Упр.8.6 ГДЗ Мордкович 10-11 класс (Алгебра)
Переведём углы из градусов в радианы и воспользуемся значениями тригонометрических функций для углов $$30^\circ$$, $$150^\circ$$, $$210^\circ$$ и $$240^\circ$$.
$$30^\circ=\frac{30^\circ\cdot \pi}{180^\circ}=\frac{\pi}{6}$$
Тогда
$$\sin \alpha=\sin \frac{\pi}{6}=\frac{1}{2}, \qquad \cos \alpha=\cos \frac{\pi}{6}=\frac{\sqrt{3}}{2}$$
$$\tg \alpha=\frac{\sin \alpha}{\cos \alpha}=\frac{1/2}{\sqrt{3}/2}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}$$
$$\ctg \alpha=\frac{\cos \alpha}{\sin \alpha}=\frac{\sqrt{3}/2}{1/2}=\sqrt{3}$$
$$150^\circ=\frac{150^\circ\cdot \pi}{180^\circ}=\frac{5\pi}{6}$$
Тогда
$$\sin \alpha=\sin \frac{5\pi}{6}=\frac{1}{2}, \qquad \cos \alpha=\cos \frac{5\pi}{6}=-\frac{\sqrt{3}}{2}$$
$$\tg \alpha=\frac{\sin \alpha}{\cos \alpha}=\frac{1/2}{-\sqrt{3}/2}=-\frac{1}{\sqrt{3}}=-\frac{\sqrt{3}}{3}$$
$$\ctg \alpha=\frac{\cos \alpha}{\sin \alpha}=\frac{-\sqrt{3}/2}{1/2}=-\sqrt{3}$$
$$210^\circ=\frac{210^\circ\cdot \pi}{180^\circ}=\frac{7\pi}{6}$$
Тогда
$$\sin \alpha=\sin \frac{7\pi}{6}=-\frac{1}{2}, \qquad \cos \alpha=\cos \frac{7\pi}{6}=-\frac{\sqrt{3}}{2}$$
$$\tg \alpha=\frac{\sin \alpha}{\cos \alpha}=\frac{-1/2}{-\sqrt{3}/2}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}$$
$$\ctg \alpha=\frac{\cos \alpha}{\sin \alpha}=\frac{-\sqrt{3}/2}{-1/2}=\sqrt{3}$$
$$240^\circ=\frac{240^\circ\cdot \pi}{180^\circ}=\frac{4\pi}{3}$$
Тогда
$$\sin \alpha=\sin \frac{4\pi}{3}=-\frac{\sqrt{3}}{2}, \qquad \cos \alpha=\cos \frac{4\pi}{3}=-\frac{1}{2}$$
$$\tg \alpha=\frac{\sin \alpha}{\cos \alpha}=\frac{-\sqrt{3}/2}{-1/2}=\sqrt{3}$$
$$\ctg \alpha=\frac{\cos \alpha}{\sin \alpha}=\frac{-1/2}{-\sqrt{3}/2}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}$$
Ответ: а) $$\frac{1}{2},\ \frac{\sqrt{3}}{2},\ \frac{\sqrt{3}}{3},\ \sqrt{3}$$; б) $$\frac{1}{2},\ -\frac{\sqrt{3}}{2},\ -\frac{\sqrt{3}}{3},\ -\sqrt{3}$$; в) $$-\frac{1}{2},\ -\frac{\sqrt{3}}{2},\ \frac{\sqrt{3}}{3},\ \sqrt{3}$$; г) $$-\frac{\sqrt{3}}{2},\ -\frac{1}{2},\ \sqrt{3},\ \frac{\sqrt{3}}{3}$$.
