1.66 Is the number $\sin \frac{\pi}{18} \sin \frac{3 \pi}{18} \sin \frac{5 \pi}{18} \sin \frac{7 \pi}{18} \sin \frac{9 \pi}{18}$ a rational number?
［Solution］ From the double angle formula, we get
$$\begin{aligned}
\sin \frac{8 \pi}{18} & =2 \sin \frac{4 \pi}{18} \cos \frac{4 \pi}{18}=4 \sin \frac{2 \pi}{18} \cos \frac{2 \pi}{18} \cos \frac{4 \pi}{18} \\
& =8 \sin \frac{\pi}{18} \cos \frac{\pi}{18} \cos \frac{2 \pi}{18} \cos \frac{4 \pi}{18}
\end{aligned}$$

Noting that $\cos \alpha=\sin \left(\frac{\pi}{2}-\alpha\right)$, we know
$$\sin \frac{8 \pi}{18}=8 \sin \frac{\pi}{18} \sin \frac{8 \pi}{18} \sin \frac{7 \pi}{18} \sin \frac{5 \pi}{18}$$

Therefore, $\sin \frac{\pi}{18} \sin \frac{5 \pi}{18} \sin \frac{7 \pi}{18}=\frac{1}{8}$.
Since $\sin \frac{3 \pi}{18}=\sin \frac{\pi}{6}=\frac{1}{2}, \sin \frac{9 \pi}{18}=\sin \frac{\pi}{2}=1$, we have
$$\sin \frac{\pi}{18} \sin \frac{3 \pi}{18} \sin \frac{5 \pi}{18} \sin \frac{7 \pi}{18} \sin \frac{9 \pi}{18}=\frac{1}{16}$$

Thus, the number in the question is a rational number.