求证: $\frac{\sin 2 \alpha+\beta}{\sin \alpha}-2 \cos (\alpha+\beta)=\frac{\sin \beta}{\sin \alpha}$.
$\because \sin (2 \alpha+\beta)-2 \cos (\alpha+\beta) \sin \alpha$$=\sin [(\alpha+\beta)+\alpha]-2 \cos (\alpha+\beta) \sin \alpha$$=\sin (\alpha+\beta) \cos \alpha+\cos (\alpha+\beta) \sin \alpha-2 \cos (\alpha+\beta) \sin \alpha$$=\sin (\alpha+\beta) \cos \alpha-\cos (\alpha+\beta) \sin \alpha$$=\sin [(\alpha+\beta)-\alpha]=\sin \beta$,两边同除以 $\sin \alpha$ 得 $\frac{\sin 2 \alpha+\beta}{\sin \alpha}-2 \cos (\alpha+\beta)=\frac{\sin \beta}{\sin \alpha}$.