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]

Question

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]

maheshmeghwal9 · Answer

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.

maheshmeghwal9 · Answer

@ujjwal @zepp @zzr0ck3r Plz help:)

ujjwal · Answer

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

ujjwal · Answer

So, the second part goes accordingly!

anonymous · Answer

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

ujjwal · Answer

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

anonymous · Answer

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

maheshmeghwal9 · Answer

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

ujjwal · Answer

*n
In my last reply zero=n..

anonymous · Answer

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

ujjwal · Answer

I know that is informal and there must be a very formal way to derive that relation.. And @Ishaan94 is giving it to you.

maheshmeghwal9 · Answer

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