Q (14) Prove the following

\small \sin2x +2\sin4x + \sin6x = 4\cos^{2}x\sin4x

Answers (1)
S safeer

We know that 

\sin A+ \sin B = 2\sin \left ( \frac{A+B}{2} \right )\cos\left ( \frac{A-B}{2} \right )
We are using this identity 
sin2x + 2sin4x + sin6x = (sin2x + sin6x) + 2sin4x 

sin2x + sin6x = 2sin4xcos(-2x) =  2sin4xcos(2x)          (\because cos(-x) = cos x)

So, our equation becomes
sin2x + 2sin4x + sin6x = 2sin4xcos(2x) + 2sin4x
Now, take the 2sin4x common
sin2x + 2sin4x + sin6x = 2sin4x(cos2x +1)           (  \because \cos2x = 2\cos^{2}x - 1 )
                                     =2sin4x(2\cos^{2}x - 1 +1 )
                                     =2sin4x(2\cos^{2}x)
                                     =4\sin4x\cos^{2}x
                                                                          R.H.S.

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