2 \(\cos \left ( A-B\right ) \) and \(\sin \left ( A-B\right ) \)

This can be derived in similar way to the above using \(e^{i\left ( A-B\right ) }=\cos \left ( A-B\right ) +i\sin \left ( A-B\right ) \) and so on. But more easily, it can be derived from (3,4) directly by just changing replacing \(B\) by \(-B\) everywhere and then changing \(\sin \left ( -B\right ) \) to \(-\sin B\) and leaving \(\cos B\) the same since \(\cos \left ( -B\right ) =\cos B\). This is because \(\cos \) is even and \(\sin \) is odd, then (3) becomes

\begin{align} \cos \left ( A-B\right ) & =\cos A\cos B+\sin A\sin B\tag {3A}\\ \sin \left ( A-B\right ) & =-\cos A\sin B+\sin A\cos B\tag {4A}\end{align}

So we really just need to find (3) to find the 4 formulas for addition and subtractions of angles.