Question Number 181998 by Tolmasbek last updated on 03/Dec/22
Answered by a.lgnaoui last updated on 03/Dec/22
![α=3((π/(18))+((πn)/9)) tan α=((tan ((π/(18))+((πn)/9))[3−tan^2 ((π/(18))+((πn)/9))])/(1−3tan^2 ((π/(18))+((πn)/9)))) =((tan ((π/9)((1/2)+n)[3−tan^2 ((π/9)((1/2)+n)])/(1−3tan^2 ((π/9)((1/2)+n))) { (((π/9)((1/2)+n)≠(π/2)+kπ)),((1−3tan^2 ((π/9)((1/2)+n))≠0)) :} { (( n ≠4+9k (k∈Z^+ ))),((n≠1+k and n≠9k−2 )) :} donc n≠2k k∈Z k>1 So, α is not also true](https://www.tinkutara.com/question/Q182021.png)
$$\alpha=\mathrm{3}\left(\frac{\pi}{\mathrm{18}}+\frac{\pi{n}}{\mathrm{9}}\right) \\ $$$$\mathrm{tan}\:\alpha=\frac{\mathrm{tan}\:\left(\frac{\pi}{\mathrm{18}}+\frac{\pi{n}}{\mathrm{9}}\right)\left[\mathrm{3}−\mathrm{tan}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{18}}+\frac{\pi{n}}{\mathrm{9}}\right)\right]}{\mathrm{1}−\mathrm{3tan}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{18}}+\frac{\pi{n}}{\mathrm{9}}\right)} \\ $$$$=\frac{\mathrm{tan}\:\left(\frac{\pi}{\mathrm{9}}\left(\frac{\mathrm{1}}{\mathrm{2}}+{n}\right)\left[\mathrm{3}−\mathrm{tan}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{9}}\left(\frac{\mathrm{1}}{\mathrm{2}}+{n}\right)\right]\right.\right.}{\mathrm{1}−\mathrm{3tan}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{9}}\left(\frac{\mathrm{1}}{\mathrm{2}}+{n}\right)\right.} \\ $$$$\begin{cases}{\frac{\pi}{\mathrm{9}}\left(\frac{\mathrm{1}}{\mathrm{2}}+{n}\right)\neq\frac{\pi}{\mathrm{2}}+{k}\pi}\\{\mathrm{1}−\mathrm{3tan}\:^{\mathrm{2}} \left(\frac{\pi}{\mathrm{9}}\left(\frac{\mathrm{1}}{\mathrm{2}}+{n}\right)\right)\neq\mathrm{0}}\end{cases} \\ $$$$\begin{cases}{\:{n}\:\neq\mathrm{4}+\mathrm{9}{k}\:\:\:\left({k}\in\mathbb{Z}^{+} \right)}\\{{n}\neq\mathrm{1}+{k}\:\:{and}\:\:{n}\neq\mathrm{9}{k}−\mathrm{2}\:\:\:\:\:}\end{cases} \\ $$$${donc}\:\:\:\:{n}\neq\mathrm{2}{k}\:\:\:{k}\in\mathbb{Z}\:\:{k}>\mathrm{1} \\ $$$${So},\:\:\:\:\:\alpha\:\:\:{is}\:{not}\:{also}\:{true} \\ $$