $OM$ and $P''$ the reflection of $P'$ in $ON$ and we have 
$$
\angle POM=\angle P'OM
$$
$$
\angle P'ON=\angle P''ON
$$
and $\angle P'ON+\angle P'OM= \angle MON$, so $\angle POP''=2 \angle MON$
A: We can assume that the two lines intersect at the origin. The matrix that rotates the plane by angle $\theta$ is 
$$\text{Rot}(\theta)=\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}.$$
The matrix that reflects the plane in the line $y=x\tan\theta$ is
$$\text{Ref}(\theta)=\begin{bmatrix}\cos2\theta&\sin2\theta\\\sin2\theta&-\cos2\theta\end{bmatrix}.$$
Now with use of the compound angle formulae and matrix multiplication, it is not all that difficult to show that
$$\text{Ref}(\theta)\text{Ref}(\phi)=\text{Rot}(2(\theta-\phi)).$$
A: Let $L_1, L_2$ be the lines through which the reflections $F_1, F_2$ (respectively) reflect the plane.
The reflection $F_1$ does not "move" any point on the line $L_1$, that is, each of those points is its own image under $F_1$, and $F_2$ does not "move" any point on the line $L_2$.
In other words, if $P \in L_1$ ($P$ is on line $L_1$)
then $F_1(P) = P$ (the reflection $F_1$ does not "move" $P$), 
and if $Q \in L_2$ then $F_2(Q) = Q$.
Let the point $O$ be the intersection of $L_1$ and $L_2$.
Then $F_1(O) = O$ (because $O \in L_1$) and $F_2(O) = O$ (because $O \in L_2$).
Therefore $F_2(F_1(O)) = F_2(O) = O$,
which is just a way of saying in an equation that the point $O$
never "moved" as we first applied the reflection $F_1$ and then
applied the reflection $F_2$.
But if $L_1 \neq L_2$, every other point in the plane (not including their point of intersection) "moves" under the transformation composed of $F_1$ followed by $F_2$.
Let $L_1 \neq L_2$, and consider any point $P$. 
Reflect through $L_1$ to point $P'$, and consider the image of $P'$
under each of the two reflections:
$F_1(P')$, the reflection of $P'$ through $L_1$,
and $F_2(P')$, the reflection of $P'$ through $L_2$.
The reflection $F_1$ takes $P'$ back to $P$, so $F_1(P') = P$,
and the statement $F_2(F_1(P)) = P$ is equivalent to $F_2(P') = F_1(P')$.
But $F_1(P')$ is on a line $M_1$ through $P'$ perpendicular to $L_1$,
and $F_2(P')$ is on a line $M_2$ through $P'$ perpendicular to $L_2$.
Since $L_1$ and $L_2$ are not parallel, $M_1$ and $M_2$ are not parallel;
so in order to have $F_2(P') = F_1(P')$ where $F_1(P')$ is on $M_1$
and $F_2(P')$ is on $M_2$, $F_2(P')$ must be at the intersection of
$M_1$ and $M_2$, that is, it must be $P'$; and likewise for $F_1(P')$.
That is, neither of the reflections "moves" $P'$.
Therefore $P'$ must be on $L_1$ and on $L_1$, that is,
$F_1(P') = P' = O$, and $P = F_1(P') = O$.
That is, if we have a point that is returned to its starting position
under the composition of the two reflections, that point must be $O$.
So if the two reflections compose to a rotation,
the intersection of the two lines is the only possible point
that could be the center of the rotation.
To find the angle of rotation, start with any point $P$ (other than $O$)
on line $L_1$. The image of $P$ under $F_1$ is just $P$.
To find $F_2(P)$, we find line $M_2$ through $P$ perpendicular to $L_2$.
Let $A$ be the intersection of $M_2$ with $L_2$.
Then $F_2(P) = P''$ such that $\triangle POA$ and $\triangle P''OA$
are congruent (but mirror-imaged) triangles, and 
$\angle POP'' = \angle POA + \angle AOP'' = 2 \angle POA$,
that is, the angle of rotation $\angle POP''$ is exactly twice the
angle $\angle POA$ between the two lines.
To confirm that every point not on $L_1$ also is rotated by twice the
angle between the lines is just a slightly more complicated exercise
involving two pairs of congruent triangles.
