Упр.15.6 ГДЗ Мордковича 10 класс профильный уровень (Алгебра)

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

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

   $$\sin a=\sin\frac{\pi}{6}=\frac12$$

   $$\cos a=\cos\frac{\pi}{6}=\frac{\sqrt3}{2}$$

   $$\tg a=\frac{\sin a}{\cos a}=\frac{1/2}{\sqrt3/2}=\frac1{\sqrt3}$$

   $$\ctg a=\frac{\cos a}{\sin a}=\frac{\sqrt3/2}{1/2}=\sqrt3$$

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

   $$\sin a=\sin\frac{5\pi}{6}=\frac12$$

   $$\cos a=\cos\frac{5\pi}{6}=-\frac{\sqrt3}{2}$$

   $$\tg a=\frac{\sin a}{\cos a}=\frac{1/2}{-\sqrt3/2}=-\frac1{\sqrt3}$$

   $$\ctg a=\frac{\cos a}{\sin a}=\frac{-\sqrt3/2}{1/2}=-\sqrt3$$

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

   $$\sin a=\sin\frac{7\pi}{6}=-\frac12$$

   $$\cos a=\cos\frac{7\pi}{6}=-\frac{\sqrt3}{2}$$

   $$\tg a=\frac{\sin a}{\cos a}=\frac{-1/2}{-\sqrt3/2}=\frac1{\sqrt3}$$

   $$\ctg a=\frac{\cos a}{\sin a}=\frac{-\sqrt3/2}{-1/2}=\sqrt3$$

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

   $$\sin a=\sin\frac{4\pi}{3}=-\frac{\sqrt3}{2}$$

   $$\cos a=\cos\frac{4\pi}{3}=-\frac12$$

   $$\tg a=\frac{\sin a}{\cos a}=\frac{-\sqrt3/2}{-1/2}=\sqrt3$$

   $$\ctg a=\frac{\cos a}{\sin a}=\frac{-1/2}{-\sqrt3/2}=\frac1{\sqrt3}$$

Ответ:

а) $$\sin a=\frac12,\ \cos a=\frac{\sqrt3}{2},\ \tg a=\frac1{\sqrt3},\ \ctg a=\sqrt3$$;
б) $$\sin a=\frac12,\ \cos a=-\frac{\sqrt3}{2},\ \tg a=-\frac1{\sqrt3},\ \ctg a=-\sqrt3$$;
в) $$\sin a=-\frac12,\ \cos a=-\frac{\sqrt3}{2},\ \tg a=\frac1{\sqrt3},\ \ctg a=\sqrt3$$;
г) $$\sin a=-\frac{\sqrt3}{2},\ \cos a=-\frac12,\ \tg a=\sqrt3,\ \ctg a=\frac1{\sqrt3}$$.