Question Number 31496 by abdo imad last updated on 09/Mar/18
![let a∈]0,π[ and A(x)= x^(2n) −2cos(na)x^n +1 1)factorize inside C[x] A(x) 2) factorize inside R[x] A(x).](Q31496.png)
$$\left.{let}\:{a}\in\right]\mathrm{0},\pi\left[\:\:\:{and}\:{A}\left({x}\right)=\:{x}^{\mathrm{2}{n}} \:−\mathrm{2}{cos}\left({na}\right){x}^{{n}} \:+\mathrm{1}\right. \\ $$$$\left.\mathrm{1}\right){factorize}\:{inside}\:{C}\left[{x}\right]\:{A}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{factorize}\:{inside}\:{R}\left[{x}\right]\:{A}\left({x}\right). \\ $$
Commented by abdo imad last updated on 14/Mar/18
![first let put x^n =z ⇔ x=^n (√z) ⇒ A(x)=z^2 −2cos(na)z +1 Δ^′ =cos^2 (na)−1=−sin^2 (na)=(isin(na))^2 z_1 = cos(na) +isin(na)=e^(ina) and z_2 =e^(−ina) ⇒ x=z^(1/n) ⇒ x=e^(ia) or x=e^(−a) ⇒ A(x)=(z−e^(ina) )(z−e^(−ina) )=(x^n −e^(ina) )(x^n −e^(−ina) ) A(x)=0 ⇔ x^n = e^(ina) or x^n =e^(−ina) ⇔ (x e^(−ia) )^n =1x or (x e^(ia) )^n =1 ⇔ x e^(−ia) =e^(i((2kπ)/n)) or x e^(ia) =e^(i((2kπ)/n)) ⇒ x=e^(i(((2kπ)/n)+a)) or x= e^(i(((2kπ)/n)−a)) so the roots of A(x) are x_k =e^(i(((2kπ)/n)+a)) and t_k = e^(i(((2kπ)/n) −a)) and k∈[[0,n−1]].and A(x)= Π_(k=0) ^(n−1) (x −x_k )(x−t_k ) A(x)= Π_(k=0) ^(n−1) (x−e^(i(((2kπ)/n)+a)) ) Π_(k=0) ^(n−1) (x −e^(i(((2kπ)/(n ))−a)) ) be continued...](Q31783.png)
$${first}\:{let}\:{put}\:{x}^{{n}} \:={z}\:\:\Leftrightarrow\:{x}=^{{n}} \sqrt{{z}}\:\Rightarrow\:{A}\left({x}\right)={z}^{\mathrm{2}} \:−\mathrm{2}{cos}\left({na}\right){z}\:+\mathrm{1} \\ $$$$\Delta^{'} \:={cos}^{\mathrm{2}} \left({na}\right)−\mathrm{1}=−{sin}^{\mathrm{2}} \left({na}\right)=\left({isin}\left({na}\right)\right)^{\mathrm{2}} \\ $$$${z}_{\mathrm{1}} =\:{cos}\left({na}\right)\:+{isin}\left({na}\right)={e}^{{ina}} \:\:{and}\:{z}_{\mathrm{2}} ={e}^{−{ina}} \:\Rightarrow \\ $$$${x}={z}^{\frac{\mathrm{1}}{{n}}} \:\Rightarrow\:{x}={e}^{{ia}} \:{or}\:\:{x}={e}^{−{a}} \:\Rightarrow \\ $$$${A}\left({x}\right)=\left({z}−{e}^{{ina}} \right)\left({z}−{e}^{−{ina}} \right)=\left({x}^{{n}} \:−{e}^{{ina}} \right)\left({x}^{{n}} \:−{e}^{−{ina}} \right) \\ $$$${A}\left({x}\right)=\mathrm{0}\:\:\Leftrightarrow\:{x}^{{n}} \:=\:{e}^{{ina}} \:{or}\:{x}^{{n}} ={e}^{−{ina}} \:\Leftrightarrow\:\left({x}\:{e}^{−{ia}} \right)^{{n}} =\mathrm{1}{x} \\ $$$${or}\:\left({x}\:{e}^{{ia}} \right)^{{n}} =\mathrm{1}\:\Leftrightarrow\:{x}\:{e}^{−{ia}} ={e}^{{i}\frac{\mathrm{2}{k}\pi}{{n}}} \:\:{or}\:\:\:{x}\:{e}^{{ia}} \:={e}^{{i}\frac{\mathrm{2}{k}\pi}{{n}}} \:\Rightarrow \\ $$$${x}={e}^{{i}\left(\frac{\mathrm{2}{k}\pi}{{n}}+{a}\right)} \:{or}\:{x}=\:{e}^{{i}\left(\frac{\mathrm{2}{k}\pi}{{n}}−{a}\right)} \:{so}\:{the}\:{roots}\:{of}\:{A}\left({x}\right)\:{are} \\ $$$${x}_{{k}} ={e}^{{i}\left(\frac{\mathrm{2}{k}\pi}{{n}}+{a}\right)} \:\:{and}\:{t}_{{k}} =\:{e}^{{i}\left(\frac{\mathrm{2}{k}\pi}{{n}}\:−{a}\right)} \:{and}\:{k}\in\left[\left[\mathrm{0},{n}−\mathrm{1}\right]\right].{and} \\ $$$${A}\left({x}\right)=\:\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\left({x}\:−{x}_{{k}} \right)\left({x}−{t}_{{k}} \right) \\ $$$${A}\left({x}\right)=\:\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left({x}−{e}^{{i}\left(\frac{\mathrm{2}{k}\pi}{{n}}+{a}\right)} \right)\:\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\left({x}\:−{e}^{{i}\left(\frac{\mathrm{2}{k}\pi}{{n}\:}−{a}\right)} \right)\:\:{be}\:{continued}... \\ $$