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maheshmeghwal9

  • 3 years 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]

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  1. maheshmeghwal9
    • 3 years ago
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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.

  2. maheshmeghwal9
    • 3 years ago
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    @ujjwal @zepp @zzr0ck3r Plz help:)

  3. ujjwal
    • 3 years ago
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    \[x+\frac{1}{x}=2\cos \alpha\]only when x=1 and \(\alpha\)=0 This relation is not satisfied by any other values.

  4. ujjwal
    • 3 years ago
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    So, the second part goes accordingly!

  5. Ishaan94
    • 3 years ago
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    \[r\left(e^{i\theta} + e^{- i \theta}\right) = 2\cos \alpha\]

  6. ujjwal
    • 3 years ago
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    Contd. \(x^n\)=1 and n\(\alpha\)=0 for any value of zero Remember we already have x=1 and \(\alpha\)=0

  7. Ishaan94
    • 3 years ago
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    \[x^n = e^{ni\theta}\]

  8. maheshmeghwal9
    • 3 years ago
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    I gt that but isn't there logical way @ujjwal ?

  9. ujjwal
    • 3 years ago
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    *n In my last reply zero=n..

  10. Ishaan94
    • 3 years ago
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    \[x^n + \frac1{x^n} = e^{ni\theta}+e^{n-i\theta} = \cos n \theta + i\sin n\theta +\cos n \theta - i\sin n\theta = 2\cos n \theta \]

  11. ujjwal
    • 3 years ago
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    I know that is informal and there must be a very formal way to derive that relation.. And @Ishaan94 is giving it to you.

  12. maheshmeghwal9
    • 3 years ago
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    oh ok i see :) thanx to all for the help :)

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