9. The sequence $\left\{\alpha_{n}\right\}$ is an arithmetic sequence, $\beta$ is the common difference, the sequence $\left\{\sin \alpha_{n}\right\}$ is a geometric sequence, with common ratio $q$, and $\alpha_{1}, \beta \in \mathbf{R}$, then $q=$ $\qquad$
9. -1 or 1

Given $\sin ^{2} \alpha_{2}=\sin \alpha_{1} \sin \alpha_{3}$, i.e., $\frac{1-\cos 2 \alpha_{2}}{2}=-\frac{1}{2}\left[\cos \left(\alpha_{1}+\alpha_{3}\right)-\cos \left(\alpha_{1}-\alpha_{3}\right)\right]$, and $\alpha_{3}-\alpha_{1}=2 \beta, \alpha_{1}+$ $\alpha_{3}=2 \alpha_{2}$, we have $1-\cos 2 \alpha_{2}=-\cos 2 \alpha_{2}+\cos 2 \beta \Rightarrow \beta=k \pi, k \in \mathbf{Z}$.
Therefore, $q=\frac{\sin \alpha_{2}}{\sin \alpha_{1}}=\frac{\sin \left(\alpha_{1}+k \pi\right)}{\sin \alpha_{1}}=(-1)^{k}$, hence $q=1$ or -1.