Question

2arcsin1+arccos(-√2/2)-3arctg(-1)

Answer 1

$\text{arcsin 1} = \frac{\pi}{2}$

$\text{arccos}( -\frac{\sqrt 2}{2} ) = \pi - \text{arrcos}\frac{\sqrt2}{2} = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$

$\text{arctg(-1)} = -\text{arctg1} = -\frac{\pi}{4}$

$2\cdot \text{arcsin 1} +\text{arccos}( -\frac{\sqrt 2}{2} ) - 3\cdot\text{arctg(-1)} = 2\cdot\frac{\pi}{2}+\frac{3\pi}{4}+\frac{3\pi}{4} = \frac{5\pi}{2} = 2,5\pi$