Question Number 202325 by Ikbal last updated on 24/Dec/23
Answered by AST last updated on 24/Dec/23
![Let x=re^(iθ) =r(cosθ+isin(θ)); (1/x)=(e^(−iθ) /r)=(1/r)(cos(θ)−isin(θ)) ⇒x^n =r^n [cos(nθ)+isin(nθ)] (1/x^n )=(1/r^n )[cos(nθ)−isin(nθ)] x+(1/x)=2cos(θ)⇒(r+(1/r))cos(θ)+(r−(1/r))isin(θ) ⇒r+(1/r)=2 ∧ (r−(1/r))sin(θ)=0 ⇒r^2 −2r+1=0⇒r=1 x^n +(1/x^n )=(r^n +(1/r^n ))cos(nθ)+(r^n −(1/r^n ))isin(nθ) =2cos(nθ)](https://www.tinkutara.com/question/Q202338.png)
$${Let}\:{x}={re}^{{i}\theta} ={r}\left({cos}\theta+{isin}\left(\theta\right)\right); \\ $$$$\frac{\mathrm{1}}{{x}}=\frac{{e}^{−{i}\theta} }{{r}}=\frac{\mathrm{1}}{{r}}\left({cos}\left(\theta\right)−{isin}\left(\theta\right)\right) \\ $$$$\Rightarrow{x}^{{n}} ={r}^{{n}} \left[{cos}\left({n}\theta\right)+{isin}\left({n}\theta\right)\right] \\ $$$$\frac{\mathrm{1}}{{x}^{{n}} }=\frac{\mathrm{1}}{{r}^{{n}} }\left[{cos}\left({n}\theta\right)−{isin}\left({n}\theta\right)\right] \\ $$$${x}+\frac{\mathrm{1}}{{x}}=\mathrm{2}{cos}\left(\theta\right)\Rightarrow\left({r}+\frac{\mathrm{1}}{{r}}\right){cos}\left(\theta\right)+\left({r}−\frac{\mathrm{1}}{{r}}\right){isin}\left(\theta\right) \\ $$$$\Rightarrow{r}+\frac{\mathrm{1}}{{r}}=\mathrm{2}\:\:\wedge\:\:\left({r}−\frac{\mathrm{1}}{{r}}\right){sin}\left(\theta\right)=\mathrm{0} \\ $$$$\Rightarrow{r}^{\mathrm{2}} −\mathrm{2}{r}+\mathrm{1}=\mathrm{0}\Rightarrow{r}=\mathrm{1} \\ $$$${x}^{{n}} +\frac{\mathrm{1}}{{x}^{{n}} }=\left({r}^{{n}} +\frac{\mathrm{1}}{{r}^{{n}} }\right){cos}\left({n}\theta\right)+\left({r}^{{n}} −\frac{\mathrm{1}}{{r}^{{n}} }\right){isin}\left({n}\theta\right) \\ $$$$=\mathrm{2}{cos}\left({n}\theta\right) \\ $$