how can i prove this trigonometry equation I need help on proving the following:
$$\frac{\cos {7x} - \cos {x} + \sin {3x}}{ \sin {7x} + \sin {x} - \cos {3x} }= -\tan {3x}$$
So far I've only gotten to this step:
$$\frac{-2 \sin {4x} \sin {3x} + \sin {x}}{ 2 \sin {4x} \cos {3x} - \cos {x}}$$
Any help would be appreciated as trigonometry is not my forte.

Notice, the following formula $$\color{blue}{\cos A-\cos B=2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{B-A}{2}\right)}$$ & $$\color{blue}{\sin A+\sin B=2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)}$$ Now, we have 
$$\frac{\cos 7x-\cos x+{\sin 3x}}{\sin 7x+\sin x-\cos 3x}=-\tan 3x$$
$$\implies LHS=\frac{(\cos 7x-\cos x)+{\sin 3x}}{(\sin 7x+\sin x)-\cos 3x}=\frac{2\sin 4x\sin (-3x) +\sin 3x}{2\sin 4x\cos 3x-\cos 3x}$$
$$= \frac{-\sin 3x(2\sin 4x-1)}{\cos 3x(2\sin 4x-1)}=-\frac{\sin 3x}{\cos 3x}=-\tan 3x=RHS$$
