How to prove $\sin 3A+\cos 3A=(\cos A-\sin A)(1+2\sin 2A)$ How to prove the following equation?
\begin{eqnarray}
\sin 3A+\cos 3A&amp;=&amp;\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&amp;=&amp;sin 3A+\cos 3A\\
&amp;=&amp;\sin \left(2A+A\right)+\cos \left(2A+A\right)\\
&amp;=&amp;\sin 2A\cos A+\cos 2A\sin A+\cos 2A\cos A-\sin 2A\sin A\\
&amp;=&amp;\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\\
&amp;=&amp;\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&amp;=&amp;\left(\cos A-\sin A\right)\left(1+2\sin 2A\right)\\
&amp;=&amp;\cos A-\sin A+2\sin 2A\cos A-2\sin 2A\sin A\\
&amp;=&amp;\cos A-\sin A+4\sin A\cos ^2A-4\sin ^2A\cos A\\
&amp;=&amp;?\\
\end{eqnarray}

Maybe there are something wrong.

Anyone can tell me what I should do?

Thank you for your attention.
Using$$\sin3A=3\sin A-4\sin^3A,\cos3A=4\cos^3A-3\cos A$$

$$\sin3A+\cos3A=3\sin A-4\sin^3A+4\cos^3A-3\cos A=3(\sin A-\cos A)-4(\sin^3A-\cos^3A)$$

Now, $\displaystyle\sin^3A-\cos^3A=(\sin A-\cos A)(\sin^2A+\sin A\cos A+\cos^2A)=(\sin A-\cos A)\left(1+\frac{\sin2A}2\right)$
LHS
sin(2A+A)+cos(2A+A)
= sin2A.cosA + cos2A.sinA + cos2A.cosA- sin2A.sinA
= 2sin A.cos^2A +(cos^2A-sin^2A).sinA + (cos^2A-sin^2A).cosA -2sin^2A.cosA
=cos^3A-sin^3A + sinA.cosA(2cosA+cosA- sinA-2sinA)
=(cosA-sinA)(cos^2A+cosA.sinA+sin^2A) +sinA.cosA(3cosA-3sinA)
=(cosA-sinA)(1+cosA.sinA)+3sinAcosA(cosA-sinA)
=(cosA-sinA)(1+4cosA.sinA)
=(cosA-sinA)(1+2sin2A)
=RHS