Упр.8.6 ГДЗ Мордкович 10-11 класс (Алгебра)

Переведём углы в радианы и воспользуемся значениями тригонометрических функций для углов $$30^\circ$$, $$150^\circ$$, $$210^\circ$$ и $$240^\circ$$.

а)

$$30^\circ=\frac{30^\circ\cdot\pi}{180^\circ}=\frac{\pi}{6}$$

$$\sin 30^\circ=\sin \frac{\pi}{6}=\frac12,$$
$$\cos 30^\circ=\cos \frac{\pi}{6}=\frac{\sqrt3}{2},$$
$$\tg 30^\circ=\frac{\sin 30^\circ}{\cos 30^\circ}=\frac{1/2}{\sqrt3/2}=\frac{1}{\sqrt3}=\frac{\sqrt3}{3},$$
$$\ctg 30^\circ=\frac{\cos 30^\circ}{\sin 30^\circ}=\frac{\sqrt3/2}{1/2}=\sqrt3.$$

б)

$$150^\circ=\frac{150^\circ\cdot\pi}{180^\circ}=\frac{5\pi}{6}$$

$$\sin 150^\circ=\sin \frac{5\pi}{6}=\frac12,$$
$$\cos 150^\circ=\cos \frac{5\pi}{6}=-\frac{\sqrt3}{2},$$
$$\tg 150^\circ=\frac{\sin 150^\circ}{\cos 150^\circ}=\frac{1/2}{-\sqrt3/2}=-\frac{1}{\sqrt3}=-\frac{\sqrt3}{3},$$
$$\ctg 150^\circ=\frac{\cos 150^\circ}{\sin 150^\circ}=\frac{-\sqrt3/2}{1/2}=-\sqrt3.$$

в)

$$210^\circ=\frac{210^\circ\cdot\pi}{180^\circ}=\frac{7\pi}{6}$$

$$\sin 210^\circ=\sin \frac{7\pi}{6}=-\frac12,$$
$$\cos 210^\circ=\cos \frac{7\pi}{6}=-\frac{\sqrt3}{2},$$
$$\tg 210^\circ=\frac{\sin 210^\circ}{\cos 210^\circ}=\frac{-1/2}{-\sqrt3/2}=\frac{1}{\sqrt3}=\frac{\sqrt3}{3},$$
$$\ctg 210^\circ=\frac{\cos 210^\circ}{\sin 210^\circ}=\frac{-\sqrt3/2}{-1/2}=\sqrt3.$$

г)

$$240^\circ=\frac{240^\circ\cdot\pi}{180^\circ}=\frac{4\pi}{3}$$

$$\sin 240^\circ=\sin \frac{4\pi}{3}=-\frac{\sqrt3}{2},$$
$$\cos 240^\circ=\cos \frac{4\pi}{3}=-\frac12,$$
$$\tg 240^\circ=\frac{\sin 240^\circ}{\cos 240^\circ}=\frac{-\sqrt3/2}{-1/2}=\sqrt3,$$
$$\ctg 240^\circ=\frac{\cos 240^\circ}{\sin 240^\circ}=\frac{-1/2}{-\sqrt3/2}=\frac{1}{\sqrt3}=\frac{\sqrt3}{3}.$$

Ответ

а) $$\frac12,\ \frac{\sqrt3}{2},\ \frac{\sqrt3}{3},\ \sqrt3$$;
б) $$\frac12,\ -\frac{\sqrt3}{2},\ -\frac{\sqrt3}{3},\ -\sqrt3$$;
в) $$-\frac12,\ -\frac{\sqrt3}{2},\ \frac{\sqrt3}{3},\ \sqrt3$$;
г) $$-\frac{\sqrt3}{2},\ -\frac12,\ \sqrt3,\ \frac{\sqrt3}{3}$$.