Упр.12 ГДЗ Колмогоров 10-11 класс (Алгебра)

а) Приведём аргументы к промежутку $$\left(0;\frac{\pi}{2}\right):$$

$$\sin\frac{7\pi}{8}=\sin\left(\pi-\frac{\pi}{8}\right)=\sin\frac{\pi}{8}$$
$$\cos\left(-\frac{5\pi}{3}\right)=\cos\frac{5\pi}{3}=\cos\left(2\pi-\frac{\pi}{3}\right)=\cos\frac{\pi}{3}$$
$$\tg 0{,}6\pi=\tg\left(0{,}5\pi+0{,}1\pi\right)=-\ctg 0{,}1\pi$$
$$\ctg(-1{,}2\pi)=\ctg\left(-\pi-0{,}2\pi\right)=\ctg(-0{,}2\pi)=-\ctg 0{,}2\pi$$

б) Аналогично:

$$\tg\frac{6\pi}{5}=\tg\left(\pi+\frac{\pi}{5}\right)=\tg\frac{\pi}{5}$$
$$\sin\left(-\frac{5\pi}{9}\right)=-\sin\frac{5\pi}{9}=-\sin\left(\pi-\frac{4\pi}{9}\right)=-\sin\frac{4\pi}{9}$$
$$\cos 1{,}8\pi=\cos\left(2\pi-0{,}2\pi\right)=\cos 0{,}2\pi$$
$$\ctg 0{,}9\pi=\ctg\left(\pi-0{,}1\pi\right)=-\ctg 0{,}1\pi$$

Ответ:

а) $$\sin\frac{\pi}{8},\ \cos\frac{\pi}{3},\ -\ctg 0{,}1\pi,\ -\ctg 0{,}2\pi;$$

б) $$\tg\frac{\pi}{5},\ -\sin\frac{4\pi}{9},\ \cos 0{,}2\pi,\ -\ctg 0{,}1\pi.$$