Упр.530 ГДЗ Алимов 10-11 класс (Алгебра)
- Найти значение выражения:
1) $$\cos 630^\circ-\sin 1470^\circ-\operatorname{ctg}1125^\circ$$;
2) $$\operatorname{tg}1800^\circ-\sin 495^\circ+\cos 945^\circ$$;
3) $$3\cos 3660^\circ+\sin(-1560^\circ)+\cos(-450^\circ)$$;
4) $$\cos 4455^\circ-\cos(-945^\circ)+\operatorname{tg}1035^\circ-\operatorname{ctg}(-1500^\circ)$$.
1) $$\cos 630^\circ-\sin 1470^\circ-\operatorname{ctg}1125^\circ$$
$$\cos 630^\circ-\sin 1470^\circ-\operatorname{ctg}1125^\circ = \cos(360^\circ+270^\circ)-\sin(1440^\circ+30^\circ)-\operatorname{ctg}(1080^\circ+45^\circ)$$
$$= \cos 270^\circ-\sin 30^\circ-\operatorname{ctg}45^\circ = 0-\frac12-1 = -\frac32$$
Ответ: $$-\frac32$$
2) $$\operatorname{tg}1800^\circ-\sin 495^\circ+\cos 945^\circ$$
$$\operatorname{tg}1800^\circ-\sin 495^\circ+\cos 945^\circ = \operatorname{tg}(10\cdot 180^\circ)-\sin(360^\circ+135^\circ)+\cos(2\cdot 360^\circ+225^\circ)$$
$$= 0-\sin 135^\circ+\cos 225^\circ = -\frac{\sqrt2}{2}-\frac{\sqrt2}{2} = -\sqrt2$$
Ответ: $$-\sqrt2$$
3) $$3\cos 3660^\circ+\sin(-1560^\circ)+\cos(-450^\circ)$$
$$3\cos 3660^\circ+\sin(-1560^\circ)+\cos(-450^\circ) = 3\cos(3600^\circ+60^\circ)+\sin(-1440^\circ-120^\circ)+\cos(-360^\circ-90^\circ)$$
$$= 3\cos 60^\circ+\sin(-120^\circ)+\cos(-90^\circ) = 3\cdot \frac12-\sin 120^\circ+0$$
$$= \frac32-\frac{\sqrt3}{2} = \frac{3-\sqrt3}{2}$$
Ответ: $$\frac{3-\sqrt3}{2}$$
4) $$\cos 4455^\circ-\cos(-945^\circ)+\operatorname{tg}1035^\circ-\operatorname{ctg}(-1500^\circ)$$
$$\cos 4455^\circ-\cos(-945^\circ)+\operatorname{tg}1035^\circ-\operatorname{ctg}(-1500^\circ)$$
$$= \cos(4320^\circ+135^\circ)-\cos(-900^\circ-45^\circ)+\operatorname{tg}(1080^\circ-45^\circ)-\operatorname{ctg}(-1440^\circ-60^\circ)$$
$$= \cos 135^\circ-\cos 45^\circ+\operatorname{tg}(-45^\circ)-\operatorname{ctg}(-60^\circ)$$
$$= -\frac{\sqrt2}{2}-\frac{\sqrt2}{2}-1+\frac{1}{\sqrt3} = -\sqrt2-1+\frac{\sqrt3}{3}$$
Ответ: $$-\sqrt2-1+\frac{\sqrt3}{3}$$
