Упр.9.7 ГДЗ Мордкович 10-11 класс (Алгебра)
- а) $$\cos 630^\circ-\sin 1470^\circ-\operatorname{ctg}1125^\circ$$;
б) $$\sin(-7\pi)+2\cos\frac{31\pi}{3}-\operatorname{tg}\frac{7\pi}{4}$$;
в) $$\operatorname{tg}180^\circ-\sin495^\circ+\cos945^\circ$$;
г) $$\cos(-9\pi)+2\sin\left(-\frac{49\pi}{6}\right)-\operatorname{ctg}\left(-\frac{21\pi}{4}\right)$$.
а) $$\cos 630^\circ-\sin 1470^\circ-\ctg 1125^\circ$$
$$\cos 630^\circ-\sin 1470^\circ-\ctg 1125^\circ = \cos(360^\circ+270^\circ)-\sin(4\cdot 360^\circ+30^\circ)-\ctg(3\cdot 360^\circ+45^\circ)$$
$$= \cos 270^\circ-\sin 30^\circ-\ctg 45^\circ = 0-\frac12-1=-\frac32$$
Ответ: $$-\frac32$$
б) $$\sin(-7\pi)+2\cos\frac{31\pi}{3}-\tg\frac{7\pi}{4}$$
$$\sin(-7\pi)+2\cos\frac{31\pi}{3}-\tg\frac{7\pi}{4} = \sin(-7\pi+8\pi)+2\cos\left(10\pi+\frac{\pi}{3}\right)-\tg\left(2\pi-\frac{\pi}{4}\right)$$
$$= \sin\pi+2\cos\frac{\pi}{3}-\tg\left(-\frac{\pi}{4}\right) = 0+2\cdot\frac12-(-1)=2$$
Ответ: $$2$$
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) $$\cos(-9\pi)+2\sin\left(-\frac{49\pi}{6}\right)-\ctg\left(-\frac{21\pi}{4}\right)$$
$$\cos(-9\pi)+2\sin\left(-\frac{49\pi}{6}\right)-\ctg\left(-\frac{21\pi}{4}\right) = \cos 9\pi+2\sin\left(-10\pi-\frac{9\pi}{6}\right)-\ctg\left(-6\pi-\frac{3\pi}{4}\right)$$
$$= \cos 9\pi+2\sin\left(-\frac{3\pi}{2}\right)-\ctg\left(-\frac{3\pi}{4}\right) = -1+2\cdot 1-1=-1$$
Ответ: $$-1$$
