3 Вычислите \(sin \alpha\), \(cos \alpha\), \(tg \alpha\) и \(ctg \alpha\) для заданного значения угла: a) 30; 6) 240; в) \(\frac{5\pi}{6}\); г) \(\frac{7\pi}{4}\).

1. а) \(\alpha = 30^\circ\)
   \(sin 30^\circ = \frac{1}{2}\), \(cos 30^\circ = \frac{\sqrt{3}}{2}\), \(tg 30^\circ = \frac{\sqrt{3}}{3}\), \(ctg 30^\circ = \sqrt{3}\)
2. б) \(\alpha = 240^\circ\)
   \(240^\circ = 180^\circ + 60^\circ\) \(sin 240^\circ = -sin 60^\circ = -\frac{\sqrt{3}}{2}\) \(cos 240^\circ = -cos 60^\circ = -\frac{1}{2}\) \(tg 240^\circ = tg 60^\circ = \sqrt{3}\) \(ctg 240^\circ = ctg 60^\circ = \frac{\sqrt{3}}{3}\)
3. в) \(\alpha = \frac{5\pi}{6}\)
   \(\frac{5\pi}{6} = \pi - \frac{\pi}{6} = 180^\circ - 30^\circ = 150^\circ\) \(sin \frac{5\pi}{6} = sin 30^\circ = \frac{1}{2}\) \(cos \frac{5\pi}{6} = -cos 30^\circ = -\frac{\sqrt{3}}{2}\) \(tg \frac{5\pi}{6} = -tg 30^\circ = -\frac{\sqrt{3}}{3}\) \(ctg \frac{5\pi}{6} = -ctg 30^\circ = -\sqrt{3}\)
4. г) \(\alpha = \frac{7\pi}{4}\)
   \(\frac{7\pi}{4} = 2\pi - \frac{\pi}{4} = 360^\circ - 45^\circ = 315^\circ\) \(sin \frac{7\pi}{4} = -sin 45^\circ = -\frac{\sqrt{2}}{2}\) \(cos \frac{7\pi}{4} = cos 45^\circ = \frac{\sqrt{2}}{2}\) \(tg \frac{7\pi}{4} = -tg 45^\circ = -1\) \(ctg \frac{7\pi}{4} = -ctg 45^\circ = -1\)

Ответ: смотри решение выше