Упр.9.7 ГДЗ Мордкович 10-11 класс (Алгебра)

1. а) $$\cos 630-\sin 1470-\operatorname{ctg}1125$$; б) $$\sin(-7\pi)+2\cos\left(\frac{31\pi}{3}\right)-\operatorname{tg}\left(\frac{7\pi}{4}\right)$$; в) $$\operatorname{tg}1800-\sin495+\cos945$$; г) $$\cos(-9\pi)+2\sin\left(-\frac{49\pi}{6}\right)-\operatorname{ctg}\left(-\frac{21\pi}{4}\right)$$.

а) $$\cos 630^\circ-\sin 1470^\circ-\operatorname{ctg}1125^\circ$$

$$\cos 630^\circ-\sin 1470^\circ-\operatorname{ctg}1125^\circ = \cos(360^\circ+270^\circ)-\sin(4\cdot 360^\circ+30^\circ)-\operatorname{ctg}(3\cdot 360^\circ+45^\circ)$$
$$= \cos 270^\circ-\sin 30^\circ-\operatorname{ctg}45^\circ = 0-\frac12-1=-\frac32.$$

Ответ: $$-\frac32$$

б) $$\sin(-7\pi)+2\cos\frac{31\pi}{3}-\tg\frac{7\pi}{4}$$

$$\sin(-7\pi)+2\cos\frac{31\pi}{3}-\tg\frac{7\pi}{4} = \sin(-7\pi+8\pi)+2\cos\left(10\pi+\frac{\pi}{3}\right)-\tg\left(2\pi+\frac{\pi}{4}\right)$$
$$= \sin\pi+2\cos\frac{\pi}{3}-\tg\frac{\pi}{4} = 0+2\cdot\frac12-1=1.$$

Ответ: $$1$$

в) $$\tg 1800^\circ-\sin 495^\circ+\cos 945^\circ$$

$$\tg 1800^\circ-\sin 495^\circ+\cos 945^\circ = \tg(10\cdot 180^\circ)-\sin(360^\circ+135^\circ)+\cos(2\cdot 360^\circ+225^\circ)$$
$$= \tg 0^\circ-\sin 135^\circ+\cos 225^\circ = 0-\frac{\sqrt2}{2}-\frac{\sqrt2}{2}=-\sqrt2.$$

Ответ: $$-\sqrt2$$

г) $$\cos(-9\pi)+2\sin\left(-\frac{49\pi}{6}\right)-\operatorname{ctg}\left(-\frac{21\pi}{4}\right)$$

$$\cos(-9\pi)+2\sin\left(-\frac{49\pi}{6}\right)-\operatorname{ctg}\left(-\frac{21\pi}{4}\right) = \cos 9\pi+2\sin\left(-8\pi-\frac{\pi}{6}\right)-\operatorname{ctg}\left(-6\pi-\frac{3\pi}{4}\right)$$
$$= \cos 9\pi-2\sin\frac{\pi}{6}+\operatorname{ctg}\frac{\pi}{4} = -1-2\cdot\frac12+1=-1.$$

Ответ: $$-1$$