9 \(\cos \left ( \alpha \right ) -\cos \left ( \beta \right ) \)

This can be found from (3)-(3A). Let

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

Then (3)-(3A) now becomes

\begin{align} \cos \left ( \alpha \right ) -\cos \left ( \beta \right ) & =\left ( \cos A\cos B-\sin A\sin B\right ) -\left ( \cos A\cos B+\sin A\sin B\right ) \nonumber \\ & =-2\sin A\sin B\tag {11}\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 (11) gives

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