2. It is known that the numbers $x, y, z$ form an arithmetic progression in the given order with a common difference $\alpha=\arcsin \frac{\sqrt{7}}{4}$, and the numbers $\frac{1}{\sin x}, \frac{4}{\sin y}, \frac{1}{\sin z}$ also form an arithmetic progression in the given order. Find $\sin ^{2} y$.
Answer: $\frac{7}{13}$.

Solution: Using the characteristic property of an arithmetic progression, we get $y=\frac{x+z}{2}$, $\frac{8}{\sin y}=\frac{1}{\sin x}+\frac{1}{\sin z}$. Transform the second equation:

$$
\frac{8}{\sin y}=\frac{\sin x+\sin z}{\sin x \sin z} \Leftrightarrow \frac{8}{\sin y}=\frac{2 \sin \frac{x+z}{2} \cos \frac{x-z}{2}}{0.5(\cos (x-z)-\cos (x+z))}
$$

Substituting here that $x+z=2 y, x-z=-2 \alpha$, we get $\frac{2}{\sin y}=\frac{\sin y \cos \alpha}{\cos 2 \alpha-\cos 2 y}$. This equality is equivalent to the following on the domain of definition:

$$
2-4 \sin ^{2} \alpha-2+4 \sin ^{2} y=\sin ^{2} y \cos \alpha \Leftrightarrow(4-\cos \alpha) \sin ^{2} y=4 \sin ^{2} \alpha \Leftrightarrow \sin ^{2} y=\frac{4 \sin ^{2} \alpha}{4-\cos \alpha}
$$

Since $\alpha=\arcsin \frac{\sqrt{7}}{4}$, then $\sin \alpha=\frac{\sqrt{7}}{4}, \cos \alpha=\frac{3}{4}$, and finally we get $\sin ^{2} y=\frac{7}{13}$.