Преобразуйте в произведение

Решите задачу:

$\displaystyle \frac{\sqrt{3}}{2}-sin\alpha =sin60 \textdegree-sin\alpha=\\2sin(\frac{60\textdegree-\alpha}{2})cos(\frac{60\textdegree+\alpha}{2});\\(\frac{1}{2})^2-cos^2\alpha=cos^260\textdegree-cos^2\alpha=\\(cos60\textdegree-cos\alpha)(cos60\textdegree+cos\alpha)=\\-2sin\frac{60\textdegree+\alpha}{2}sin\frac{60\textdegree-\alpha}{2}*2cos\frac{60\textdegree+\alpha}{2}cos\frac{60\textdegree-\alpha}{2}=$

$\displaystyle -(2sin\frac{60\textdegree+\alpha}{2}cos\frac{60\textdegree+\alpha}{2})*(2sin\frac{60\textdegree-\alpha}{2}cos\frac{60\textdegree-\alpha}{2})=\\-sin(60\textdegree+\alpha)sin(60\textdegree-\alpha);$