11 \(\sin \left ( A\right ) \cos \left ( B\right ) \)

Adding (4)+(4A) gives

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

Hence

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