Question Number 222743 by hardmath last updated on 06/Jul/25
$$\mathrm{Compare}: \\ $$$$\boldsymbol{\mathrm{a}}\:=\:\mathrm{arcctg}\:\sqrt{\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{b}}\:=\:\mathrm{arccos}\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{c}}\:=\:\mathrm{arctg}\:\sqrt{\mathrm{2}} \\ $$
Answered by mr W last updated on 06/Jul/25
$${a}=\mathrm{cot}^{−\mathrm{1}} \sqrt{\mathrm{2}}=\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$${b}=\mathrm{cos}^{−\mathrm{1}} \frac{\sqrt{\mathrm{2}}}{\mathrm{2}}=\frac{\pi}{\mathrm{4}}=\mathrm{tan}^{−\mathrm{1}} \mathrm{1} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}<\mathrm{1}<\sqrt{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}<\mathrm{tan}^{−\mathrm{1}} \mathrm{1}<\mathrm{tan}^{−\mathrm{1}} \sqrt{\mathrm{2}} \\ $$$$\Rightarrow{a}<{b}<{c}\:\:\checkmark \\ $$
Commented by hardmath last updated on 06/Jul/25
$$\mathrm{cool}\:\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professor} \\ $$