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Question Number 143150 by ZiYangLee last updated on 10/Jun/21

If α and β are the roots of the equation   determinant (((x−cos θ),(−sin θ)),((sin θ),(x−cos θ))),  find the value of α^n +β^n , where n∈N.

$$\mathrm{If}\:\alpha\:\mathrm{and}\:\beta\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\begin{vmatrix}{{x}−\mathrm{cos}\:\theta}&{−\mathrm{sin}\:\theta}\\{\mathrm{sin}\:\theta}&{{x}−\mathrm{cos}\:\theta}\end{vmatrix}, \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\alpha^{{n}} +\beta^{{n}} ,\:\mathrm{where}\:{n}\in\mathbb{N}. \\ $$

Commented by mr W last updated on 10/Jun/21

 determinant (((x−cos θ),(−sin θ)),((sin θ),(x−cos θ))) is not an equation.  it′s only a value.  maybe you mean:   determinant (((x−cos θ),(−sin θ)),((sin θ),(x−cos θ)))=0

$$\begin{vmatrix}{{x}−\mathrm{cos}\:\theta}&{−\mathrm{sin}\:\theta}\\{\mathrm{sin}\:\theta}&{{x}−\mathrm{cos}\:\theta}\end{vmatrix}\:{is}\:{not}\:{an}\:{equation}. \\ $$$${it}'{s}\:{only}\:{a}\:{value}. \\ $$$${maybe}\:{you}\:{mean}: \\ $$$$\begin{vmatrix}{{x}−\mathrm{cos}\:\theta}&{−\mathrm{sin}\:\theta}\\{\mathrm{sin}\:\theta}&{{x}−\mathrm{cos}\:\theta}\end{vmatrix}=\mathrm{0} \\ $$

Commented by Dwaipayan Shikari last updated on 10/Jun/21

(x−cosθ)^2 +sin^2 θ=0  ⇒x=cosθ±isinθ =e^(±iθ)   α^n +β^n =e^(inθ) +e^(−inθ) =2cos(nθ)

$$\left({x}−{cos}\theta\right)^{\mathrm{2}} +{sin}^{\mathrm{2}} \theta=\mathrm{0} \\ $$$$\Rightarrow{x}={cos}\theta\pm{isin}\theta\:={e}^{\pm{i}\theta} \\ $$$$\alpha^{{n}} +\beta^{{n}} ={e}^{{in}\theta} +{e}^{−{in}\theta} =\mathrm{2}{cos}\left({n}\theta\right) \\ $$

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