Prove that if \[x+\frac{1}{x}=2 \cos \alpha,\] then \[x^n+\frac{1}{x^n}= 2\cos n \alpha.\]. [This problem is based on DeMoivre's theorem]

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Prove that if \[x+\frac{1}{x}=2 \cos \alpha,\] then \[x^n+\frac{1}{x^n}= 2\cos n \alpha.\]. [This problem is based on DeMoivre's theorem]

Mathematics
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I have done like this: - Let \[x= e^{i \theta}\]&\[\frac{1}{x}=e^{-i \theta}\]This approach is also given in my book. But now how to do further ? This is my actual problem.
\[x+\frac{1}{x}=2\cos \alpha\]only when x=1 and \(\alpha\)=0 This relation is not satisfied by any other values.

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So, the second part goes accordingly!
\[r\left(e^{i\theta} + e^{- i \theta}\right) = 2\cos \alpha\]
Contd. \(x^n\)=1 and n\(\alpha\)=0 for any value of zero Remember we already have x=1 and \(\alpha\)=0
\[x^n = e^{ni\theta}\]
I gt that but isn't there logical way @ujjwal ?
*n In my last reply zero=n..
\[x^n + \frac1{x^n} = e^{ni\theta}+e^{n-i\theta} = \cos n \theta + i\sin n\theta +\cos n \theta - i\sin n\theta = 2\cos n \theta \]
I know that is informal and there must be a very formal way to derive that relation.. And @Ishaan94 is giving it to you.
oh ok i see :) thanx to all for the help :)

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