# Product-to-sum trigonometry identity

See:

I cannot find this anywhere on the web - does anybody know a reference? Certainly the one wiki gives does not cover it.

I'm guessing it is used for something like $\cos(x)\cos(5x)$ (easier done with the other identities mentioned) to convert into a sum but I'm not sure really. I don't understand the $S=\{1,-1\}^n$ bit.

• $S$ is the set of all $n$-tuple with coordinates $-1$ or $1$. – enzotib Jul 20 '14 at 12:36

Note that the Wikipedia formula you are referring to contains a wrong extra factor of $\frac12$ (this can be seen already by setting $n=1$). More formal proof in the general case: \begin{align} 2^n\prod_{k=1}^{n}\cos\theta_k&=\prod_{k=1}^{n}\left(e^{i\theta_k}+e^{-i\theta_k}\right)=\prod_{k=1}^{n}\left(\sum_{\epsilon_k=\pm 1}e^{i\epsilon_k\theta_k}\right)=\\&=\sum_{\epsilon_1,\ldots,\epsilon_n=\pm 1}e^{i\left(\epsilon_1\theta_1+\ldots+\epsilon_n\theta_n\right)}=\\ &=\frac12\sum_{\epsilon_1,\ldots,\epsilon_n=\pm 1}\left(e^{i\left(\epsilon_1\theta_1+\ldots+\epsilon_n\theta_n\right)}+e^{-i\left(\epsilon_1\theta_1+\ldots+\epsilon_n\theta_n\right)}\right)=\\ &=\sum_{\epsilon_1,\ldots,\epsilon_n=\pm 1}\cos\left(\epsilon_1\theta_1+\ldots+\epsilon_n\theta_n\right). \end{align}
• @onepound: One thing that may add some intuition as to why O.L. suggests this is that the factor of $2^{n-1}$ is suggestive: by multiplying both sides by $2^n$, the only functions involved are all of the form $2 \cos \phi$. – Semiclassical Jul 20 '14 at 12:52
• @onepound if you substitute that ($n=2$, $\theta_1=x$, $\theta_2=5x$) then in the corrected version on the right you will have $$\frac1{2^2} \left[\cos (+x+5x)+\cos (+x-5x)+\cos (-x+5x)+\cos (-x-5x)\right]= \frac{\cos6x+\cos4x}{2}$$ – Start wearing purple Jul 20 '14 at 15:18