Question Number 30598 by abdo imad last updated on 23/Feb/18
$${prove}\:{that}\:{it}\:{exist}\:{one}\:{polynomial}\:{p}/ \\ $$$${p}\left({cosx}\right)={cos}\left({nx}\right)\:{find}\:{the}\:{roots}\:{of}\:{p}\left({x}\right)\:. \\ $$
Commented by abdo imad last updated on 27/Feb/18
![we have by moivre formula cos(nx) +isin(nx)=(cosx +isinx)^n = Σ_(k=0) ^n C_n ^k (isinx)^k (cosx)^(n−k) = Σ_(p=0) ^([(n/2)]) C_n ^(2p) (isinx)^(2p) (cosx)^(n−2p) +Σ_(p=0) ^([((n−1)/2)]) = C_n ^(2p+1) (isinx)^(2p+1) (cosx)^(n−2p−1) cos(nx)=Re(e^(inx) )= Σ_(p=0) ^([(n/2)]) (−1)^p C_n ^(2p) (1−cos^2 x)^p (cosx)^(n−2p) =p(cosx) /p(x)= Σ_(p=0) ^([(n/2)]) (−1)^p C_n ^(2p) (1−x^2 )^p x^(n−2p) . 2) X root of p (x)⇔ p(X)=0 let put X=cosθ p(X)=0 ⇔p(cosθ)=0 ⇔cos(nθ)=0 ⇔ nθ= (π/2) +kπ ⇔ θ=(π/(2n)) +((kπ)/n)=(((2k+1)π)/(2n)) but wecan chow that deg p=n ⇒ the roits of p(x) are X_k = cos(θ_k ) =cos((2k+1)(π/(2n))) with k∈[[0,n−1]].](https://www.tinkutara.com/question/Q30848.png)
$${we}\:{have}\:{by}\:{moivre}\:{formula}\: \\ $$$${cos}\left({nx}\right)\:+{isin}\left({nx}\right)=\left({cosx}\:+{isinx}\right)^{{n}} \\ $$$$=\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\:{C}_{{n}} ^{{k}} \:\left({isinx}\right)^{{k}} \:\left({cosx}\right)^{{n}−{k}} \\ $$$$=\:\sum_{{p}=\mathrm{0}} ^{\left[\frac{{n}}{\mathrm{2}}\right]} \:\:{C}_{{n}} ^{\mathrm{2}{p}} \:\left({isinx}\right)^{\mathrm{2}{p}} \:\left({cosx}\right)^{{n}−\mathrm{2}{p}} \:+\sum_{{p}=\mathrm{0}} ^{\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]} =\:{C}_{{n}} ^{\mathrm{2}{p}+\mathrm{1}} \left({isinx}\right)^{\mathrm{2}{p}+\mathrm{1}} \left({cosx}\right)^{{n}−\mathrm{2}{p}−\mathrm{1}} \\ $$$${cos}\left({nx}\right)={Re}\left({e}^{{inx}} \right)=\:\sum_{{p}=\mathrm{0}} ^{\left[\frac{{n}}{\mathrm{2}}\right]} \:\:\left(−\mathrm{1}\right)^{{p}} \:{C}_{{n}} ^{\mathrm{2}{p}} \:\left(\mathrm{1}−{cos}^{\mathrm{2}} {x}\right)^{{p}} \left({cosx}\right)^{{n}−\mathrm{2}{p}} \\ $$$$={p}\left({cosx}\right)\:/{p}\left({x}\right)=\:\sum_{{p}=\mathrm{0}} ^{\left[\frac{{n}}{\mathrm{2}}\right]} \:\left(−\mathrm{1}\right)^{{p}} \:{C}_{{n}} ^{\mathrm{2}{p}} \:\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{{p}} \:{x}^{{n}−\mathrm{2}{p}} \:. \\ $$$$\left.\mathrm{2}\right)\:{X}\:{root}\:{of}\:{p}\:\left({x}\right)\Leftrightarrow\:{p}\left({X}\right)=\mathrm{0}\:\:{let}\:{put}\:{X}={cos}\theta \\ $$$${p}\left({X}\right)=\mathrm{0}\:\Leftrightarrow{p}\left({cos}\theta\right)=\mathrm{0}\:\Leftrightarrow{cos}\left({n}\theta\right)=\mathrm{0} \\ $$$$\Leftrightarrow\:\:{n}\theta=\:\frac{\pi}{\mathrm{2}}\:+{k}\pi\:\:\Leftrightarrow\:\theta=\frac{\pi}{\mathrm{2}{n}}\:+\frac{{k}\pi}{{n}}=\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{\mathrm{2}{n}}\:{but}\:{wecan}\:{chow} \\ $$$${that}\:{deg}\:{p}={n}\:\Rightarrow\:{the}\:{roits}\:{of}\:{p}\left({x}\right)\:{are} \\ $$$${X}_{{k}} =\:{cos}\left(\theta_{{k}} \right)\:={cos}\left(\left(\mathrm{2}{k}+\mathrm{1}\right)\frac{\pi}{\mathrm{2}{n}}\right)\:{with}\:{k}\in\left[\left[\mathrm{0},{n}−\mathrm{1}\right]\right]. \\ $$