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.