I was playing around with the sum $\sum_{k = 1}^{n} \sin k$, and using very loose rigour I arrived at the following:
Proposition. Let $n \equiv n_0 \pmod {44}$ and $n_0 \equiv n_1 \pmod {6}$. Then $$\sum_{k = 1}^{n} \sin k \sim \frac {1} {2} \left ( 3 - |3 - n_1| \right ).$$
Does heuristics support this claim?
Note: I can provide my "proof" on request.
I think it appropriate to present my rigour here now that it created sufficient curiosity.
Consider $$f (n) = \sum_{k = 1}^{n} \sin k.$$ Let $a_k = \left [ \frac {2k} {\pi} \right ]$ for $1 \leqslant k \leqslant n$ and $r_1, r_2, \cdots, r_n$ be real numbers in the interval $[0, 1]$. Then, denote $$\sin k = \begin {cases} r_k, & \text {if } a_k \equiv 0 \pmod {4} \\ 1 - r_k, & \text {if } a_k \equiv 1 \pmod {4} \\ -r_k, & \text {if } a_k \equiv 2 \pmod {4} \\ r_k - 1, & \text {if } a_k \equiv 3 \pmod {4} \\ \end {cases}$$ Here $r_{\theta}$ is the side of the right triangle just in front of the angle $\theta$. By Weyl's criterion, we see that $r_k$ is equidistributed modulo $1$, so the average of real numbers $r_k$ is $\frac {1} {2}$. Hence, $$\sin k \approx \begin {cases} \frac {1} {2}, & \text {if } a_k \equiv 0 \pmod {4} \\ \frac {1} {2}, & \text {if } a_k \equiv 1 \pmod {4} \\ - \frac {1} {2}, & \text {if } a_k \equiv 2 \pmod {4} \\ - \frac {1} {2}, & \text {if } a_k \equiv 3 \pmod {4} \\ \end {cases}$$ Since $\frac {2mk} {\pi}$ is close to an integer when $m = 11 t$ and the period of $\sin k$ with respect to $a_k$ is $4$, it is safe to assume $f (44t + n_0) \approx f (n_0)$ for any integer $t$. Let $n = 44 t + n_1$. Also, since $a_{6j + i} \equiv a_i \pmod {4}$ for any integers $i, j$, we have by setting $i = n_1$ the desired result.
This is a slightly simplified and fast version of what I'd done to arrive at the ridiculous expression that my question was about.