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.
 A: You are almost there (I think you have a typo in the expression you got) : 
Using
$$\cos A-\cos B=-2\sin\frac{A+B}{2}\sin\frac{A-B}{2}$$
$$\sin A+\sin B=2\sin\frac{A+B}{2}\cos\frac{A-B}{2}$$
we have
$$\begin{align}\frac{\cos(7x)-\cos x+\sin(3x)}{\sin(7x)+\sin x-\cos(3x)}&=\frac{-2\sin(4x)\sin(3x)+\sin(3x)}{2\sin(4x)\cos(3x)-\cos(3x)}\\&=\frac{\sin(3x)(-2\sin(4x)+1)}{\cos(3x)(2\sin(4x)-1)}\\&=\frac{\sin(3x)}{\cos(3x)}\cdot \frac{-2\sin(4x)+1}{-(-2\sin(4x)+1)}\end{align}$$
A: 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 
$$\frac{\cos 7x-\cos x+{\sin 3x}}{\sin 7x+\sin x-\cos 3x}=-\tan 3x$$
$$\implies LHS=\frac{(\cos 7x-\cos x)+{\sin 3x}}{(\sin 7x+\sin x)-\cos 3x}=\frac{2\sin 4x\sin (-3x) +\sin 3x}{2\sin 4x\cos 3x-\cos 3x}$$
$$= \frac{-\sin 3x(2\sin 4x-1)}{\cos 3x(2\sin 4x-1)}=-\frac{\sin 3x}{\cos 3x}=-\tan 3x=RHS$$
A: The LHS can seen as $$\frac{\cos 7x -\cos x + \sin 3x}{\sin 7x + \sin x - \cos 3x} = \frac{-2\sin 4x \sin 3x + \sin 3x}{2\sin 4x\cos 3x - \cos 3x} = \tan 3x \frac{1-2\sin 4x}{-1+2\sin 4x} = -\tan 3x.$$
