## maheshmeghwal9 4 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]

1. maheshmeghwal9

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

@ujjwal @zepp @zzr0ck3r Plz help:)

3. ujjwal

$x+\frac{1}{x}=2\cos \alpha$only when x=1 and $$\alpha$$=0 This relation is not satisfied by any other values.

4. ujjwal

So, the second part goes accordingly!

5. anonymous

$r\left(e^{i\theta} + e^{- i \theta}\right) = 2\cos \alpha$

6. ujjwal

Contd. $$x^n$$=1 and n$$\alpha$$=0 for any value of zero Remember we already have x=1 and $$\alpha$$=0

7. anonymous

$x^n = e^{ni\theta}$

8. maheshmeghwal9

I gt that but isn't there logical way @ujjwal ?

9. ujjwal

*n In my last reply zero=n..

10. anonymous

$x^n + \frac1{x^n} = e^{ni\theta}+e^{n-i\theta} = \cos n \theta + i\sin n\theta +\cos n \theta - i\sin n\theta = 2\cos n \theta$

11. ujjwal

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

oh ok i see :) thanx to all for the help :)

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