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Question Number 28371 by abdo imad last updated on 24/Jan/18
prove that x^2 −2x cosθ +1 divide x^(2n)  −2x^n cos(nθ)+1
$${prove}\:{that}\:{x}^{\mathrm{2}} −\mathrm{2}{x}\:{cos}\theta\:+\mathrm{1}\:{divide}\:{x}^{\mathrm{2}{n}} \:−\mathrm{2}{x}^{{n}} {cos}\left({n}\theta\right)+\mathrm{1} \\ $$
Commented by abdo imad last updated on 25/Jan/18
let put   p_n (x)= x^(2n)  −2x^n cos(nθ)+1 and t(x)= x^2 −2xcosθ +1  roots of t(x)?   Δ=4 cos^2 θ −4=−4sin^2 θ =(2isinθ)^2   z_1 =((2cosθ +2isinθ)/2)= e^(iθ)   and z_(2 ) = e^(−iθ)   let prove that  p_n (z_1 )=0 and p_n (z_2 )=0  we have  p_n (z_1 )= e^(i2nθ)  −2e^(inθ) ((e^(inθ)  +e^(−inθ) )/2)+1  = e^(i2nθ)  −e^(i2nθ)  −1+1=0  p_n (z_2 )= e^(−i2nθ)  −2 e^(−inθ)  ((e^(inθ)  +e^(−inθ) )/2) +1  = e^(−i2nθ)  −1 −e^(−i2nθ)    +1 =0 all roots of t(x) are roots of  p_n (x) ⇒t(x) divide p_n (x).
$${let}\:{put}\:\:\:{p}_{{n}} \left({x}\right)=\:{x}^{\mathrm{2}{n}} \:−\mathrm{2}{x}^{{n}} {cos}\left({n}\theta\right)+\mathrm{1}\:{and}\:{t}\left({x}\right)=\:{x}^{\mathrm{2}} −\mathrm{2}{xcos}\theta\:+\mathrm{1} \\ $$$${roots}\:{of}\:{t}\left({x}\right)?\:\:\:\Delta=\mathrm{4}\:{cos}^{\mathrm{2}} \theta\:−\mathrm{4}=−\mathrm{4}{sin}^{\mathrm{2}} \theta\:=\left(\mathrm{2}{isin}\theta\right)^{\mathrm{2}} \\ $$$${z}_{\mathrm{1}} =\frac{\mathrm{2}{cos}\theta\:+\mathrm{2}{isin}\theta}{\mathrm{2}}=\:{e}^{{i}\theta} \:\:{and}\:{z}_{\mathrm{2}\:} =\:{e}^{−{i}\theta} \:\:{let}\:{prove}\:{that} \\ $$$${p}_{{n}} \left({z}_{\mathrm{1}} \right)=\mathrm{0}\:{and}\:{p}_{{n}} \left({z}_{\mathrm{2}} \right)=\mathrm{0}\:\:{we}\:{have} \\ $$$${p}_{{n}} \left({z}_{\mathrm{1}} \right)=\:{e}^{{i}\mathrm{2}{n}\theta} \:−\mathrm{2}{e}^{{in}\theta} \frac{{e}^{{in}\theta} \:+{e}^{−{in}\theta} }{\mathrm{2}}+\mathrm{1} \\ $$$$=\:{e}^{{i}\mathrm{2}{n}\theta} \:−{e}^{{i}\mathrm{2}{n}\theta} \:−\mathrm{1}+\mathrm{1}=\mathrm{0} \\ $$$${p}_{{n}} \left({z}_{\mathrm{2}} \right)=\:{e}^{−{i}\mathrm{2}{n}\theta} \:−\mathrm{2}\:{e}^{−{in}\theta} \:\frac{{e}^{{in}\theta} \:+{e}^{−{in}\theta} }{\mathrm{2}}\:+\mathrm{1} \\ $$$$=\:{e}^{−{i}\mathrm{2}{n}\theta} \:−\mathrm{1}\:−{e}^{−{i}\mathrm{2}{n}\theta} \:\:\:+\mathrm{1}\:=\mathrm{0}\:{all}\:{roots}\:{of}\:{t}\left({x}\right)\:{are}\:{roots}\:{of} \\ $$$${p}_{{n}} \left({x}\right)\:\Rightarrow{t}\left({x}\right)\:{divide}\:{p}_{{n}} \left({x}\right). \\ $$$$ \\ $$

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