2. Найти cosa, tga, sin2a, cos2a, если sin a = 7 κ π < α < 3π 25 u 2

Ответ:

• Дано: \[\sin \alpha = -\frac{7}{25}\] и \[\pi < \alpha < \frac{3\pi}{2}\] (3 четверть)
• Находим \[\cos \alpha\]: \[\cos^2 \alpha = 1 - \sin^2 \alpha = 1 - \left(-\frac{7}{25}\right)^2 = 1 - \frac{49}{625} = \frac{576}{625}\] Т.к. \[\alpha\] в 3 четверти, \[\cos \alpha < 0\]: \[\cos \alpha = -\sqrt{\frac{576}{625}} = -\frac{24}{25}\]
• Находим \[\operatorname{tg} \alpha\]: \[\operatorname{tg} \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{-\frac{7}{25}}{-\frac{24}{25}} = \frac{7}{24}\]
• Находим \[\sin 2\alpha\]: \[\sin 2\alpha = 2 \sin \alpha \cos \alpha = 2 \cdot \left(-\frac{7}{25}\right) \cdot \left(-\frac{24}{25}\right) = \frac{336}{625}\]
• Находим \[\cos 2\alpha\]: \[\cos 2\alpha = \cos^2 \alpha - \sin^2 \alpha = \left(-\frac{24}{25}\right)^2 - \left(-\frac{7}{25}\right)^2 = \frac{576}{625} - \frac{49}{625} = \frac{527}{625}\]

Ответ: