已知 $\cos \alpha=\frac{1}{7}, \cos (\alpha+\beta)=-\frac{11}{14}$, 且 $\alpha \in\left(0, \frac{\pi}{2}\right), \alpha+\beta \in\left(\frac{\pi}{2}, \pi\right)$,求 $\beta$ 的值.
因为 $\alpha \in\left(0, \frac{\pi}{2}\right), \alpha+\beta \in\left(\frac{\pi}{2}, \pi\right)$, 所以 $\beta \in(0, \pi)$,

因为 $\cos \alpha=\frac{1}{7}, \alpha \in\left(0, \frac{\pi}{2}\right)$, 所以 $\sin \alpha=\sqrt{1-\cos ^{2} \alpha}=\frac{4}{7} \sqrt{3}$,

又因为 $\cos (\alpha+\beta)=-\frac{11}{14}, \alpha+\beta \in\left(\frac{\pi}{2}, \pi\right)$, 所以 $\sin (\alpha+\beta)=\sqrt{1-\cos ^{2}(\alpha+\beta)}=\frac{5}{14} \sqrt{3}$,

$\cos \beta=\cos [(\alpha+\beta)-\alpha]=\cos (\alpha+\beta) \cos \alpha+\sin (\alpha+\beta) \sin \alpha=\frac{1}{7} \times\left(-\frac{11}{14}\right)+\frac{4 \sqrt{3}}{7} \times \frac{5 \sqrt{3}}{14}=\frac{1}{2}$,

因为 $\beta \in(0, \pi)$, 所以 $\beta=\frac{\pi}{3}$.