I want to show $$f(x)=\sin \left(\frac{2 x}{17}\right)-8 \sin \left(\frac{4 x}{17}\right)+27 \sin \left(\frac{6 x}{17}\right)-64 \sin \left(\frac{8 x}{17}\right)+125 \sin \left(\frac{10 x}{17}\right)-216 \sin \left(\frac{12 x}{17}\right)+343 \sin \left(\frac{14 x}{17}\right)-512 \sin \left(\frac{16 x}{17}\right)<0,\quad x\in(0,\pi/2).$$ This trigonometric inequality has been verified by Mathematica using the Plot commend. I found $f$ can be rewritten
$$f(x)=\sin \left(\frac{2 x}{17}\right)-2^3 \sin \left(\frac{2*2 x}{17}\right)+3^3 \sin \left(\frac{2*3 x}{17}\right)-4^3 \sin \left(\frac{2*4 x}{17}\right)+5^3 \sin \left(\frac{2*5 x}{17}\right)-6^3 \sin \left(\frac{2*6 x}{17}\right)+7^3 \sin \left(\frac{2*7 x}{17}\right)-8^3 \sin \left(\frac{2*8 x}{17}\right)<0,\quad x\in(0,\pi/2).$$ However, I cannot give a rigorous proof of it. Any suggestion, idea, or comment is welcome, thanks!
All right, that is easier than expected. If we define $$ C_n(x) = \sum_{k=1}^{n}(-1)^k k^3\sin\left(\frac{2kx}{2n+1}\right)\tag{1} $$ through $\sin(\theta)=\frac{e^{i\theta}-e^{-i\theta}}{2i}$ we have: $$\small{ C_n(x) = -\frac{i (-1)^n}{2z^n (1+z)^4}\left(3 n^2 z (1+z)^2 \left(-1+z^{2 n}\right)+z (1-4z+z^2)\left(-1+z^{2 n}\right)-3 n z \left(-1+z^2\right) \left(1+z^{2 n}\right)+n^3 (1+z)^3 \left(-1+z^{1+2 n}\right)\right)}\tag{2} $$ with $z=\exp\left(\frac{2ix}{2n+1}\right)$. Such explicit formula gives that $C_n(x)$ and $(-1)^n$ have the same sign on the interval $(0,\pi)$. So $(1)$ really is an uglier but simpler version of the Fejer-Jackson inequality.