№10: Найдите $tg 2\alpha$, если $sin \alpha = -\frac{\sqrt{39}}{8}$ и $\pi < \alpha < \frac{3\pi}{2}$.

$\sin \alpha = -\frac{\sqrt{39}}{8}$ $\pi < \alpha < \frac{3\pi}{2}$ (3-я четверть) Так как $\alpha$ находится в 3-ей четверти, $\cos \alpha < 0$. $\cos^2 \alpha = 1 - \sin^2 \alpha = 1 - \left(-\frac{\sqrt{39}}{8}\right)^2 = 1 - \frac{39}{64} = \frac{64 - 39}{64} = \frac{25}{64}$ $\cos \alpha = -\sqrt{\frac{25}{64}} = -\frac{5}{8}$ $\operatorname{tg} \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{-\frac{\sqrt{39}}{8}}{-\frac{5}{8}} = \frac{\sqrt{39}}{5}$ $\operatorname{tg} 2\alpha = \frac{2 \operatorname{tg} \alpha}{1 - \operatorname{tg}^2 \alpha} = \frac{2 \cdot \frac{\sqrt{39}}{5}}{1 - \left(\frac{\sqrt{39}}{5}\right)^2} = \frac{\frac{2\sqrt{39}}{5}}{1 - \frac{39}{25}} = \frac{\frac{2\sqrt{39}}{5}}{\frac{25 - 39}{25}} = \frac{\frac{2\sqrt{39}}{5}}{-\frac{14}{25}} = \frac{2\sqrt{39}}{5} \cdot \left(-\frac{25}{14}\right) = -\frac{5\sqrt{39}}{7}$ Ответ: $-\frac{5\sqrt{39}}{7}$