5. A fixed light ray falls on a mirror, forming an acute angle $\alpha$ with its projection on the mirror's plane. The mirror is rotated around the indicated projection by an acute angle $\beta$. Find the angle between the two reflected rays obtained before and after the rotation.
Answer: $\arccos \left(1-2 \sin ^{2} \alpha \sin ^{2} \beta\right)$.

Solution. Let $A O$ be the incident ray, $O N_{1}$ be the normal to the initial position of the mirror, and $O N_{2}$ be the normal to the second position of the mirror. Consider the trihedral angle $O A N_{1} N_{2}$. Its dihedral angles are equal to: $\angle A O N_{1}=\frac{\pi}{2}-\alpha, \angle N_{1} O N_{2}=\beta$, denote $\angle A O N_{2}=\gamma$. Let $\varphi-$ be the linear angle of the dihedral angle with edge $A O$ (see Fig. 1). By the cosine theorem for the trihedral angle, we get

$$
\cos \varphi=\frac{\cos \beta-\cos \left(\frac{\pi}{2}-\alpha\right) \cos \gamma}{\sin \left(\frac{\pi}{2}-\alpha\right) \sin \gamma}=\frac{\cos \beta-\sin \alpha \cos \gamma}{\cos \alpha \sin \gamma}
$$

Since the dihedral angle $\angle N_{1}$ with edge $O N_{1}$ is a right angle (the normal $O N_{2}$ rotates in a plane perpendicular to the line $\ell$ - the projection of the ray onto the plane of the initial position of the mirror), again from the cosine theorem for the trihedral angle $O A N_{1} N_{2}$ we get

$$
0=\cos \angle N_{1}=\frac{\cos \gamma-\cos \left(\frac{\pi}{2}-\alpha\right) \cos \beta}{\sin \left(\frac{\pi}{2}-\alpha\right) \sin \beta}
$$

from which

$$
\cos \gamma=\sin \alpha \cos \beta
$$

Let $O A_{1}$ be the ray reflected from the initial position of the mirror, and $O A_{2}$ be the ray reflected from the second position of the mirror. Consider the trihedral angle $O A A_{1} A_{2}$. Its dihedral angles are equal to: $\angle A O A_{1}=\pi-2 \alpha, \angle A O A_{2}=2 \gamma, \angle A_{1} O A_{2}-$ is the desired angle. By the cosine theorem for the trihedral angle $O A A_{1} A_{2}$

$$
\cos \varphi=\frac{\cos \angle A_{1} O A_{2}-\cos (\pi-2 \alpha) \cos 2 \gamma}{\sin (\pi-2 \alpha) \sin 2 \gamma}=\frac{\cos \angle A_{1} O A_{2}+\cos 2 \alpha \cos 2 \gamma}{\sin 2 \alpha \sin 2 \gamma}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_024d7be6fbdc002fb8c6g-3.jpg?height=366&width=651&top_left_y=1870&top_left_x=268)

Fig. 1

![](https://cdn.mathpix.com/cropped/2024_05_06_024d7be6fbdc002fb8c6g-3.jpg?height=379&width=665&top_left_y=1855&top_left_x=958)

Fig. 2

From formulas (1)-(3) we get

$$
\begin{array}{r}
\cos \angle A_{1} O A_{2}=4 \cos ^{2} \beta \cos ^{2} \alpha \sin ^{2} \alpha-\left(2 \cos ^{2} \alpha-1\right)\left(2 \cos ^{2} \gamma-1\right)=2 \cos ^{2} \alpha+2 \sin ^{2} \alpha \cos ^{2} \beta-1= \\
=2-2 \sin ^{2} \alpha+2 \sin ^{2} \alpha \cos ^{2} \beta-1=1-2 \sin ^{2} \alpha \sin ^{2} \beta
\end{array}
$$