Помогите пожалуйста, 40 баллов

Решите задачу:

$\frac{\sin4\alpha}{\cos3\alpha-\cos\alpha}=\frac{2\sin2\alpha\cos2\alpha}{\cos(2\alpha+\alpha)-\cos\alpha}=\frac{4\sin\alpha\cos\alpha\cos2\alpha}{\cos2\alpha\cos\alpha-\sin2\alpha\sin\alpha-\cos\alpha}=\\$
$=\frac{4\sin\alpha\cos\alpha\cos2\alpha}{\cos2\alpha\cos\alpha-2\cos\alpha\sin^2\alpha-\cos\alpha}=4\frac{\cos\alpha}{\cos\alpha}\cdot\frac{\sin\alpha\cos2\alpha}{\cos2\alpha-2\sin^2\alpha-1}=\\ =4\frac{\sin\alpha\cos2\alpha}{1-2\sin^2\alpha-2\sin^2\alpha-1}=\frac{4}{-4}\cdot\frac{\sin\alpha\cos2\alpha}{\sin^2\alpha}=-\frac{\cos2\alpha}{\sin\alpha}=\\ =-\frac{\cos^2\alpha-\sin^2\alpha}{\sin\alpha}=\sin\alpha-\cos\alpha\cdot ctg\alpha$