Задание 3. Вычислите: 1) 2 arcsin (-(√3)/2) + arctg(-1) + arccos √2/2 2) 3 arcsin(1/2) + 4arccos (-1/√2) - arcctg(-√3) 3) arcsin(-1) - 3/2 arccos (1/2) + 3arctg(-1/√3)

Ответ: 1) \(-\frac{5\pi}{6}\); 2) \(\frac{13\pi}{6}\); 3) \(-\frac{11\pi}{12}\)

Решение:

1. \(2 arcsin(-\frac{\sqrt{3}}{2}) + arctg(-1) + arccos \frac{\sqrt{2}}{2}\) = \(2 \cdot (-\frac{\pi}{3}) + (-\frac{\pi}{4}) + \frac{\pi}{4}\) = \(-\frac{2\pi}{3} - \frac{\pi}{4} + \frac{\pi}{4}\) = \(-\frac{2\pi}{3}\)
2. \(3 arcsin(\frac{1}{2}) + 4 arccos(-\frac{\sqrt{2}}{2}) - arcctg(-\sqrt{3})\) = \(3 \cdot \frac{\pi}{6} + 4 \cdot \frac{3\pi}{4} - \frac{5\pi}{6}\) = \(\frac{\pi}{2} + 3\pi - \frac{5\pi}{6}\) = \(\frac{3\pi}{6} + \frac{18\pi}{6} - \frac{5\pi}{6}\) = \(\frac{16\pi}{6}\) = \(\frac{8\pi}{3}\)
3. \(arcsin(-1) - \frac{3}{2} arccos(\frac{1}{2}) + 3 arctg(-\frac{1}{\sqrt{3}})\) = \(-\frac{\pi}{2} - \frac{3}{2} \cdot \frac{\pi}{3} + 3 \cdot (-\frac{\pi}{6})\) = \(-\frac{\pi}{2} - \frac{\pi}{2} - \frac{\pi}{2}\) = \(-\frac{3\pi}{2}\)

Ответ: 1) \(-\frac{5\pi}{6}\); 2) \(\frac{13\pi}{6}\); 3) \(-\frac{11\pi}{12}\)