Bound of a certain sum of cosines Let $N$ be a sufficiently large natural number
and let $k \in \mathbb{N}$ such that $k | N$. Suppose I have
a sequence $\{ \alpha_j \}_{j=1}^N \subseteq [0,1)$, which satisfies
$$
\# \{ j \in \{1, ..., N\} : \alpha_j \in [l/k, (l+1)/k) \} = N/k
$$ 
for $l = 0, ..., k-1$.
What is a bound for 
$$
\sum_{j=1}^N cos(2 \pi \alpha_j) ?
$$
I was guessing that if $k$ is large enough, then perhaps there would be enough cancelation to make the sum small. I would appreciate any help with it. Thank you!  

Let 
$$I_l=\left\{j\in\{1,\ldots,N\}:\alpha_j\in\left[\frac{l}{k},\frac{l+1}{k}\right]\right\},\quad\hbox{for $l=0,1,\ldots k-1$.}$$
and let $b_l=\dfrac{2l+1}{2k}$; the midpoint of the interval $\left[\frac{l}{k},\frac{l+1}{k}\right]$.
Now, clearly we have
$$\sum_{j\in I_l}\vert\cos(2\pi\alpha_j)-\cos(2\pi b_l)\vert
\leq \#(I_l)(2\pi)\max_{j\in I_l}|\alpha_j-b_l|\leq\frac{\pi}{k}\#(I_l)=\frac{\pi N}{k^2}
$$
Adding these inequalities, for $l=0,1,\ldots k-1$, we get
$$
\left|\sum_{j=1}^N\cos(2\pi\alpha_j)-\frac{N}{k}\sum_{l=0}^{k-1}\cos(2\pi b_l)\right|
\leq \frac{\pi N}{k}
$$
Finally, it is an easy exercise to prove that 
$$\sum_{l=0}^{k-1}\cos(2\pi b_l)=0,$$
Therefore,
$$
\left|\sum_{j=1}^N\cos(2\pi\alpha_j)\right|
\leq \frac{\pi N}{k}.
$$
This confirms the fact that when $k$ is big, a lot of cancellation intervene to reduce the sum.$\qquad\square$
