$\sum_{i=1}^{89} \sin^{2n} (\frac{\pi}{180}i)$ 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 $n,k$?
 A: 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. 
A: 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.
