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