Упр.7.20 ГДЗ Мордкович Семенов 11 класс (Алгебра)

Задача

а) cos(21?/4) — sin(15?/4) — ctg(33?/4);
б) cos(31?/3) — 2tg(13?/3) — sin(-26?/3);
в) cos(11?/6) + ctg(-31?/6) — 2sin(-29?/6);
г) sin(31?/4) — cos(25?/4) + tg(17?/4);
д) sin(37?/3) — 2ctg(-29?/3) + cos(-19?/3);
е) cos(23?/6) — 3tg(-17?/6) + sin(35?/6).

Подробный ответ

$$\cos \frac{21\pi}{4}-\sin \frac{15\pi}{4}-\ctg \frac{33\pi}{4}=\cos \frac{5\pi}{4}-\sin \frac{7\pi}{4}-\ctg \frac{\pi}{4}$$
$$=-\cos \frac{\pi}{4}-\left(-\sin \frac{\pi}{4}\right)-1=-\frac{\sqrt2}{2}+\frac{\sqrt2}{2}-1=-1.$$
$$\cos \frac{31\pi}{3}-2\tg \frac{13\pi}{3}-\sin \left(-\frac{26\pi}{3}\right)=\cos \frac{\pi}{3}-2\tg \frac{\pi}{3}-\sin \frac{2\pi}{3}$$
$$=\frac12-2\sqrt3-\frac{\sqrt3}{2}=\frac{1-5\sqrt3}{2}.$$
$$\cos \frac{11\pi}{6}+\ctg \left(-\frac{31\pi}{6}\right)-2\sin \left(-\frac{29\pi}{6}\right)=\cos \frac{11\pi}{6}-\ctg \frac{7\pi}{6}+2\sin \frac{5\pi}{6}$$
$$=\frac{\sqrt3}{2}-\sqrt3+2\cdot \frac12=\frac{2-\sqrt3}{2}.$$
$$\sin \frac{31\pi}{4}-\cos \frac{25\pi}{4}+\tg \frac{17\pi}{4}=\sin \frac{7\pi}{4}-\cos \frac{\pi}{4}+\tg \frac{\pi}{4}$$
$$=-\frac{\sqrt2}{2}-\frac{\sqrt2}{2}+1=1-\sqrt2.$$
$$\sin \frac{37\pi}{3}-2\ctg \left(-\frac{29\pi}{3}\right)+\cos \left(-\frac{19\pi}{3}\right)=\sin \frac{\pi}{3}+2\ctg \frac{5\pi}{3}+\cos \frac{\pi}{3}$$
$$=\frac{\sqrt3}{2}+2\cdot \left(-\frac{\sqrt3}{3}\right)+\frac12=\frac{3-\sqrt3}{6}.$$
$$\cos \frac{23\pi}{6}-3\tg \left(-\frac{17\pi}{6}\right)+\sin \frac{35\pi}{6}=\cos \frac{11\pi}{6}+3\tg \frac{5\pi}{6}+\sin \frac{11\pi}{6}$$
$$=\frac{\sqrt3}{2}+3\cdot \left(-\frac{\sqrt3}{3}\right)-\frac12=-\frac{1+\sqrt3}{2}.$$