$\sum_{i=1}^{89} \sin^{2n} (\frac{\pi}{180}i)$ is a dyadic rational

Last year's Euclid contest had a problem asking for the rational value of $\sum_{i=1}^{89} \sin^{6} (\frac{\pi}{180}i)$. I tested this sum for different even powers, and the result was always a dyadic rational (meaning it is of the form $\frac{n}{2^m}$ for positive integers $n$ and $m$). Can someone prove that $\sum_{i=1}^{89} \sin^{2n} (\frac{\pi}{180}i)$ is a dyadic rational for all positive integers $n$, or find a counterexample?

More generally, is $\sum_{i=1}^{k} \sin^{2n} (\frac{\pi}{k}i)$ always a dyadic rational for positive integers $n,k$?

Let $t=p/q$ be rational. Then $2\sin\pi t$ is an algebraic integer and its conjugates are the numbers $2\sin\pi kt$, so $\sum_k(2\sin\pi kt)^r$ is a rational integer for all positive integers $r$. So $\sum_k(\sin\pi kt)^r$ is a dyadic rational.

The details are best understood in the context of Galois Theory and Algebraic Number Theory.

Gerry Myerson posted a nice answer using Galois theory, but I thought I would post an elementary answer thanks to Greg Martin's suggestion of replacing sin(x) with its exponential expression.

$$ \sum_{j=0}^{k-1} \sin^{2n}(\frac{j\pi}{k}) = \sum_{j=0}^{k-1} \left( \frac{e^{\frac{j i \pi}{k}} - e^{-\frac{ji\pi}{k}}}{2i}\right)^{2n} = \frac{1}{(2i)^{2n}} \sum_{j=0}^{k-1} \sum_{m=0}^{2n} {2n\choose m}\left(e^{\frac{j i \pi}{k}}\right)^{m} \left(-e^{-\frac{j i \pi}{k}}\right)^{2n-m} $$ $$ = \frac{1}{(-4)^n}\sum_{m=0}^{2n} {2n\choose m}(-1)^m \sum_{j=0}^{k-1} e^{\frac{j(2m-2n)i\pi}{k}} $$

That last sum (over $j$) is a sum of roots of unity - specifically, if $d = GCD(m-n,k)$, the last sum is the sum of the $\frac{k}{d}$ roots of unity $d$ times. So the sum is equal to 0, except when $d=k$, in which case all the terms are $1$ and the sum is $k$. This implies that the summand of the sum over $m$ is an integer, so the entire expression is an integer divided by $4^n$.

In the case that $n < k$, we get the simple expression

$$ \frac{1}{(-4)^n} {2n\choose n}(-1)^n k = \frac{1}{4^n} {2n\choose n} k $$

For the original problem I posed, we have

$$ 2 \sum_{j=1}^{89} \sin^{2n}\left(\frac{\pi i}{180}\right) + 1 = \sin^{2n}(0) + \sum_{j=1}^{89} \sin^{2n}\left(\frac{\pi i}{180}\right) + \sin^{2n}(\frac{90\pi}{180}) + \sum_{j=91}^{179} \sin^{2n}\left(\frac{\pi i}{180}\right) $$ $$ =\sum_{j=0}^{179} \sin^{2n}\left(\frac{\pi i}{180}\right) $$

and we have proven that the last sum is a dyadic ratioanl, so $\sum_{j=1}^{89} \sin^{2n}\left(\frac{\pi i}{180}\right)$ is a dyadic rational as well.