Упр.8.6 ГДЗ Мордкович 10-11 класс (Алгебра)
а) 30 градусов;
б) 150 градусов;
в) 210 градусов;
г) 240 градусов.
$$30^\circ=\frac{30^\circ\cdot \pi}{180^\circ}=\frac{\pi}{6}$$
$$\sin 30^\circ=\sin \frac{\pi}{6}=\frac{1}{2}$$
$$\cos 30^\circ=\cos \frac{\pi}{6}=\frac{\sqrt{3}}{2}$$
$$\tg 30^\circ=\frac{\sin 30^\circ}{\cos 30^\circ}=\frac{1/2}{\sqrt{3}/2}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}$$
$$\ctg 30^\circ=\frac{\cos 30^\circ}{\sin 30^\circ}=\frac{\sqrt{3}/2}{1/2}=\sqrt{3}$$
$$150^\circ=\frac{150^\circ\cdot \pi}{180^\circ}=\frac{5\pi}{6}$$
$$\sin 150^\circ=\sin \frac{5\pi}{6}=\frac{1}{2}$$
$$\cos 150^\circ=\cos \frac{5\pi}{6}=-\frac{\sqrt{3}}{2}$$
$$\tg 150^\circ=\frac{\sin 150^\circ}{\cos 150^\circ}=\frac{1/2}{-\sqrt{3}/2}=-\frac{1}{\sqrt{3}}=-\frac{\sqrt{3}}{3}$$
$$\ctg 150^\circ=\frac{\cos 150^\circ}{\sin 150^\circ}=\frac{-\sqrt{3}/2}{1/2}=-\sqrt{3}$$
$$210^\circ=\frac{210^\circ\cdot \pi}{180^\circ}=\frac{7\pi}{6}$$
$$\sin 210^\circ=\sin \frac{7\pi}{6}=-\frac{1}{2}$$
$$\cos 210^\circ=\cos \frac{7\pi}{6}=-\frac{\sqrt{3}}{2}$$
$$\tg 210^\circ=\frac{\sin 210^\circ}{\cos 210^\circ}=\frac{-1/2}{-\sqrt{3}/2}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}$$
$$\ctg 210^\circ=\frac{\cos 210^\circ}{\sin 210^\circ}=\frac{-\sqrt{3}/2}{-1/2}=\sqrt{3}$$
$$240^\circ=\frac{240^\circ\cdot \pi}{180^\circ}=\frac{4\pi}{3}$$
$$\sin 240^\circ=\sin \frac{4\pi}{3}=-\frac{\sqrt{3}}{2}$$
$$\cos 240^\circ=\cos \frac{4\pi}{3}=-\frac{1}{2}$$
$$\tg 240^\circ=\frac{\sin 240^\circ}{\cos 240^\circ}=\frac{-\sqrt{3}/2}{-1/2}=\sqrt{3}$$
$$\ctg 240^\circ=\frac{\cos 240^\circ}{\sin 240^\circ}=\frac{-1/2}{-\sqrt{3}/2}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}$$
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) $$-\frac{\sqrt{3}}{2},\ -\frac{1}{2},\ \sqrt{3},\ \frac{\sqrt{3}}{3}$$.
