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Question Number 28371 by abdo imad last updated on 24/Jan/18

$$provethat{x}^2-2xcos\theta +1divide{x}^{2n}-2x^{n}cos\left(n\theta \right)+1$$

Commented by abdo imad last updated on 25/Jan/18

$$letput{p}_n\left(x\right)=x^{2n}-2x^{n}cos\left(n\theta \right)+1andt\left(x\right)=x^2-2xcos\theta +1$$$$rootsoft\left(x\right)?\mathrm{\Delta }=4{cos}^2\theta -4=-4{sin}^2\theta ={\left(2isin\theta \right)}^2$$$$z_1=\frac{2cos\theta +2isin\theta }{2}=e^{i\theta }and{z}_2=e^{-i\theta }letprovethat$$$$p_n\left(z_1\right)=0and{p}_n\left(z_2\right)=0wehave$$$$p_n\left(z_1\right)=e^{i2n\theta }-2e^{in\theta }\frac{e^{in\theta }+e^{-in\theta }}{2}+1$$$$=e^{i2n\theta }-e^{i2n\theta }-1+1=0$$$$p_n\left(z_2\right)=e^{-i2n\theta }-2e^{-in\theta }\frac{e^{in\theta }+e^{-in\theta }}{2}+1$$$$=e^{-i2n\theta }-1-e^{-i2n\theta }+1=0allrootsoft\left(x\right)arerootsof$$$$p_n\left(x\right)\Rightarrow t\left(x\right)divide{p}_n\left(x\right).$$$$$$

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