<s>[INST] 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.
 [/INST] 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
$$\begin{align}\frac{\cos(7x)-\cos x+\sin(3x)}{\sin(7x)+\sin x-\cos(3x)}&=\frac{-2\sin(4x)\sin(3x)+\sin(3x)}{2\sin(4x)\cos(3x)-\cos(3x)}\\&=\frac{\sin(3x)(-2\sin(4x)+1)}{\cos(3x)(2\sin(4x)-1)}\\&=\frac{\sin(3x)}{\cos(3x)}\cdot \frac{-2\sin(4x)+1}{-(-2\sin(4x)+1)}\end{align}$$
</s>