Calculate $\sin\frac{3\pi}{14}-\sin\frac{\pi}{14}-\sin\frac{5\pi}{14}$ I have interesting trigonometric expression for professionals in mathematical science. So, here it is:
$$\sin\dfrac{3\pi}{14}-\sin\dfrac{\pi}{14}-\sin\dfrac{5\pi}{14};$$
Okay! I attempt calculate it:
\begin{gather}
\sin\dfrac{3\pi}{14}-\left(\sin\dfrac{\pi}{14}+\sin\dfrac{5\pi}{14}\right)=\\
=\sin\dfrac{3\pi}{14}-\left(2\sin\dfrac{3\pi}{14}\cdot\cos\dfrac{\pi}{7}\right)=\\
=\sin\dfrac{3\pi}{14}\left[1-2\cdot\cos\dfrac{\pi}{7}\right]=...
\end{gather}
Tried everything... Here deadlock. I really do not know what to do next. Help somebody, please.
 A: For simplicity, let $x = \frac{\pi}{14}$, then we want to simplify:
$$\sin 3x-\sin x -\sin 5x$$
Multiply by $\cos x$ to get:
$$\color{blue}{\sin 3x\cos x}-\color{green}{\sin x\cos x} -\color{red}{\sin 5x\cos x} \quad (*)$$
With $\sin\alpha\cos\beta = \tfrac{1}{2}\left( \sin(\alpha+\beta)+\sin(\alpha-\beta) \right)$, you have:
$$\color{blue}{\sin 3x\cos x = \tfrac{1}{2}\left( \sin 4x +\sin 2x \right)} \quad \mbox{and} \quad \color{red}{\sin 5x\cos x = \tfrac{1}{2}\left( \sin 6x +\sin 4x \right)} $$
and $\color{green}{\sin x\cos x = \tfrac{1}{2}\sin 2x}$; so:
$$\require{cancel} (*) \quad 
\tfrac{1}{2}\left( \cancel{\sin 4x} +\bcancel{\sin 2x} \right) - \bcancel{\tfrac{1}{2}\sin 2x} - \tfrac{1}{2}\left( \sin 6x +\cancel{\sin 4x} \right) =  - \tfrac{1}{2}\sin 6x
$$
Divide again by $\cos x$:
$$- \tfrac{1}{2}\frac{\sin 6x}{\cos x} = - \tfrac{1}{2}\frac{\sin \frac{6\pi}{14}}{\cos \frac{\pi}{14}}= - \tfrac{1}{2}\frac{\cos\left( \frac{\pi}{2}-\frac{6\pi}{14}\right)}{\cos \frac{\pi}{14}} =- \tfrac{1}{2}\frac{\cos\frac{\pi}{14}}{\cos \frac{\pi}{14}} = -\frac{1}{2}$$
A: Let $14x=\pi$
$$S=\sin3x-\sin x-\sin5x=\sin3x+\sin(-x)+\sin(-5x)$$
Using How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression? ,
$$\sin(-2x)\cdot S=\cos5x-\cos7x$$
$$-\sin2x\cdot S=\cos5x=\sin2x$$
As $\cos7x=0,\cos5x=\sin2x$ as $5x+2x=\dfrac\pi2$
Can you take it from here?
