An amazing identity!

Hello! I'd like to share with you an identity I found in one of my old question papers. It says that \[\text{If}~x+\frac{1}{x}=2\cos\theta,~\text{then}~x^n+\frac{1}{x^n}=2\cos n\theta\] \(\forall n\in\mathbb{Z}\). It can be proved very easily. The proof goes like this.

Let x=cosθ+isinθx=\cos\theta+i\sin\theta. Then 1x=1cosθ+isinθ=cosθisinθ\frac{1}{x}=\frac{1}{cos\theta+i\sin\theta}=\cos\theta-i\sin\theta

Now, (cosθ±isinθ)n(\cos\theta\pm i\sin\theta)^n

=(1,±θ)n      (polar coordinates)= (1,\pm\theta)^n~~~~~~(\text{polar}~\text{coordinates})

=(1n,±nθ)= (1^n,\pm n\theta)

=(1,±nθ)= (1,\pm n\theta)

=cosnθ±isinnθ=\cos n\theta\pm i\sin n\theta

xn+1xn\therefore x^{n}+\frac{1}{x^{n}}

=cosnθ+isinnθ+cosnθisinnθ=\cos n\theta+i\sin n\theta + \cos n\theta-i\sin n\theta

=2cosnθ= 2\cos n\theta

This identity comes very handy in solving sums like this, this and other sums which I've collected in this set. If you have any other problems which can be solved by this identity, comment a link to it, and I'll add it to the set.

The only problem is that its use is very limited. It requires x+1x2<1\frac{\left\lvert x+\frac{1}{x} \right\rvert}{2} <1, as that is the condition for 2cosθ2\cos\theta. If you have any other additions to this identity, or methods to cross these limitations, comment! This wiki may come handy. Maybe it could help cross these limitations.

Note by Omkar Kulkarni
6 years, 4 months ago

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1 vote

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And yes, there should have been an extra condition specified that nZn \in \mathbb{Z}.

By the way, nice note.

Kishlaya Jaiswal - 6 years, 3 months ago

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Yeah, you're right. I added that. Thanks!

Omkar Kulkarni - 6 years, 3 months ago

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Thanks @Omkar Kulkarni for sharing this note :)

A Former Brilliant Member - 6 years, 4 months ago

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Welcome :D

Omkar Kulkarni - 6 years, 4 months ago

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