已知 $\sin \alpha=\frac{2 \sqrt{2}}{3}, \cos (\alpha+\beta)=-\frac{1}{3}$, 且 $\alpha, \beta \in\left(0, \frac{\pi}{2}\right)$, 则 $\sin (\alpha-\beta)$ 的值等于
由于 $\sin \alpha=\frac{2 \sqrt{2}}{3}$, 所以 $\cos \alpha=\frac{1}{3}, \sin 2 \alpha=\frac{4 \sqrt{2}}{9}, \cos 2 \alpha=-\frac{7}{9}$, 由于 $\cos (\alpha+\beta)=-\frac{1}{3}$,
$\sin (\alpha+\beta)=\frac{2 \sqrt{2}}{3}$,

$\sin (\alpha-\beta)=\sin [2 \alpha-(\alpha+\beta)]=\sin 2 \alpha \cos (\alpha+\beta)-\cos 2 \alpha \sin (\alpha+\beta)=\frac{10 \sqrt{2}}{27}$.