(7) If $\sin \alpha=-\frac{\sqrt{2}}{2}$, and $\cos (\alpha-\beta)=\frac{1}{2}(\beta>0)$, then the minimum value of $\beta$ that satisfies the above conditions is $\qquad$ .
$7 \frac{\pi}{12}$ Hint: Draw the terminal sides of angles $\alpha$ and $\alpha-\beta$. Since $\beta>0$, the minimum value of $\beta$ is $\frac{\pi}{4}-\frac{\pi}{6}=\frac{\pi}{12}$.