1 \(\cos \left ( A+B\right ) \) and \(\sin \left ( A+B\right ) \)

\begin{equation} e^{i\left ( A+B\right ) }=\cos \left ( A+B\right ) +i\sin \left ( A+B\right ) \tag {1}\end{equation}

But \(e^{i\left ( A+B\right ) }=e^{iA}e^{iB}\) therefore

\begin{align} e^{iA}e^{iB} & =\left ( \cos A+i\sin A\right ) \left ( \cos B+i\sin B\right ) \nonumber \\ & =\cos A\cos B+i\cos A\sin B+i\sin A\cos B-\sin A\sin B\nonumber \\ & =\left ( \cos A\cos B-\sin A\sin B\right ) +i\left ( \cos A\sin B+\sin A\cos B\right ) \tag {2}\end{align}

Now (1) is the same as (2). Hence the real part and the imaginary parts must be the same. Therefore

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