Showing that $R= \frac{1}{n}\sum_{k=1}^n\cos(\theta_k - \overline{\theta})$ given a system of equations

Question

Let angles $\theta_1,\dots,\theta_n$ be given and define $C = \frac{1}{n}\sum_{k=1}^n\cos(\theta_k)$ and $S = \frac{1}{n}\sum_{k=1}^n\sin(\theta_k)$. Then, there exists a value $\overline{\theta}$ such that $C = R\cos(\overline{\theta})$ and $S = R\sin(\overline{\theta})$ when $R := \sqrt{C^2 + S^2}$. My source material claims that it follows immediately from these equalities that $\sum_{k=1}^n\cos(\theta_k - \overline{\theta}) = R$ and $\sum_{k=1}^n\sin(\theta_k - \overline{\theta}) = 0$. I've managed to prove that $\sum_{k=1}^n\sin(\theta_k - \overline{\theta}) = 0$, but I'm not sure how I should show that $\sum_{k=1}^n\cos(\theta_k - \overline{\theta}) = R$.

Answer 1

$Re^{i\overline {\theta}}=C+iS=\frac 1 n \sum\limits_{k=1}^{n} e^{i\theta_k}$. Multiply both side by $e^{-i\overline {\theta}}$ and take real part.