1. Вычислите: a) sin 3π/4; б) arctg (-1) + 2 arccos (√2/2) - arcsin (1/2)

Решение:

1. \( \sin(\frac{3\pi}{4}) = \sin(\pi - \frac{\pi}{4}) = \sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} \)
2. \( \mathrm{arctg}(-1) = -\frac{\pi}{4} \)
3. \( \mathrm{arccos}(\frac{\sqrt{2}}{2}) = \frac{\pi}{4} \)
4. \( \mathrm{arcsin}(\frac{1}{2}) = \frac{\pi}{6} \)
5. \( -\frac{\pi}{4} + 2 \cdot \frac{\pi}{4} - \frac{\pi}{6} = -\frac{\pi}{4} + \frac{\pi}{2} - \frac{\pi}{6} = \frac{-3\pi + 6\pi - 2\pi}{12} = \frac{\pi}{12} \)

Ответ: a) \( \frac{\sqrt{2}}{2} \); б) \( \frac{\pi}{12} \).