8 \(\sin \left ( \alpha \right ) -\sin \left ( \beta \right ) \)

This can be found from (4)-(4A). Let

\begin{align} A+B & =\alpha \tag {7}\\ A-B & =\beta \nonumber \end{align}

Then (4)-(4A) now becomes

\begin{align} \sin \left ( \alpha \right ) -\sin \left ( \beta \right ) & =\left ( \cos A\sin B+\sin A\cos B\right ) +\cos A\sin B-\sin A\cos B\nonumber \\ & =2\cos A\sin B\tag {10}\end{align}

Now we solve for \(A,B\) from (7). Which gives

\begin{align*} A & =\frac {\alpha +\beta }{2}\\ B & =\frac {\alpha -\beta }{2}\end{align*}

Substituting the above in (10) gives

\[ \sin \left ( \alpha \right ) -\sin \left ( \beta \right ) =2\cos \left ( \frac {\alpha +\beta }{2}\right ) \sin \left ( \frac {\alpha -\beta }{2}\right ) \]