Упр.549 ГДЗ Алимов 10-11 класс (Алгебра)

2) sin 25пи/3 — tg 10пи/3;
3) 3 cos 3660° + sin (-1560°);
4) cos (-945°) + tg 1035°.

1. $$\cos \frac{23\pi}{4}-\sin \frac{15\pi}{4}=\cos \left(6\pi-\frac{\pi}{4}\right)-\sin \left(4\pi-\frac{\pi}{4}\right)=\cos \frac{\pi}{4}+\sin \frac{\pi}{4}$$

   $$\cos \frac{\pi}{4}+\sin \frac{\pi}{4}=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}=\sqrt{2}$$

2. $$\sin \frac{25\pi}{3}-\tg \frac{10\pi}{3}=\sin \left(8\pi+\frac{\pi}{3}\right)-\tg \left(3\pi+\frac{\pi}{3}\right)=\sin \frac{\pi}{3}-\tg \frac{\pi}{3}$$

   $$\sin \frac{\pi}{3}-\tg \frac{\pi}{3}=\frac{\sqrt{3}}{2}-\sqrt{3}=\frac{\sqrt{3}-2\sqrt{3}}{2}=-\frac{\sqrt{3}}{2}$$

3. $$3\cos 3660^\circ+\sin(-1560^\circ)=3\cos(10\cdot 360^\circ+60^\circ)+\sin(-5\cdot 360^\circ+240^\circ)$$

   $$=3\cos 60^\circ+\sin 240^\circ=3\cdot \frac12+\sin(180^\circ+60^\circ)$$

   $$=\frac32-\sin 60^\circ=\frac32-\frac{\sqrt{3}}{2}=\frac{3-\sqrt{3}}{2}$$

4. $$\cos(-945^\circ)+\tg 1035^\circ=\cos(-3\cdot 360^\circ+135^\circ)+\tg(5\cdot 180^\circ+135^\circ)$$

   $$=\cos 135^\circ+\tg 135^\circ=\cos(90^\circ+45^\circ)+\tg(90^\circ+45^\circ)$$

   $$=-\sin 45^\circ-\ctg 45^\circ=-\frac{\sqrt{2}}{2}-1=-\frac{\sqrt{2}+2}{2}$$

Ответ

1) $$\sqrt{2}$$; 2) $$-\frac{\sqrt{3}}{2}$$; 3) $$\frac{3-\sqrt{3}}{2}$$; 4) $$-\frac{\sqrt{2}+2}{2}$$.