若 $\sin (-\alpha)=\frac{1}{3}, \alpha \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, 则 $\cos (\pi+\alpha)=$
$\because \sin (-\alpha)=\frac{1}{3}, \therefore \sin \alpha=-\frac{1}{3} \because \alpha \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right), \therefore \cos \alpha=\sqrt{1-\left(-\frac{1}{3}\right)}=\frac{2 \sqrt{2}}{3}, \therefore \cos (\pi+\alpha)=-\cos \alpha$ $=-\frac{2 \sqrt{2}}{3}$.