Упр.15 ГДЗ Мерзляк Полонский 9 класс (Геометрия)

Задача

Найдите значение выражения:
1) $$2\sin(120^\circ)+4\cos(150^\circ)-2\tg(135^\circ)$$;
2) $$\cos(120^\circ)-8\sin^2(150^\circ)+3\cos(90^\circ)\cos(162^\circ)$$;
3) $$\cos(180^\circ)\left(\sin(135^\circ)\tg(60^\circ)-\cos(135^\circ)\right)^2$$;
4) $$2\sin^2(150^\circ)+\cos^2(60^\circ)+\sin^2(45^\circ)+\tg^2(120^\circ)-\ctg^2(30^\circ)$$.

Подробный ответ

1) $$2\sin 120^\circ+4\cos 150^\circ-2\tg 135^\circ$$

$$=2\sin 60^\circ-4\cos 30^\circ+2\tg 45^\circ$$
$$=2\cdot \frac{\sqrt{3}}{2}-4\cdot \frac{\sqrt{3}}{2}+2\cdot 1$$
$$=\sqrt{3}-2\sqrt{3}+2=2-\sqrt{3}.$$

2) $$\cos 120^\circ-8\sin^2 150^\circ+3\cos 90^\circ\cos 162^\circ$$

$$=-\cos 60^\circ-8\sin^2 30^\circ+3\cdot 0\cdot \cos 162^\circ$$
$$=-\frac{1}{2}-8\cdot \left(\frac{1}{2}\right)^2+0$$
$$=-\frac{1}{2}-2=-2\frac{1}{2}.$$

3) $$\cos 180^\circ\left(\sin 135^\circ\cdot \tg 60^\circ-\cos 135^\circ\right)^2$$

$$=-1\cdot \left(\sin 45^\circ\cdot \sqrt{3}+\cos 45^\circ\right)^2$$
$$=-\left(\frac{\sqrt{2}}{2}\cdot \sqrt{3}+\frac{\sqrt{2}}{2}\right)^2$$
$$=-\left(\frac{\sqrt{6}+\sqrt{2}}{2}\right)^2$$
$$=-\frac{6+2\sqrt{12}+2}{4}=-\frac{8+4\sqrt{3}}{4}=-2-\sqrt{3}.$$

4) $$2\sin^2 150^\circ+\cos^2 60^\circ+\sin^2 45^\circ+\tg^2 120^\circ-\ctg^2 30^\circ$$

$$=2\sin^2 30^\circ+\left(\frac{1}{2}\right)^2+\left(\frac{\sqrt{2}}{2}\right)^2+\tg^2 60^\circ-\left(\sqrt{3}\right)^2$$
$$=2\cdot \left(\frac{1}{2}\right)^2+\frac{1}{4}+\frac{1}{2}+3-3$$
$$=\frac{1}{2}+\frac{1}{4}+\frac{1}{2}=1\frac{1}{4}.$$