Find $4\sin (x)\,\big(\sin(3x)+\sin (7x)+\sin(11x)+\sin(15x)\big)$ when $x=\frac{\pi}{34}$. 
If $x=\frac{\pi}{34}$, then find the value of
  $$S=4\sin (x)\,\big(\sin(3x)+\sin (7x)+\sin(11x)+\sin(15x)\big).$$

Source: Joszef Wildt International Math Competition Problem
My attempt:
$$\sin(3x)+\sin(15x)=2\sin(9x)\cos(6x)$$
$$\sin(7x)+\sin(11x)=2\sin(9x)\cos(2x)$$
So $$S=8\sin(x)\sin(9x)\big(\cos(6x)+\cos(2x)\big)$$
$$S=16\sin(x)\sin(9x)\cos(4x)\cos(2x)$$
$$S=16\sin(x)\sin(9x)\sin(13x)\sin(15x)$$
any clue here?
 A: Alternatively, telescope the expression with
\begin{align}
&4\sin x(\sin 3x+\sin 7x+\sin 11x+\sin 15x)\\
=&\frac{2\sin2x}{\cos x}(\sin 3x+\sin 7x+\sin 11x+\sin 15x)\\
=&\frac 1{\cos x} [ (\cos x-\cos5x) + (\cos 5x-\cos9x) + (\cos 9x-\cos13x) + (\cos 13x-\cos17x)]\\
=& \frac{1}{\cos x}(\cos x - \cos 17x)
=1-\frac{\cos\frac\pi2}{\cos\frac{\pi}{34}}=1
\end{align}
A: as You simplified and as @WE Tutorial School pointed out,
$$S=16\cos\frac{\pi}{17}\cos\frac{2\pi}{17}\cos\frac{4\pi}{17}\cos\frac{8\pi}{17}.$$ 
Applying $\sin 2x = 2 \sin x \cos x$ 4 times,
$$ S \cdot \sin \frac \pi{17}  = \sin \frac {16\pi}{17}$$
$$ S = \frac {\sin \left( \pi - \frac \pi{17} \right)}{\sin \frac \pi{17}}$$
$$S=1$$
A: As you calculated, we have
$$S=16\sin(x)\sin(9x)\cos(4x)\cos(2x).$$
Using $\sin x=\cos(\tfrac{\pi}{2}-x)$, we get
$$S\left(\frac{\pi}{34}\right)=16\cos\frac{\pi}{17}\cos\frac{2\pi}{17}\cos\frac{4\pi}{17}\cos\frac{8\pi}{17}.$$
Let $\zeta=e^{2\pi i/17}$. Noting that $\zeta^k+\zeta^{-k}=2\cos \frac{2k\pi}{17}$ and $\cos \frac{\pi}{17}=-\cos \frac{16\pi}{17}$, we have
$$S\left(\dfrac{\pi}{34}\right)=-(\zeta+\zeta^{-1})(\zeta^2+\zeta^{-2})(\zeta^4+\zeta^{-4})(\zeta^8+\zeta^{-8})=\boxed{1}$$
using $\sum_{k=1}^{16} \zeta^k=-1$.
