Упр.42.50 ГДЗ Мерзляк 10 класс Базовый уровень (Алгебра)

1) cos(arccos(v3/2)+(1/2)arcsin(v3/2)); 3) tg(2arctg(-v3/3));
2) sin(arcsin(1/2)+arccos(1/2)); 4) ctg(3arccos(-v3/2)).

1. $$\cos\left(\arccos\frac{\sqrt{3}}{2}+\frac12\arcsin\frac{\sqrt{3}}{2}\right)$$
   $$\arccos\frac{\sqrt{3}}{2}=\frac{\pi}{6}, \qquad \arcsin\frac{\sqrt{3}}{2}=\frac{\pi}{3}$$
   $$\cos\left(\frac{\pi}{6}+\frac12\cdot\frac{\pi}{3}\right)=\cos\left(\frac{\pi}{6}+\frac{\pi}{6}\right)=\cos\frac{\pi}{3}=\frac12.$$
2. $$\sin\left(\arcsin\frac12+\arccos\frac12\right)$$
   $$\arcsin\frac12=\frac{\pi}{6}, \qquad \arccos\frac12=\frac{\pi}{3}$$
   $$\sin\left(\frac{\pi}{6}+\frac{\pi}{3}\right)=\sin\frac{\pi}{2}=1.$$
3. $$\tg\left(2\arctg\left(-\frac{\sqrt{3}}{3}\right)\right)$$
   $$\arctg\left(-\frac{\sqrt{3}}{3}\right)=-\frac{\pi}{6}$$
   $$\tg\left(2\cdot\left(-\frac{\pi}{6}\right)\right)=\tg\left(-\frac{\pi}{3}\right)=-\sqrt{3}.$$
4. $$\ctg\left(3\arccos\left(-\frac{\sqrt{3}}{2}\right)\right)$$
   $$\arccos\left(-\frac{\sqrt{3}}{2}\right)=\frac{5\pi}{6}$$
   $$\ctg\left(3\cdot\frac{5\pi}{6}\right)=\ctg\frac{5\pi}{2}=\ctg\left(2\pi+\frac{\pi}{2}\right)=\ctg\frac{\pi}{2}=0.$$

Ответ

1) $$\frac12$$; 2) $$1$$; 3) $$-\sqrt{3}$$; 4) $$0$$.