Упр.16.4 ГДЗ Мордкович 10-11 класс (Алгебра)

a) arccos (-1/2) + arcsin (-1/2);
6) arccos (-корень(2)/2) — arcsin (-1);
в) arccos (-корень(3)/2) + arcsin (-корень(3)/2);
Рі) arccos корень(2)/2 — arcsin (-корень(3)/2).

1. $$\arccos\left(-\frac12\right)+\arcsin\left(-\frac12\right)$$

   $$\arccos\left(-\frac12\right)=\frac{2\pi}{3}, \qquad \arcsin\left(-\frac12\right)=-\frac{\pi}{6}$$

   $$\frac{2\pi}{3}-\frac{\pi}{6}=\frac{4\pi}{6}-\frac{\pi}{6}=\frac{3\pi}{6}=\frac{\pi}{2}$$

2. $$\arccos\left(-\frac{\sqrt2}{2}\right)-\arcsin(-1)$$

   $$\arccos\left(-\frac{\sqrt2}{2}\right)=\frac{3\pi}{4}, \qquad \arcsin(-1)=-\frac{\pi}{2}$$

   $$\frac{3\pi}{4}-\left(-\frac{\pi}{2}\right)=\frac{3\pi}{4}+\frac{2\pi}{4}=\frac{5\pi}{4}$$

3. $$\arccos\left(-\frac{\sqrt3}{2}\right)+\arcsin\left(-\frac{\sqrt3}{2}\right)$$

   $$\arccos\left(-\frac{\sqrt3}{2}\right)=\frac{5\pi}{6}, \qquad \arcsin\left(-\frac{\sqrt3}{2}\right)=-\frac{\pi}{3}$$

   $$\frac{5\pi}{6}-\frac{\pi}{3}=\frac{5\pi}{6}-\frac{2\pi}{6}=\frac{3\pi}{6}=\frac{\pi}{2}$$

4. $$\arccos\left(\frac{\sqrt2}{2}\right)-\arcsin\left(-\frac{\sqrt3}{2}\right)$$

   $$\arccos\left(\frac{\sqrt2}{2}\right)=\frac{\pi}{4}, \qquad \arcsin\left(-\frac{\sqrt3}{2}\right)=-\frac{\pi}{3}$$

   $$\frac{\pi}{4}-\left(-\frac{\pi}{3}\right)=\frac{\pi}{4}+\frac{\pi}{3}=\frac{3\pi}{12}+\frac{4\pi}{12}=\frac{7\pi}{12}$$

Ответ

а) $$\frac{\pi}{2}$$; б) $$\frac{5\pi}{4}$$; в) $$\frac{\pi}{2}$$; г) $$\frac{7\pi}{12}$$.