Question Number 196008 by universe last updated on 15/Aug/23
Commented by universe last updated on 15/Aug/23
$${prove}\:{that} \\ $$
Commented by York12 last updated on 15/Aug/23
$${what}\:{is}\:{the}\:{source}\:{of}\:{those} \\ $$$$ \\ $$
Answered by MM42 last updated on 17/Aug/23
![According to the “ de moivre” sin(2n+1)α= (((2n+1)),(( 1)) )(cosα)^(2n) (sinα)− (((2n+1)),(( 3)) )(cosα)^(2n−2) (sinα)^3 + (((2n+1)),(( 5)) )(cosα)^(2n−3) (sinα)^5 +... =(cosα)^(2n) (sinα)[ (((2n+1)),(( 1)) )− (((2n+1)),(( 3)) )(tanα)^2 + (((2n+1)),(( 5)) )(tanα)^4 − (((2n+1)),(( 7)) )(tanα)^6 +...] for α_k =((kπ)/(2n+1)) ; ∀ 1≤k≤n ⇒ sin(2n+1)α=0 ⇒ (((2n+1)),(( 1)) )− (((2n+1)),(( 3)) )(tanα_k )^2 + (((2n+1)),(( 5)) )(tanα_k )^4 −....=0 ⇒x_k =(tanα_k )^2 ; 1≤k≤n the roots of equation are below x^n − (((2n+1)),((2n−1)) )x^(n−1) + (((2n+1)),((2n−3)) )x^(n−2) −....=0 the sume of the roots “ s= (((2n+1)),((2n−1)) ) ” ⇒Σ_(k=1) ^n (tan((kπ)/(2n+1)))^2 =(((2n+1)!)/((2n−1)!))=n(2n+1) ✓ the second part is similary proved](https://www.tinkutara.com/question/Q196075.png)
$${According}\:{to}\:{the}\:“\:{de}\:{moivre}'' \\ $$$${sin}\left(\mathrm{2}{n}+\mathrm{1}\right)\alpha=\begin{pmatrix}{\mathrm{2}{n}+\mathrm{1}}\\{\:\:\:\:\:\:\mathrm{1}}\end{pmatrix}\left({cos}\alpha\right)^{\mathrm{2}{n}} \left({sin}\alpha\right)−\begin{pmatrix}{\mathrm{2}{n}+\mathrm{1}}\\{\:\:\:\:\:\:\mathrm{3}}\end{pmatrix}\left({cos}\alpha\right)^{\mathrm{2}{n}−\mathrm{2}} \left({sin}\alpha\right)^{\mathrm{3}} +\begin{pmatrix}{\mathrm{2}{n}+\mathrm{1}}\\{\:\:\:\:\:\:\mathrm{5}}\end{pmatrix}\left({cos}\alpha\right)^{\mathrm{2}{n}−\mathrm{3}} \left({sin}\alpha\right)^{\mathrm{5}} +… \\ $$$$=\left({cos}\alpha\right)^{\mathrm{2}{n}} \left({sin}\alpha\right)\left[\begin{pmatrix}{\mathrm{2}{n}+\mathrm{1}}\\{\:\:\:\:\:\:\mathrm{1}}\end{pmatrix}−\begin{pmatrix}{\mathrm{2}{n}+\mathrm{1}}\\{\:\:\:\:\:\mathrm{3}}\end{pmatrix}\left({tan}\alpha\right)^{\mathrm{2}} +\begin{pmatrix}{\mathrm{2}{n}+\mathrm{1}}\\{\:\:\:\:\:\:\mathrm{5}}\end{pmatrix}\left({tan}\alpha\right)^{\mathrm{4}} −\begin{pmatrix}{\mathrm{2}{n}+\mathrm{1}}\\{\:\:\:\:\:\:\mathrm{7}}\end{pmatrix}\left({tan}\alpha\right)^{\mathrm{6}} +…\right] \\ $$$${for}\:\alpha_{{k}} =\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\:\:\:\:;\:\forall\:\:\:\mathrm{1}\leqslant{k}\leqslant{n}\:\Rightarrow\:{sin}\left(\mathrm{2}{n}+\mathrm{1}\right)\alpha=\mathrm{0} \\ $$$$\Rightarrow\begin{pmatrix}{\mathrm{2}{n}+\mathrm{1}}\\{\:\:\:\:\:\:\mathrm{1}}\end{pmatrix}−\begin{pmatrix}{\mathrm{2}{n}+\mathrm{1}}\\{\:\:\:\:\:\:\mathrm{3}}\end{pmatrix}\left({tan}\alpha_{{k}} \right)^{\mathrm{2}} +\begin{pmatrix}{\mathrm{2}{n}+\mathrm{1}}\\{\:\:\:\:\:\:\mathrm{5}}\end{pmatrix}\left({tan}\alpha_{{k}} \right)^{\mathrm{4}} −….=\mathrm{0} \\ $$$$\Rightarrow{x}_{{k}} =\left({tan}\alpha_{{k}} \right)^{\mathrm{2}} \:\:\:\:;\:\mathrm{1}\leqslant{k}\leqslant{n}\:\:\:{the}\:{roots}\:{of}\:{equation}\:{are}\:{below} \\ $$$${x}^{{n}} −\begin{pmatrix}{\mathrm{2}{n}+\mathrm{1}}\\{\mathrm{2}{n}−\mathrm{1}}\end{pmatrix}{x}^{{n}−\mathrm{1}} +\begin{pmatrix}{\mathrm{2}{n}+\mathrm{1}}\\{\mathrm{2}{n}−\mathrm{3}}\end{pmatrix}{x}^{{n}−\mathrm{2}} −….=\mathrm{0} \\ $$$${the}\:{sume}\:{of}\:{the}\:{roots}\:“\:{s}=\begin{pmatrix}{\mathrm{2}{n}+\mathrm{1}}\\{\mathrm{2}{n}−\mathrm{1}}\end{pmatrix}\:\:'' \\ $$$$\Rightarrow\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\:\left({tan}\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)^{\mathrm{2}} =\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)!}{\left(\mathrm{2}{n}−\mathrm{1}\right)!}={n}\left(\mathrm{2}{n}+\mathrm{1}\right)\:\checkmark \\ $$$$ \\ $$$${the}\:{second}\:{part}\:{is}\:{similary}\:{proved} \\ $$$$ \\ $$