9.已知 $\cos (\alpha-\beta)=-\frac{12}{13}, \cos (\alpha+\beta)=\frac{12}{13}$, 且 $\alpha-\beta \in\left(\frac{\pi}{2}, \pi\right), \alpha+\beta \in\left(\frac{3 \pi}{2}, 2 \pi\right)$. 求 $\cos 2 \alpha, \cos 2 \beta$ 及角 $\beta$ 的值. [思路分析] 讨论角的范围时, $\alpha-\beta$ 一般看作 $\alpha+(-\beta)$, 先求出 $-\beta$ 的范围, 再求 $\alpha+(-\beta)$ 的范围.
由 $\alpha-\beta \in\left(\frac{\pi}{2}, \pi\right)$, 且 $\cos (\alpha-\beta)=-\frac{12}{13}$, 得 $\sin (\alpha-\beta)=\frac{5}{13}$.

由 $\alpha+\beta \in\left(\frac{3 \pi}{2}, 2 \pi\right)$, 且 $\cos (\alpha+\beta)=\frac{12}{13}$.

得 $\sin (\alpha+\beta)=-\frac{5}{13}$.

$\therefore \cos 2 \alpha=\cos [(\alpha+\beta)+(\alpha-\beta)]$

$=\cos (\alpha+\beta) \cos (\alpha-\beta)-\sin (\alpha+\beta) \sin (\alpha-\beta)$

$=-\frac{12}{13} \times \frac{12}{13}-\left(-\frac{5}{13}\right) \times \frac{5}{13}=-\frac{119}{169}$.

$\cos 2 \beta=\cos [(\alpha+\beta)-(\alpha-\beta)]$

$=\cos (\alpha+\beta) \cos (\alpha-\beta)+\sin (\alpha+\beta) \sin (\alpha-\beta)$

$=-\frac{12}{13} \times \frac{12}{13}+\left(-\frac{5}{13}\right) \times \frac{5}{13}=-1$.

又 $\because \alpha+\beta \in\left(\frac{3}{2} \pi, 2 \pi\right)$,

$\alpha-\beta \in\left(\frac{\pi}{2}, \pi\right) \Rightarrow \beta-\alpha \in\left(-\pi,-\frac{\pi}{2}\right)$

$\Rightarrow 2 \beta \in\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)$.

$\therefore 2 \beta=\pi$, 则 $\beta=\frac{\pi}{2}$.