Найдите tg 2a, если cosa = rac{2√6}{5} и rac{3π}{2} < α < 2π.

Дано: \[\cos \alpha = \frac{2\sqrt{6}}{5}, \frac{3\pi}{2} < \alpha < 2\pi\] Нужно найти: \[\tan 2\alpha\] Сначала найдем \(\sin \alpha\). Так как \(\alpha\) находится в четвертой четверти, то \(\sin \alpha < 0\). \[\sin^2 \alpha + \cos^2 \alpha = 1\] \[\sin^2 \alpha = 1 - \cos^2 \alpha\] \[\sin^2 \alpha = 1 - \left(\frac{2\sqrt{6}}{5}\right)^2 = 1 - \frac{4 \cdot 6}{25} = 1 - \frac{24}{25} = \frac{1}{25}\] \[\sin \alpha = -\sqrt{\frac{1}{25}} = -\frac{1}{5}\] Теперь найдем \(\tan \alpha\): \[\tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{-\frac{1}{5}}{\frac{2\sqrt{6}}{5}} = -\frac{1}{2\sqrt{6}} = -\frac{\sqrt{6}}{12}\] Теперь найдем \(\tan 2\alpha\): \[\tan 2\alpha = \frac{2 \tan \alpha}{1 - \tan^2 \alpha} = \frac{2 \cdot \left(-\frac{\sqrt{6}}{12}\right)}{1 - \left(-\frac{\sqrt{6}}{12}\right)^2} = \frac{-\frac{\sqrt{6}}{6}}{1 - \frac{6}{144}} = \frac{-\frac{\sqrt{6}}{6}}{1 - \frac{1}{24}} = \frac{-\frac{\sqrt{6}}{6}}{\frac{23}{24}} = -\frac{\sqrt{6}}{6} \cdot \frac{24}{23} = -\frac{4\sqrt{6}}{23}\] Ответ: \(-\frac{4\sqrt{6}}{23}\)