8. If the real numbers $\alpha, \beta, \gamma$ form a geometric sequence with a common ratio of 2, and $\sin \alpha, \sin \beta, \sin \gamma$ form a geometric sequence, then $\cos \alpha=$ $\qquad$
8. $-\frac{1}{2}$.

Let $\beta=2 \alpha, \gamma=4 \alpha$.
$$
\begin{array}{l}
\text { From } \frac{\sin \beta}{\sin \alpha}=\frac{\sin \gamma}{\sin \beta} \\
\Rightarrow \frac{\sin 2 \alpha}{\sin \alpha}=\frac{\sin 4 \alpha}{\sin 2 \alpha} \\
\Rightarrow \frac{2 \sin \alpha \cdot \cos \alpha}{\sin \alpha}=\frac{2 \sin 2 \alpha \cdot \cos 2 \alpha}{\sin 2 \alpha} \\
\Rightarrow 2 \cos ^{2} \alpha-\cos \alpha-1=0 \\
\Rightarrow \cos \alpha=-\frac{1}{2} \text { or } 1 .
\end{array}
$$

If $\cos \alpha=1$, then $\sin \alpha=0$ (discard).
Thus, $\cos \alpha=-\frac{1}{2}$.