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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]
 one year ago
 one year ago
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]
 one year ago
 one year ago

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maheshmeghwal9Best ResponseYou've already chosen the best response.0
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.
 one year ago

maheshmeghwal9Best ResponseYou've already chosen the best response.0
@ujjwal @zepp @zzr0ck3r Plz help:)
 one year ago

ujjwalBest ResponseYou've already chosen the best response.1
\[x+\frac{1}{x}=2\cos \alpha\]only when x=1 and \(\alpha\)=0 This relation is not satisfied by any other values.
 one year ago

ujjwalBest ResponseYou've already chosen the best response.1
So, the second part goes accordingly!
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.2
\[r\left(e^{i\theta} + e^{ i \theta}\right) = 2\cos \alpha\]
 one year ago

ujjwalBest ResponseYou've already chosen the best response.1
Contd. \(x^n\)=1 and n\(\alpha\)=0 for any value of zero Remember we already have x=1 and \(\alpha\)=0
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.2
\[x^n = e^{ni\theta}\]
 one year ago

maheshmeghwal9Best ResponseYou've already chosen the best response.0
I gt that but isn't there logical way @ujjwal ?
 one year ago

ujjwalBest ResponseYou've already chosen the best response.1
*n In my last reply zero=n..
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.2
\[x^n + \frac1{x^n} = e^{ni\theta}+e^{ni\theta} = \cos n \theta + i\sin n\theta +\cos n \theta  i\sin n\theta = 2\cos n \theta \]
 one year ago

ujjwalBest ResponseYou've already chosen the best response.1
I know that is informal and there must be a very formal way to derive that relation.. And @Ishaan94 is giving it to you.
 one year ago

maheshmeghwal9Best ResponseYou've already chosen the best response.0
oh ok i see :) thanx to all for the help :)
 one year ago
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