Question Number 196086 by MM42 last updated on 17/Aug/23
![Answer to the question “196008” Σ_(k=1) ^n tan^2 (((kπ)/(2n+1)))=n(2n+1) Ans) according “de moivre” sin(2n+1)α= (((2n+1)),(( 1)) )(cosα)^(2n) sinα− (((2n+1)),(( 3)) )(cosα)^(2n−2) (sinα)^3 +.... =(cosα)^(2n) (sinα)[ (((2n+1)),(( 1)) )− (((2n+1)),(( 3)) ) tan^2 α+...] for “α_k =((kπ)/(2n+1)) ; 1≤k≤n ⇒sin(2n+1)α_k =0 ⇒∀ 1≤k≤n→ (((2n+1)),(( 1)) )− (((2n+1)),(( 3)) ) tan^2 α_k +...=0 therefore “ x_k =tan^2 α_k ” thr roots of the equation are blowe x^n − (((2n+1)),((2n−1)) ) x^n + (((2n+1)),((2n−3)) )x^(n−1) −...=0 the sume of the roots of the equation is “ s= (((2n+1)),((2n−1)) ) ” ⇒s=Σ_(k=1) ^n tan^2 (((kπ)/(2n+1)))=n(2n+1)✓ the proof of the seconf part is done similarly](Q196086.png)
$${Answer}\:{to}\:{the}\:{question}\:``\mathrm{196008}'' \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{tan}^{\mathrm{2}} \left(\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)={n}\left(\mathrm{2}{n}+\mathrm{1}\right) \\ $$$$\left.{Ans}\right) \\ $$$${according}\:\:``{de}\:{moivre}'' \\ $$$${sin}\left(\mathrm{2}{n}+\mathrm{1}\right)\alpha=\begin{pmatrix}{\mathrm{2}{n}+\mathrm{1}}\\{\:\:\:\:\:\:\mathrm{1}}\end{pmatrix}\left({cos}\alpha\right)^{\mathrm{2}{n}} {sin}\alpha−\begin{pmatrix}{\mathrm{2}{n}+\mathrm{1}}\\{\:\:\:\:\:\mathrm{3}}\end{pmatrix}\left({cos}\alpha\right)^{\mathrm{2}{n}−\mathrm{2}} \left({sin}\alpha\right)^{\mathrm{3}} +.... \\ $$$$=\left({cos}\alpha\right)^{\mathrm{2}{n}} \left({sin}\alpha\right)\left[\begin{pmatrix}{\mathrm{2}{n}+\mathrm{1}}\\{\:\:\:\:\:\:\mathrm{1}}\end{pmatrix}−\begin{pmatrix}{\mathrm{2}{n}+\mathrm{1}}\\{\:\:\:\:\:\mathrm{3}}\end{pmatrix}\:{tan}^{\mathrm{2}} \alpha+...\right] \\ $$$${for}\:\:``\alpha_{{k}} =\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\:\:\:\:\:;\:\:\mathrm{1}\leqslant{k}\leqslant{n}\:\Rightarrow{sin}\left(\mathrm{2}{n}+\mathrm{1}\right)\alpha_{{k}} =\mathrm{0} \\ $$$$\Rightarrow\forall\:\:\mathrm{1}\leqslant{k}\leqslant{n}\rightarrow\:\begin{pmatrix}{\mathrm{2}{n}+\mathrm{1}}\\{\:\:\:\:\:\mathrm{1}}\end{pmatrix}−\begin{pmatrix}{\mathrm{2}{n}+\mathrm{1}}\\{\:\:\:\:\:\:\mathrm{3}}\end{pmatrix}\:{tan}^{\mathrm{2}} \alpha_{{k}} +...=\mathrm{0} \\ $$$${therefore}\:\:``\:{x}_{{k}} ={tan}^{\mathrm{2}} \alpha_{{k}} \:''\:{thr}\:{roots}\:{of}\:\:{the}\:{equation}\:{are}\:{blowe} \\ $$$${x}^{{n}} −\begin{pmatrix}{\mathrm{2}{n}+\mathrm{1}}\\{\mathrm{2}{n}−\mathrm{1}}\end{pmatrix}\:{x}^{{n}} +\begin{pmatrix}{\mathrm{2}{n}+\mathrm{1}}\\{\mathrm{2}{n}−\mathrm{3}}\end{pmatrix}{x}^{{n}−\mathrm{1}} −...=\mathrm{0} \\ $$$${the}\:{sume}\:{of}\:{the}\:{roots}\:{of}\:{the}\:{equation}\:{is}\:``\:{s}=\begin{pmatrix}{\mathrm{2}{n}+\mathrm{1}}\\{\mathrm{2}{n}−\mathrm{1}}\end{pmatrix}\:'' \\ $$$$\Rightarrow{s}=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\:{tan}^{\mathrm{2}} \left(\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)={n}\left(\mathrm{2}{n}+\mathrm{1}\right)\checkmark \\ $$$${the}\:{proof}\:{of}\:{the}\:{seconf}\:{part}\:{is}\:{done}\:{similarly} \\ $$$$ \\ $$$$ \\ $$