Упр.100 ГДЗ Колмогоров 10-11 класс (Алгебра)

Используем периодичность и формулы приведения.

1. $$\tg \frac{18\pi}{5}=\tg\left(3\pi+\frac{3\pi}{5}\right)=\tg \frac{3\pi}{5}$$
   $$\sin \frac{28\pi}{3}=\sin\left(8\pi+\frac{4\pi}{3}\right)=\sin \frac{4\pi}{3}=\sin\left(\pi+\frac{\pi}{3}\right)=-\sin \frac{\pi}{3}$$
2. $$\cos\left(-\frac{15\pi}{8}\right)=\cos \frac{15\pi}{8}=\cos\left(2\pi-\frac{\pi}{8}\right)=\cos \frac{\pi}{8}$$
   $$\ctg\left(-\frac{8\pi}{5}\right)=\ctg\left(\frac{2\pi}{5}-2\pi\right)=\ctg \frac{2\pi}{5}$$
3. $$\sin\left(-\frac{14\pi}{5}\right)=-\sin \frac{14\pi}{5}=-\sin\left(3\pi-\frac{\pi}{5}\right)=-\sin \frac{\pi}{5}$$
   $$\tg \frac{15\pi}{8}=\tg\left(2\pi-\frac{\pi}{8}\right)=\tg\left(-\frac{\pi}{8}\right)=-\tg \frac{\pi}{8}$$
4. $$\cos \frac{20\pi}{7}=\cos\left(2\pi+\frac{6\pi}{7}\right)=\cos \frac{6\pi}{7}=-\cos \frac{\pi}{7}$$
   $$\ctg \frac{35\pi}{9}=\ctg\left(4\pi-\frac{\pi}{9}\right)=\ctg\left(-\frac{\pi}{9}\right)=-\ctg \frac{\pi}{9}$$

Ответ: а) $$\tg \frac{3\pi}{5},\ -\sin \frac{\pi}{3}$$; б) $$\cos \frac{\pi}{8},\ \ctg \frac{2\pi}{5}$$; в) $$-\sin \frac{\pi}{5},\ -\tg \frac{\pi}{8}$$; г) $$-\cos \frac{\pi}{7},\ -\ctg \frac{\pi}{9}$$.