How to prove the following equation?
\begin{eqnarray}
\sin 3A+\cos 3A&=&\left(\cos A-\sin A\right)\left(1+2\sin 2A\right)\\
\end{eqnarray}
Let's start with the left hand side.
\begin{eqnarray}
LHS&=&sin 3A+\cos 3A\\
&=&\sin \left(2A+A\right)+\cos \left(2A+A\right)\\
&=&\sin 2A\cos A+\cos 2A\sin A+\cos 2A\cos A-\sin 2A\sin A\\
&=&\left(2\sin A\cos A\right)\cos A+\left(\cos ^2A-\sin ^2A\right)\sin A+\left(\cos ^2A-\sin ^2A\right)\cos A-\left(2\sin A\cos A\right)\sin A\\
&=&\cos ^3A-\sin ^3A-\sin ^2A\cos A+\sin A\cos ^2A-2\sin ^2A\cos A+2\sin A\cos ^2A\\
\end{eqnarray}
Than, I have no idea what I should do.
Just try to expand the RHS.
\begin{eqnarray}
RHS&=&\left(\cos A-\sin A\right)\left(1+2\sin 2A\right)\\
&=&\cos A-\sin A+2\sin 2A\cos A-2\sin 2A\sin A\\
&=&\cos A-\sin A+4\sin A\cos ^2A-4\sin ^2A\cos A\\
&=&?\\
\end{eqnarray}
Maybe there are something wrong.
Anyone can tell me what I should do?
Thank you for your attention.
Best Answer
Following your method we have
$$\sin3A+\cos3A=\sin2A(\cos A-\sin A)+\cos2A(\cos A+\sin A)$$
Now,$$\cos2A(\cos A+\sin A)=(\cos^2A-\sin^2A)(\cos A+\sin A)=(\cos A-\sin A)(\cos A+\sin A)^2=(\cos A-\sin A)(1+\sin2A)$$ as $\sin2A=2\sin A\cos A$
Can you take it from here?