Упр.531 ГДЗ Алимов 10-11 класс (Алгебра)
Вычислить:
- $$\cos\frac{23\pi}{4}-\sin\frac{15\pi}{4}-\operatorname{ctg}\left(-\frac{11\pi}{2}\right)$$;
- $$\sin\frac{25\pi}{3}-\cos\left(-\frac{17\pi}{2}\right)-\operatorname{tg}\frac{10\pi}{3}$$;
- $$\sin(-7\pi)-2\cos\frac{31\pi}{3}-\operatorname{tg}\frac{7\pi}{4}$$;
- $$\cos(-9\pi)+2\sin\left(-\frac{49\pi}{6}\right)-\operatorname{ctg}\left(-\frac{21\pi}{4}\right)$$.
1) $$\cos \frac{23\pi}{4}-\sin \frac{15\pi}{4}-\ctg\left(-\frac{11\pi}{2}\right)$$
$$\cos \frac{23\pi}{4}=\cos\left(6\pi-\frac{\pi}{4}\right)=\cos \frac{\pi}{4}=\frac{\sqrt2}{2}$$
$$\sin \frac{15\pi}{4}=\sin\left(4\pi-\frac{\pi}{4}\right)=-\sin \frac{\pi}{4}=-\frac{\sqrt2}{2}$$
$$\ctg\left(-\frac{11\pi}{2}\right)=\ctg\left(-6\pi+\frac{\pi}{2}\right)=\ctg \frac{\pi}{2}=0$$
$$\frac{\sqrt2}{2}-\left(-\frac{\sqrt2}{2}\right)-0=\sqrt2$$
2) $$\sin \frac{25\pi}{3}-\cos\left(-\frac{17\pi}{2}\right)-\tg \frac{10\pi}{3}$$
$$\sin \frac{25\pi}{3}=\sin\left(8\pi+\frac{\pi}{3}\right)=\sin \frac{\pi}{3}=\frac{\sqrt3}{2}$$
$$\cos\left(-\frac{17\pi}{2}\right)=\cos\left(-8\pi-\frac{\pi}{2}\right)=\cos\left(-\frac{\pi}{2}\right)=0$$
$$\tg \frac{10\pi}{3}=\tg\left(3\pi+\frac{\pi}{3}\right)=\tg \frac{\pi}{3}=\sqrt3$$
$$\frac{\sqrt3}{2}-0-\sqrt3=-\frac{\sqrt3}{2}$$
3) $$\sin(-7\pi)-2\cos \frac{31\pi}{3}-\tg \frac{7\pi}{4}$$
$$\sin(-7\pi)=0$$
$$\cos \frac{31\pi}{3}=\cos\left(10\pi+\frac{\pi}{3}\right)=\cos \frac{\pi}{3}=\frac12$$
$$\tg \frac{7\pi}{4}=\tg\left(2\pi-\frac{\pi}{4}\right)=-1$$
$$0-2\cdot \frac12-(-1)=0$$
4) $$\cos(-9\pi)+2\sin\left(-\frac{49\pi}{6}\right)-\ctg\left(-\frac{21\pi}{4}\right)$$
$$\cos(-9\pi)=\cos 9\pi=\cos \pi=-1$$
$$\sin\left(-\frac{49\pi}{6}\right)=\sin\left(-8\pi-\frac{\pi}{6}\right)=-\sin \frac{\pi}{6}=-\frac12$$
$$\ctg\left(-\frac{21\pi}{4}\right)=\ctg\left(-5\pi-\frac{\pi}{4}\right)=\ctg\left(-\frac{\pi}{4}\right)=-\ctg \frac{\pi}{4}=-1$$
$$-1+2\cdot\left(-\frac12\right)-(-1)=-1$$
Ответ
1) $$\sqrt2$$; 2) $$-\frac{\sqrt3}{2}$$; 3) $$0$$; 4) $$-1$$.
