A trigonometric identity.

Question

I have determined that there is a trigonometric identity (in radians) that goes as follows:

$$(2\cos(n))^k=\cos(nk)+\sum_{i=1}^{\infty}\frac{k!}{i!(k-i)!}(\cos(n(k-2i)))$$

For $n,k\in\mathbb{C}$.

The derivation is found here for those who are interested.

I was wondering if this is a well known identity and if there are anything I should note, such as whether or not it will converge.

Also, as a bonus, I would be happy to see if someone could simplify this into something simpler. (Someone had mentioned "De Moivre" expansion, though I don't know what that means here...)

Some links would be nice, as I would like to read about this.

Update:

I am rewriting $\cos(nk)=2^k\cos^k(n)-\sum_{i=1}^{\infty}\frac{k!}{i!(k-i)!}(\cos(n(k-2i)))$ so that I may have $\cos(nk)=P_n(\cos(k))$ where $P_n$ is a polynomial, possibly with an infinite amount of terms if $n$ is not a whole number.

There is a lot of material on De Moivre's Theorem. You shouild be able to look it up on your own. The identity ! as it is written gives hint of De Moivre's being quite useful in establishing it. — Shailesh, Feb 18 '16 at 01:18
Convergence is not an issue. The sum actually terminates at $i=k$ because of the $(k-i)!$ in the denominator. — Barry Cipra, Feb 18 '16 at 01:19
@BarryCipra Actually, it won't if $k$ isn't a whole number, since $i$ must be a whole number. — Simply Beautiful Art, Feb 18 '16 at 01:20
@Shailesh Thanks, I found it. The general idea behind it is very similar to mine and can be shown to become mine... but I have the problem of $n,k$ being irrational and possibly complex. How shall I proceed with De Moivre's method in those cases? — Simply Beautiful Art, Feb 18 '16 at 01:21
@SimpleArt, ah, I didn't notice the conditions on $n$ and $k$. — Barry Cipra, Feb 18 '16 at 01:27
@SimpleArt See here for irrational and complex powers. You have to be careful of the multiple branches. — Shailesh, Feb 18 '16 at 05:23

A trigonometric identity.

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