Showing that $\sum_{n=1}^{\infty}\left(\frac{\sin(22n)}{7n}\right)^3=\frac{1}{2}\left(\pi-\frac{22}{7}\right)^3$ How to show that? 
$$\sum_{n=1}^{\infty}\left(\frac{\sin(22n)}{7n}\right)^3=\frac{1}{2}\left(\pi-\frac{22}{7}\right)^3$$
I have no ideas to prove it, but it seems correct via Wolfram's calculator
 A: Too long for a comment.
I think that we could make amazing identities for the more general case of
$$S_{a,b}=\sum_{n=1}^{\infty}\left(\frac{\sin(an)}{bn}\right)^3=\frac 1{b^3}\sum_{n=1}^{\infty}\left(\frac{\sin(an)}{n}\right)^3$$ Using first
$$\sin ^3(a n)=\frac{3}{4} \sin (a n)-\frac{1}{4} \sin (3 a n)$$
$$S_{a,b}=\frac{3}{4b^3}\sum_{n=1}^{\infty}\frac{\sin(an)}{n^3}-\frac{1}{4b^3}\sum_{n=1}^{\infty}\frac{\sin(3an)}{n^3}$$ which is the imaginary part of 
$$T_{a,b}=\frac{3}{4b^3}\sum_{n=1}^{\infty}\frac{e^{ian}}{n^3}-\frac{1}{4b^3}\sum_{n=1}^{\infty}\frac{e^{3ian}}{n^3}$$ and now use the fact that
$$\sum_{n=1}^{\infty}\frac{e^{ikn}}{n^3}=\text{Li}_3\left(e^{i k}\right)$$
As a result
$$S_{a,b}=\frac{i}{8 b^3} \left(3 \text{Li}_3\left(e^{-i a}\right)-3 \text{Li}_3\left(e^{i
   a}\right)-\text{Li}_3\left(e^{-3 i a}\right)+\text{Li}_3\left(e^{3 i
   a}\right)\right)$$
Now, for the present case, 
$$i \left(\text{Li}_3\left(e^{-22 i}\right)-\text{Li}_3\left(e^{22 i}\right)\right)=-\frac{2}{3} (3 \pi -11) (4 \pi -11) (7 \pi -22)$$
$$i \left(\text{Li}_3\left(e^{-66 i}\right)-\text{Li}_3\left(e^{66 i}\right)\right)=-22 (\pi -3) (7 \pi -22) (10 \pi -33)$$ make
$$S_{22,b}=\frac{(7 \pi -22)^3}{2 b^3}=\frac 12\left(\frac{7\pi}b-\frac {22} b \right)^3$$
In fact, exploring the cases where
$$i\left(3 \text{Li}_3\left(e^{-i a}\right)-3 \text{Li}_3\left(e^{i
   a}\right)-\text{Li}_3\left(e^{-3 i a}\right)+\text{Li}_3\left(e^{3 i
   a}\right)\right)$$ is a multiple of a perfect cube, up to $a=100$ is found the sequence
$$\{3,4,9,10,15,16,21,\color{red}{22},23,28,29,34,35,40,41,47,48,53,54,59,60,65,66,67,72,73,78,79,84,
   85,91,92,97,98\}$$
