6. 若角 $\alpha$ 终边经过点 $P(-\sqrt{3}, y)$, 且 $\sin \alpha=\frac{\sqrt{3}}{4} y(y \neq 0)$, 则 $\cos \alpha=$
$\because$ 过点 $P(-\sqrt{3}, y), \therefore \sin \alpha=\frac{y}{\sqrt{3+y^{2}}}=\frac{\sqrt{3}}{4} y$.

又 $y \neq 0, \therefore \frac{1}{\sqrt{3+y^{2}}}=\frac{\sqrt{3}}{4}$,

$\therefore|O P|=\sqrt{3+y^{2}}=\frac{4}{\sqrt{3}}=\frac{4 \sqrt{3}}{3}=r$,

$\therefore \cos \alpha=\frac{x}{r}=\frac{-\sqrt{3}}{\frac{4 \sqrt{3}}{3}}=-\frac{3}{4}$.

【答案】 $-\frac{3}{4}$