Trigonometric relation between angles in billiard dynamics Introduction: In the article Spectral Gap for a Class of Random Billiards by Renato Feres and Hong-Kun Zhang (link: https://www.math.wustl.edu/~feres/spectralgapnew.pdf ), proposition 8 (page 20) does not have a proof. In particular I am interested in the proof of equation (3.2). This proposition and the particular equation can be understood as follows without having to read the article:
Figure 1
Setup (see figure 1, the text is just a description of the figure): Consider a region bounded by four curves $\Gamma_0,\Gamma_1,\Gamma_2$ and $\Gamma_3$. $\Gamma_1$ and $\Gamma_2$ are quarter circles with curvature $K \geq 2$ or radius $0 < \frac{1}{K} \leq \frac{1}{2}$, $\Gamma_0$ is a line tangent to both quarter circles of length 1 and $\Gamma_3$ is a line connecting the other endpoints of the quarter circles. Suppose that a point particle passes through $\Gamma_0$ with an angle $\theta$ (angle with respect to the right side of $\Gamma_0$) and has a strict collision with the quarter circle $\Gamma_1$ at the point $q$ (''strict'' meaning that it is not just grazing, and where ''grazing'' means that the trajectory would be tangent to $\Gamma_1$). Suppose that this collision is according to the standard collision law, meaning that the angle between the incoming vector $v_1$ and the normal vector $n_q$ at $q$ is equal to the angle between the outgoing vector $v_2$ and $n_q$. Suppose furthermore that the particle grazes $\Gamma_2$ (meaning that the trajectory is tangent to $\Gamma_2$) and then exists the region through $\Gamma_0$ again. It exists the region with an angle $\Theta = \Theta_{\theta,1}$ (again with respect to the right side of $\Gamma_0$).
Now equation (3.2) from proposition 8 says that the angles $\theta$ and $\Theta$ are related via:
$$K\sin \Theta = 1 - \cos \left( \frac{\theta + \Theta}{2} \right).$$
My question: How does one prove this relation?
Warning: The point on $\Gamma_2$ where the particles has a grazing collision and the point on $\Gamma_0$ where the particle exits the region are not the same, but this is not very clear from figure 1.
What I have tried (this can be ignored): In the same proposition one can find equation (3.1). This corresponds to the same setup as in figure 1, only now a grazing collision with the quarter circle $\Gamma_2$ is not required. Also one can now enter $\Gamma_0$ through some starting point $r$ in the interval $\left[0 , r_{\theta,1}\right]$. The above setup corresponds to the boundary case $r = r_{\theta,1}$. I have tried to use this equation and substitute for $r$ the boundary value $r_{\theta,1}$. Here I wanted to express $r_{\theta,1}$ in terms of $\theta, \Theta$ and $K$, but I only get very ugly expressions (by using all sorts of identities such as the sinus law and area formulas for triangle etc. with a lot of ugly steps) and these do not seem to give the equation at all.
Thanks a lot in advance!
 A: 
As you can see from our diagram, we have shifted the point of collision $C$ between $\Gamma_1$ and the incoming billiard ball up to get a better view of things going on around the point of grazing $G$ between $\Gamma_2$ and the reflected ball. This has changed the angle $\theta$ from an acute angle to an obtuse angle. However, the scenario remains the same geometrically. Note that we have introduced three angles $\alpha$, $\psi$, and $\omega$ to facilitate the derivation of the sought equation.
Point $E$ is the point of intersection between the extended $MC$ and $AB$. Since $MAE$ is a right-angles triangle, we have, $\measuredangle AEC = 90^o-\alpha$. Using the angles of $\triangle CDE$ we can express $\omega$ as $\omega= 90^o+\alpha-\theta$. Similarly, we use $\triangle CUG$ to obtain another expression for $\omega$ as $\omega=90^o-\alpha-\Theta$. When we eliminate $\omega$ from these two equations, we get,
$$\alpha = \dfrac{\theta-\Theta}{2}.\tag{1}$$
We leave it up to OP to show the following relationship between $\Theta$ and $\psi$.
$$\psi = \Theta\tag{2}$$
Now, pay your attention to the right-angles triangle $CSG$. We shall write,
$$\tan\left(\Theta\right)=\dfrac{CS}{SG}.\tag{3}$$
Next, using (2), we express $TB$ and $GT$ in terms of the radius $r$, $\Theta$ as shown below.
$$TB=NB-NT=r\left(1-\cos\left(\Theta\right)\right)\tag{4}$$
$$GT=r\sin\left(\Theta\right)\tag{5}$$
Since $SF=TB$, using (1) and (4), we shall write the following expression for $CS$.
$$\small{CS=MA-MH-TB=r\left(1-\cos\left(\dfrac{\theta-\Theta}{2}\right)- 1+\cos\left(\Theta\right)\right)=r\left(\cos\left(\Theta\right)-\cos\left(\dfrac{\theta-\Theta}{2}\right)\right)}\tag{6}$$
In similar vein, using (1) and (5), we proceed to derive an equation expressing $SG$.
$$\small{SG=AB-AF-GT=1-{\large{r}}\sin\left(\alpha\right)- {\large{r}}\sin\left(\Theta\right)=1-{\large{r}}\left(\sin\left(\dfrac{\theta-\Theta}{2}\right)+\sin\left(\Theta\right)\right)} \tag{7}$$
Finally, we use (3), (6), and (7) to derive the sought equation (3.2) given in the mentioned article.
$$\tan\left(\Theta\right)=\dfrac{\sin\left(\Theta\right)}{\cos\left(\Theta\right)}=\dfrac{r\left(\cos\left(\Theta\right)-\cos\left(\dfrac{\theta-\Theta}{2}\right)\right)}{1-r\left(\sin\left(\dfrac{\theta-\Theta}{2}\right)+\sin\left(\Theta\right)\right)}$$
We kindly invite OP to cross multiply and simplify the above equation to put the finishing touches to the derivation we started.
