Prove that

$=\frac{\cos 4 x \sin 3 x-\cos 2 x \sin x}{\sin 4 x \sin x+\cos 6 x \cos x}$

$=\frac{2 \cos 4 x \sin 3 x-2 \cos 2 x \sin x}{2 \sin 4 x \sin x+2 \cos 6 x \cos x}$

$=\frac{\sin (4 x+3 x)-\sin (4 x-3 x)-\{\sin (2 x+x)-\sin (2 x-x)\}}{\cos (4 x-x)-\cos (4 x+x)+\cos (6 x+x)+\cos (6 x-x)}$

$=\frac{\sin 7 x+\sin x-\sin 3 x+\sin x}{\cos 3 x-\cos 5 x+\cos 7 x+\cos 5 x}$

$=\frac{\sin 7 x-\sin 3 x}{\cos 3 x+\cos 7 x}$

$=\frac{2 \cos \frac{7 x+3 x}{2} \sin \frac{7 x-3 x}{2}}{2 \cos \frac{7 x+3 x}{2} \cos \frac{7 x-3 x}{2}}$

Using the formulas,

$2 \cos A \sin B=\sin (A+B)-\sin (A-B)$

$2 \cos A \cos B=\cos (A+B)+\cos (A-B)$

$2 \sin A \sin B=\cos (A-B)-\cos (A+B)$