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Question Number 30598 by abdo imad last updated on 23/Feb/18
prove that it exist one polynomial p/  p(cosx)=cos(nx) find the roots of p(x) .
provethatitexistonepolynomialp/p(cosx)=cos(nx)findtherootsofp(x).
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]].
wehavebymoivreformulacos(nx)+isin(nx)=(cosx+isinx)n=k=0nCnk(isinx)k(cosx)nk=p=0[n2]Cn2p(isinx)2p(cosx)n2p+p=0[n12]=Cn2p+1(isinx)2p+1(cosx)n2p1cos(nx)=Re(einx)=p=0[n2](1)pCn2p(1cos2x)p(cosx)n2p=p(cosx)/p(x)=p=0[n2](1)pCn2p(1x2)pxn2p.2)Xrootofp(x)p(X)=0letputX=cosθp(X)=0p(cosθ)=0cos(nθ)=0nθ=π2+kπθ=π2n+kπn=(2k+1)π2nbutwecanchowthatdegp=ntheroitsofp(x)areXk=cos(θk)=cos((2k+1)π2n)withk[[0,n1]].

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