Вычислить: 1) 2 arcsin √3/2 + 3 arcsin (-1/2); 2) arcsin (1/√2) - 4 arcsin 1; 3) arccos (-1/2) - arcsin (√3/2); 4) arccos (-1) - arcsin (-1); 5) 2 arctg 1 + 3 arctg (-1/√3); 6) 4 arctg (-1) + 3 arctg √3.

Решение:

1. 1)

   \( 2 \arcsin \frac{\sqrt{3}}{2} + 3 \arcsin \left(-\frac{1}{2}\right) = 2 \cdot \frac{\pi}{3} + 3 \cdot \left(-\frac{\pi}{6}\right) = \frac{2\pi}{3} - \frac{\pi}{2} = \frac{4\pi - 3\pi}{6} = \frac{\pi}{6} \)

2. 2)

   \( \arcsin \frac{1}{\sqrt{2}} - 4 \arcsin 1 = \frac{\pi}{4} - 4 \cdot \frac{\pi}{2} = \frac{\pi}{4} - 2\pi = \frac{\pi - 8\pi}{4} = -\frac{7\pi}{4} \)

3. 3)

   \( \arccos \left(-\frac{1}{2}\right) - \arcsin \frac{\sqrt{3}}{2} = \frac{2\pi}{3} - \frac{\pi}{3} = \frac{\pi}{3} \)

4. 4)

   \( \arccos (-1) - \arcsin (-1) = \pi - \left(-\frac{\pi}{2}\right) = \pi + \frac{\pi}{2} = \frac{3\pi}{2} \)

5. 5)

   \( 2 \arctan 1 + 3 \arctan \left(-\frac{1}{\sqrt{3}}\right) = 2 \cdot \frac{\pi}{4} + 3 \cdot \left(-\frac{\pi}{6}\right) = \frac{\pi}{2} - \frac{\pi}{2} = 0 \)

6. 6)

   \( 4 \arctan (-1) + 3 \arctan \sqrt{3} = 4 \cdot \left(-\frac{\pi}{4}\right) + 3 \cdot \frac{\pi}{3} = -\pi + \pi = 0 \)

Ответ: 1) \( \frac{\pi}{6} \); 2) \( -\frac{7\pi}{4} \); 3) \( \frac{\pi}{3} \); 4) \( \frac{3\pi}{2} \); 5) 0; 6) 0.