Question Number 134134 by mr W last updated on 28/Feb/21
$$\mathrm{If}\:\mathrm{cos}^{−\mathrm{1}} {x}_{\mathrm{1}} +\mathrm{cos}^{−\mathrm{1}} {x}_{\mathrm{2}} +…+\mathrm{cos}^{−\mathrm{1}} {x}_{{n}} ={n}\pi, \\ $$$$\mathrm{then}\:\:{x}_{\mathrm{1}} \:+\:{x}_{\mathrm{2}} ^{\mathrm{2}} \:+\:{x}_{\mathrm{3}} ^{\mathrm{3}} \:+\:…+\:{x}_{{n}} ^{{n}} \:=? \\ $$
Answered by mindispower last updated on 28/Feb/21
$${n}\in\mathbb{N} \\ $$$${cos}^{−} \left({t}\right)\in\left[\mathrm{0},\pi\right] \\ $$$$\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{cos}^{−} \left({x}_{{i}} \right)={n}\pi\Leftrightarrow{x}_{{i}} =\pi,\forall{i}\in\left[\mathrm{1},{n}\right] \\ $$$$−\mathrm{1} \\ $$$$\underset{{i}\leqslant{n}} {\sum}{x}_{{i}} ^{{i}} =\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\left(^{{i}} −\mathrm{1}\right)=−\mathrm{1}\frac{\mathrm{1}−\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}}=\begin{cases}{\mathrm{0},{n}=\mathrm{2}{k}}\\{−\mathrm{1},{n}=\mathrm{2}{k}+\mathrm{1}}\end{cases} \\ $$
Commented by mindispower last updated on 28/Feb/21
$${yes}\:{sorry}\: \\ $$
Commented by mr W last updated on 28/Feb/21
$${thanks}\:{sir}! \\ $$$${please}\:{check}: \\ $$$$\mathrm{cos}^{−\mathrm{1}} \:\left({x}_{{i}} \right)=\pi\:\Leftrightarrow\:{x}_{{i}} =−\mathrm{1}\:? \\ $$