In the book Adaptive Signal Processing by Widrow, an equation (2.20) on page 23 is presented without proof as:
$$E \left[ x_k x_{k-n} \right] = \frac{1}{N} \sum_{k=1} ^{N} \sin\left(\frac{2 \pi k}{N}\right) \sin \left(\frac{2 \pi (k-n)}{N}\right) $$
$$ = \frac12 \cos \left(\frac{2 \pi n}{N}\right);\quad n = 0, 1 $$
where $N>2$ is an integer, $n\in\{0,1\}$, and $x_k = \sin \left(\frac{2 \pi k} N\right)$.
I have been trying to figure out how the above equation was simplified, but my trig is fairly rusty... any help would be appreciated!
Spent a day reviewing one of my old textbooks:
$$ E[x_k x_{k-n}] = \frac{1}{N} \sum ^N _{k=1} \sin \left( \frac{2 \pi k}{N} \right) \sin \left( \frac{2 \pi (k-n)}{N} \right) $$
Using the product to sum identity: $$ = \frac{1}{2N} \sum ^N _{k=1} \left[ \cos \left( \frac{2 \pi n}{N} \right) - \cos \left( \frac{2 \pi k + 2 \pi (k-n)}{N} \right) \right] $$
The left term is constant, giving: $$ = \frac12 \cos \left( \frac{2 \pi n}{N} \right) - \frac{1}{2N} \sum ^N _{k=1} \cos \left( \frac{4 \pi k - 2 \pi n}{N} \right) $$
Actually I was stuck here for quite some time, which was why I posted the original question, but it seems so obvious now: as k increases, the cosine will be sampled at equidistant points along one period, for exactly one period. Therefore the whole sum is equal to 0, just as if I were to integrate under the curve, which leaves:
$$ E[x_k x_{k-n}] = \frac12 \cos \left( \frac{2 \pi n}{N} \right) $$