设 $\alpha$ 为第四象限角, 且 $\frac{\sin 3 \alpha}{\sin \alpha}=\frac{13}{5}$, 则 $\tan 2 \alpha=$
因为 $\frac{\sin 3 \alpha}{\sin \alpha}=\frac{\sin (2 \alpha+\alpha)}{\sin \alpha}$

$=\frac{\sin 2 \alpha \cos \alpha+\cos 2 \alpha \sin \alpha}{\sin \alpha}$

$=\frac{2 \sin \alpha \cos ^{2} \alpha+\left(2 \cos ^{2} \alpha-1\right) \sin \alpha}{\sin \alpha}$

$=\frac{4 \sin \alpha \cos ^{2} \alpha-\sin \alpha}{\sin \alpha}$

$=4 \cos ^{2} \alpha-1=2\left(2 \cos ^{2} \alpha-1\right)+1=2 \cos 2 \alpha+1$

$=\frac{13}{5}$, 所以 $\cos 2 \alpha=\frac{4}{5}$.

又 $\alpha$ 是第四象限角, 所以 $\sin 2 \alpha=-\frac{3}{5}, \tan 2 \alpha=-\frac{3}{4}$.