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

а) cos 630 — sin 1470 — ctg 1125;
б) sin (-7пи) + 2cos(31пи/3) — tg (7пи/4);
в) tg 1800 — sin 495 + cos 945;
г) cos (-9пи) + 2sin (-49пи/6) — ctg (-21пи/4).

1. $$\cos 630^\circ-\sin 1470^\circ-\ctg 1125^\circ$$

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

   $$0-\frac12-1=-\frac32.$$

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

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

   $$0+2\cdot\frac12-(-1)=2.$$

3. $$\tg 1800^\circ-\sin 495^\circ+\cos 945^\circ$$

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

   $$0-\frac{\sqrt2}{2}-\frac{\sqrt2}{2}=-\sqrt2.$$

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

   $$\cos(-9\pi)=\cos 9\pi=\cos \pi=-1,$$
   $$\sin\left(-\frac{49\pi}{6}\right)=-\sin\frac{49\pi}{6}=-\sin\left(8\pi+\frac{\pi}{6}\right)=-\frac12,$$
   $$\ctg\left(-\frac{21\pi}{4}\right)=-\ctg\frac{21\pi}{4}=-\ctg\left(5\pi+\frac{\pi}{4}\right)=-1.$$

   $$-1+2\cdot\left(-\frac12\right)-(-1)=-1.$$

Ответ

а) $$-\frac32$$; б) $$2$$; в) $$-\sqrt2$$; г) $$-1$$.