已知 $\frac{\pi}{2}<\beta<\alpha<\frac{3\pi}{4}$，$\cos(\alpha-\beta)=\frac{12}{13}$，$\sin(\alpha+\beta)=-\frac{3}{5}$. 求 $\sin 2\alpha$ 的值.

\begin{solution} 令 $\delta=\alpha-\beta$，则 $0<\delta<\frac{\pi}{4}$，所以 $\sin\delta=\frac{5}{13}$. 又 $\pi<\alpha+\beta<\frac{3\pi}{2}$，且 $\sin(\alpha+\beta)=-\frac{3}{5}$，故 $\cos(\alpha+\beta)=-\frac{4}{5}$. 于是 $\sin2\alpha=\sin\bigl((\alpha+\beta)+(\alpha-\beta)\bigr)=\sin(\alpha+\beta)\cos\delta+\cos(\alpha+\beta)\sin\delta= -\frac{3}{5}\cdot\frac{12}{13}-\frac{4}{5}\cdot\frac{5}{13}=-\frac{56}{65}$. \end{solution}