Question

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3 arcctg(-√3/3)+1/2 arccos√2/2

Answer 1

$3\ arcctg(-\frac{\sqrt3}{3})+\frac{1}{2}\, arccos\frac{\sqrt2}{2}=-3\cdot \frac{\pi}{3}+\frac{1}{2}\cdot \frac{\pi}{4}=-\pi +\frac{\pi}{8}=-\frac{7\pi }{8}$