Задание №2. Вычислите sin t, cos t и tg t, если: a) t = 5π/6; б) t = 5π/4; в) t = 7π/6; г) t = 7π/4.

Решение:

Для каждого значения t вычислим sin t, cos t и tg t, определяя положение угла на единичной окружности.

а) t = 5π/6

• \( \sin \frac{5\pi}{6} = \sin (\pi - \frac{\pi}{6}) = \sin \frac{\pi}{6} = \frac{1}{2} \)


• \( \cos \frac{5\pi}{6} = \cos (\pi - \frac{\pi}{6}) = -\cos \frac{\pi}{6} = -\frac{\sqrt{3}}{2} \)


• \( \operatorname{tg} \frac{5\pi}{6} = \frac{\sin \frac{5\pi}{6}}{\cos \frac{5\pi}{6}} = \frac{1/2}{-\sqrt{3}/2} = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3} \)

б) t = 5π/4

• \( \sin \frac{5\pi}{4} = \sin (\pi + \frac{\pi}{4}) = -\sin \frac{\pi}{4} = -\frac{\sqrt{2}}{2} \)


• \( \cos \frac{5\pi}{4} = \cos (\pi + \frac{\pi}{4}) = -\cos \frac{\pi}{4} = -\frac{\sqrt{2}}{2} \)


• \( \operatorname{tg} \frac{5\pi}{4} = \frac{\sin \frac{5\pi}{4}}{\cos \frac{5\pi}{4}} = \frac{-\sqrt{2}/2}{-\sqrt{2}/2} = 1 \)

в) t = 7π/6

• \( \sin \frac{7\pi}{6} = \sin (\pi + \frac{\pi}{6}) = -\sin \frac{\pi}{6} = -\frac{1}{2} \)


• \( \cos \frac{7\pi}{6} = \cos (\pi + \frac{\pi}{6}) = -\cos \frac{\pi}{6} = -\frac{\sqrt{3}}{2} \)


• \( \operatorname{tg} \frac{7\pi}{6} = \frac{\sin \frac{7\pi}{6}}{\cos \frac{7\pi}{6}} = \frac{-1/2}{-\sqrt{3}/2} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} \)

г) t = 7π/4

• \( \sin \frac{7\pi}{4} = \sin (2\pi - \frac{\pi}{4}) = -\sin \frac{\pi}{4} = -\frac{\sqrt{2}}{2} \)


• \( \cos \frac{7\pi}{4} = \cos (2\pi - \frac{\pi}{4}) = \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2} \)


• \( \operatorname{tg} \frac{7\pi}{4} = \frac{\sin \frac{7\pi}{4}}{\cos \frac{7\pi}{4}} = \frac{-\sqrt{2}/2}{\sqrt{2}/2} = -1 \)

Ответ:

• а) \( \sin \frac{5\pi}{6} = \frac{1}{2}, \cos \frac{5\pi}{6} = -\frac{\sqrt{3}}{2}, \operatorname{tg} \frac{5\pi}{6} = -\frac{\sqrt{3}}{3} \)


• б) \( \sin \frac{5\pi}{4} = -\frac{\sqrt{2}}{2}, \cos \frac{5\pi}{4} = -\frac{\sqrt{2}}{2}, \operatorname{tg} \frac{5\pi}{4} = 1 \)


• в) \( \sin \frac{7\pi}{6} = -\frac{1}{2}, \cos \frac{7\pi}{6} = -\frac{\sqrt{3}}{2}, \operatorname{tg} \frac{7\pi}{6} = \frac{\sqrt{3}}{3} \)


• г) \( \sin \frac{7\pi}{4} = -\frac{\sqrt{2}}{2}, \cos \frac{7\pi}{4} = \frac{\sqrt{2}}{2}, \operatorname{tg} \frac{7\pi}{4} = -1 \)