Question: > 
> 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?

Answer: 
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}
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,

$$ S \cdot \sin \frac \pi{17} = \sin \frac {16\pi}{17}$$

$$ S = \frac {\sin \left( \pi - \frac \pi{17} \right)}{\sin \frac \pi{17}}$$

$$S=1$$