## maheshmeghwal9 Group Title 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] 2 years ago 2 years ago

1. maheshmeghwal9 Group Title

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 Group Title

@ujjwal @zepp @zzr0ck3r Plz help:)

3. ujjwal Group Title

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

4. ujjwal Group Title

So, the second part goes accordingly!

5. Ishaan94 Group Title

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

6. ujjwal Group Title

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

7. Ishaan94 Group Title

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

8. maheshmeghwal9 Group Title

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

9. ujjwal Group Title

*n In my last reply zero=n..

10. Ishaan94 Group Title

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

11. ujjwal Group Title

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 Group Title

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