3 \(\cos \left ( 2A\right ) \) and \(\sin \left ( 2A\right ) \)

These also can be found from (3,4). By replacing \(B\) with \(A\) resulting in

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

Therefore

\begin{align} \cos \left ( 2A\right ) & =\cos ^{2}A-\sin ^{2}A\tag {3C}\\ \sin \left ( 2A\right ) & =2\cos A\sin A\tag {4C}\end{align}

Or we could use Euler formula, but the above is simpler. To use Euler formula, we write

\begin{equation} e^{i\left ( 2A\right ) }=\cos \left ( 2A\right ) +i\sin \left ( 2A\right ) \tag {5}\end{equation}

But \(e^{i\left ( 2A\right ) }=e^{iA}e^{iA}\) therefore

\begin{align} e^{iA}e^{iA} & =\left ( \cos A+i\sin A\right ) \left ( \cos A+i\sin A\right ) \nonumber \\ & =\cos ^{2}A+2i\cos A\sin A-\sin ^{2}A\nonumber \\ & =\left ( \cos ^{2}A-\sin ^{2}A\right ) +i\left ( 2\cos A\sin A\right ) \tag {6}\end{align}

Comparing (5,6) shows that

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

Which is the same as (3C,4C) above.