# A trigonometric identity.

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...)

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.
• I am consider complex and irrational values for $n,k$. – Simply Beautiful Art Feb 18 '16 at 1:14
• 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 1:19
• @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 1:20
• @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 1:21