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 \(x=\cos\theta+i\sin\theta\). Then \(\frac{1}{x}=\frac{1}{cos\theta+i\sin\theta}=\cos\theta-i\sin\theta\)

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

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

\(= (1^n,\pm n\theta)\)

\(= (1,\pm n\theta)\)

\(=\cos n\theta\pm i\sin n\theta\)

\(\therefore x^{n}+\frac{1}{x^{n}}\)

\(=\cos n\theta+i\sin n\theta + \cos n\theta-i\sin n\theta \)

\(= 2\cos n\theta\)

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 \(\frac{\left\lvert x+\frac{1}{x} \right\rvert}{2} <1\), as that is the condition for \(2\cos\theta\). 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.

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## Comments

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TopNewestAnd yes, there should have been an extra condition specified that \(n \in \mathbb{Z}\).

By the way, nice note.

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

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Thanks @Omkar Kulkarni for sharing this note :)

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Welcome :D

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