Question Number 148339 by mathdanisur last updated on 27/Jul/21
$${arccos}\:\left({cos}\:\mathrm{9}\right)\:=\:? \\ $$
Answered by Olaf_Thorendsen last updated on 27/Jul/21
![x = arccos(cos9)) x = arccos(cos(9−2π))), 9−2π ∈[0,π] x = 9−2π](https://www.tinkutara.com/question/Q148352.png)
$$\left.{x}\:=\:\mathrm{arccos}\left(\mathrm{cos9}\right)\right) \\ $$$$\left.{x}\:=\:\mathrm{arccos}\left(\mathrm{cos}\left(\mathrm{9}−\mathrm{2}\pi\right)\right)\right),\:\mathrm{9}−\mathrm{2}\pi\:\in\left[\mathrm{0},\pi\right] \\ $$$${x}\:=\:\mathrm{9}−\mathrm{2}\pi \\ $$
Commented by mathdanisur last updated on 27/Jul/21
$${Thank}\:{you}\:{Sir} \\ $$