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Question Number 31496 by abdo imad last updated on 09/Mar/18
leta∈]0,π[andA(x)=x2n−2cos(na)xn+11)factorizeinsideC[x]A(x)2)factorizeinsideR[x]A(x).
Commented by abdo imad last updated on 14/Mar/18
firstletputxn=z⇔x=nz⇒A(x)=z2−2cos(na)z+1Δ′=cos2(na)−1=−sin2(na)=(isin(na))2z1=cos(na)+isin(na)=einaandz2=e−ina⇒x=z1n⇒x=eiaorx=e−a⇒A(x)=(z−eina)(z−e−ina)=(xn−eina)(xn−e−ina)A(x)=0⇔xn=einaorxn=e−ina⇔(xe−ia)n=1xor(xeia)n=1⇔xe−ia=ei2kπnorxeia=ei2kπn⇒x=ei(2kπn+a)orx=ei(2kπn−a)sotherootsofA(x)arexk=ei(2kπn+a)andtk=ei(2kπn−a)andk∈[[0,n−1]].andA(x)=∏k=0n−1(x−xk)(x−tk)A(x)=∏k=0n−1(x−ei(2kπn+a))∏k=0n−1(x−ei(2kπn−a))becontinued...
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