17. $\alpha_{i} \in\left(0, \frac{\pi}{2}\right)(i=1,2,3, \cdots, n, n \geqslant 2)$, and $\prod_{i=1}^{n} \sin \alpha_{i}=a$ (a constant), find the maximum value of $\prod_{i=1}^{n} \cos \alpha_{i}$.
17. Let $x_{i}=\sin \alpha_{i}$, then $\cos \alpha_{i}=\sqrt{1-x_{i}^{2}}(i=1,2,3, \cdots, n), \prod_{i=1}^{n} x_{i}=a,\left(\prod_{i=1}^{n} \cos \alpha_{i}\right)^{2}=a \prod_{i=1}^{n}\left(\frac{1}{x_{i}}-x_{i}\right)$
$$
\sqrt[n]{\frac{\prod_{i=1}^{n} x_{i}}{\prod_{i=1}^{n} \frac{1}{x_{i}}}} \leqslant 1 \text {, i.e., } \sqrt[n]{\prod_{i=1}^{n}\left(\frac{1}{x_{i}}-x_{i}\right)} \leqslant \sqrt[n]{\prod_{i=1}^{n} \frac{1}{x_{i}}}-\sqrt[n]{\prod_{i=1}^{n} x_{i}} \leqslant \frac{1}{\sqrt[n]{a}}-\sqrt[n]{a} \text {, }
$$

Therefore, $\prod_{i=1}^{n} \cos \alpha_{i}=\sqrt{a \prod_{i=1}^{n}\left(\frac{1}{x_{i}}-x_{i}\right)} \leqslant \sqrt{a\left(\frac{1}{\sqrt[n]{a}}-\sqrt[n]{a}\right)^{n}}$. The equality holds if and only if $\frac{x_{i}}{\frac{1}{x_{i}}}=c_{1}$ (constant), $\frac{\frac{1}{x_{i}}-x_{i}}{\frac{1}{x_{i}}}=c_{2}$ (constant) $(i=1,2,3, \cdots)$,
i.e., $x_{1}=x_{2}=x_{3}=\cdots=x_{n}$, which means $\alpha_{1}=\alpha_{2}=\alpha_{3}=\cdots=\alpha_{n}$,
so the maximum value of $\prod_{i=1}^{n} \cos \alpha_{i}$ is $\sqrt{a\left(\frac{1}{\sqrt[n]{a}}-\sqrt[n]{a}\right)^{n}}$.