12 \(\sin \left ( A\right ) \sin \left ( B\right ) \)

(3)-(3A) gives

\begin{align*} \cos \left ( A+B\right ) & =\cos A\cos B-\sin A\sin B\\ \cos \left ( A-B\right ) & =\cos A\cos B+\sin A\sin B\\ \cos \left ( A+B\right ) -\cos \left ( A-B\right ) & =-2\sin A\sin B \end{align*}

Hence

\[ \sin A\sin B=\frac {1}{2}\left ( \cos \left ( A-B\right ) -\cos \left ( A+B\right ) \right ) \]