The base of a right prism is an isosceles trapezoid with an acute angle $\alpha$. The lateral side of the trapezoid and its smaller base are equal. Find the volume of the prism if the diagonal of the prism is $a$ and forms an angle $\beta$ with the base plane.

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Let $ABCD$ be the base of the given prism $ABCD A1B1C1D1$, and $AD$ and $BC$ be the bases of the trapezoid $ABCD$,

$$
AB = BC = CD, \angle BAD = \angle CDA = \alpha, \angle ACA1 = \beta.
$$

Since $AB = BC$, triangle $ABC$ is isosceles, therefore

$$
\angle CAD = \angle ACB = \angle BAC = \frac{\pi}{2}
$$

From the right triangle $ACA1$, we find that

$$
AA1 = A1C \sin \angle ACA1 = a \sin \beta, AC = A1C \cos \angle ACA1 = a \cos \beta
$$

Let $CK$ be the height of the trapezoid $ABCD$. Then

$$
\begin{gathered}
DK = \frac{1}{2}(AD - BC), AK = AD - DK = AD - \frac{1}{2}(AD - BC) = \frac{1}{2}(AD + BC) \\
CK = AC \sin \angle CAK = a \cos \beta \sin \frac{\frac{\pi}{2}}{2} \\
AK = AC \cos \angle CAK = a \cos \beta \cos \frac{\frac{\alpha}{2}}{2}
\end{gathered}
$$

Therefore,

$$
S_{ABCD} = \frac{1}{2}(AD + BC) \cdot CK = AK \cdot CK = a \cos \beta \sin \frac{\frac{\alpha}{2}}{2} \cdot a \cos \beta \cos \frac{\frac{\alpha}{2}}{2} = \frac{1}{2} a^2 \cos^2 \beta \sin \alpha
$$

Thus,

$$
\begin{aligned}
V_{ABCD A1B1C1D1} & = S_{ABCD} \cdot AA1 = \frac{1}{2} a^2 \cos^2 \beta \sin \alpha \cdot a \sin \beta = \\
& = \frac{1}{2} a^3 \sin \alpha \sin \beta \cos^2 \beta
\end{aligned}
$$