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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.

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up vote 1 down vote accepted

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?

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@CasperLi, how about this? – lab bhattacharjee Oct 1 '13 at 18:42
Nice, I can prove cos2A(cosA+sinA)=(cosA−sinA)(1+sin2A) – Casper LI Oct 2 '13 at 12:59
so, sin2A(cosA−sinA)+(1+sin2A)(cosA−sinA)=(cosA−sinA)(sin2A+1+sin2A), and (1+sin2A+sin2A)=(1+2sin2A),rigth? ---->Proved – Casper LI Oct 2 '13 at 13:11
@CasperLi, yes, right you are – lab bhattacharjee Oct 2 '13 at 15:22

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, $\displaystyle\sin^3A-\cos^3A=(\sin A-\cos A)(\sin^2A+\sin A\cos A+\cos^2A)=(\sin A-\cos A)\left(1+\frac{\sin2A}2\right)$

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sin 3A +cos 3A= sin(2A+A)+cos(2A+A)

= sin 2A.cos A+cos 2A.sin A+cos2A.cos A-sin A.sin 2A

= 2.sin A.cos A.cos A+(1-2sin^2 A).sin A+(2cos^2 A-1).cos A-sin A.2sin A.cos A

= 2.sin A.cos^2 A+sin A-2sin^3 A+2cos^3 A-cos A-2sin^2 A.cos A

= 2sin A(1-sin^2 A)+sin A-2sin^3 A+2cos^3 A-cos A-2(1-cos^2 A).cos A

= 3sin A-4sin^3 A+4cos^3 A-3cos A

= 3(sin A-cos A)-4(sin^3 A-cos^3 A)

Remember: sin^3 A-cos^3 A= (sin A-cos A)((sin A-cos A)^2+3.sin A.cos A)

= 3(sin A-cos A)-4(sin A-cos A)((sin A-cos A)^2+3.sin A.cos A)

= (sin A-cos A)(3-4((sin A-cos A)^2+3.sin A.cos A))

= (sin A-cos A)(3-4(1-2sin A.cos A+3 sin A.cos A))

= (sin A-cos A)(3-4(1+sin A.cos A))

= (sin A-cos A)(-1-4.sin A.cos A)

= (cos A-sin A)(1+4.sin A.cos A)--> remember sin 2A= 2.sin A.cos A

= (cos A-sin A)(1+2.sin 2A)-->proved

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we can directly use the formula of $\sin3A,\cos3A$ instead of deriving them, right? – lab bhattacharjee Oct 1 '13 at 18:22
Right, I just explain to him how to derive it, so she can understand step by step. – akusaja Oct 1 '13 at 18:23

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