Question

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3 arcctg(-√3/3)+1/2 arccos√2/2

Answer 1

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