Упр.22.8 ГДЗ Мерзляк 10 класс Базовый уровень (Алгебра)

1) $$4\cos 225^\circ-6\cos 120^\circ+3\ctg 300^\circ+\tg 240^\circ$$
$$=4\cos(180^\circ+45^\circ)-6\cos(90^\circ+30^\circ)+3\ctg(270^\circ+30^\circ)+\tg(270^\circ-30^\circ)$$
$$=4(-\cos45^\circ)-6(-\sin30^\circ)+3(-\tg30^\circ)+\ctg30^\circ$$
$$=4\left(-\frac{\sqrt2}{2}\right)-6\left(-\frac12\right)+3\left(-\frac{\sqrt3}{3}\right)+\sqrt3$$
$$=-2\sqrt2+3-\sqrt3+\sqrt3=3-2\sqrt2.$$

2) $$\frac{6\cos^2(-240^\circ)\ctg 210^\circ}{\sin(-300^\circ)\cos^2 180^\circ}$$
$$=\frac{6\cos^2(240^\circ)\ctg 210^\circ}{-\sin 300^\circ\cdot \cos^2 180^\circ}$$
$$=\frac{6\cos^2(180^\circ+60^\circ)\ctg(180^\circ+30^\circ)}{-\sin(270^\circ+30^\circ)\cdot \cos^2 0^\circ}$$
$$=\frac{6(-\cos60^\circ)^2\cdot \ctg30^\circ}{-(-\cos30^\circ)\cdot 1^2}$$
$$=\frac{6\left(\frac12\right)^2\cdot \sqrt3}{\frac{\sqrt3}{2}}=3.$$

3) $$\sin\frac{7\pi}{4}\cdot \cos\frac{2\pi}{3}\cdot \tg\frac{4\pi}{3}\cdot \ctg\frac{7\pi}{6}$$
$$=\sin\left(2\pi-\frac{\pi}{4}\right)\cdot \cos\left(\pi-\frac{\pi}{3}\right)\cdot \tg\left(\pi+\frac{\pi}{3}\right)\cdot \ctg\left(\pi+\frac{\pi}{6}\right)$$
$$=-\sin\frac{\pi}{4}\cdot \left(-\cos\frac{\pi}{3}\right)\cdot \tg\frac{\pi}{3}\cdot \ctg\frac{\pi}{6}$$
$$=-\frac{\sqrt2}{2}\cdot \left(-\frac12\right)\cdot \sqrt3\cdot \sqrt3=\frac{3\sqrt2}{4}.$$

4) $$\frac{\cos64^\circ\cdot \cos4^\circ-\cos86^\circ\cdot \cos26^\circ}{\cos71^\circ\cdot \cos41^\circ-\cos49^\circ\cdot \cos19^\circ}$$
$$=\frac{\cos64^\circ\cdot \cos4^\circ-\cos(90^\circ-4^\circ)\cdot \cos(90^\circ-64^\circ)}{\cos71^\circ\cdot \cos41^\circ-\cos(90^\circ-41^\circ)\cdot \cos(90^\circ-71^\circ)}$$
$$=\frac{\cos64^\circ\cdot \cos4^\circ-\sin4^\circ\cdot \sin64^\circ}{\cos71^\circ\cdot \cos41^\circ-\sin41^\circ\cdot \sin71^\circ}$$
$$=\frac{\cos(64^\circ+4^\circ)}{\cos(71^\circ+41^\circ)}=\frac{\cos68^\circ}{\cos112^\circ}$$
$$=\frac{\cos68^\circ}{-\cos68^\circ}=-1.$$

Ответ: 1) $$3-2\sqrt2$$; 2) $$3$$; 3) $$\frac{3\sqrt2}{4}$$; 4) $$-1$$.