May 31st, 2018, 12:50 PM  #1 
Newbie Joined: May 2018 From: Brazil Posts: 2 Thanks: 0  Transformations
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May 31st, 2018, 02:26 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 19,042 Thanks: 1618 
Using $\sin4x + \sin2x = 2\sin3x\cos x \text{ and } \cos4x + \cos2x = 2\cos3x\cos x$, LHS = $\displaystyle \frac{(2\cos x  1)\sin3x}{(2\cos x  1)\cos3x} = \tan3x$. Last edited by skipjack; May 31st, 2018 at 06:22 PM. 
May 31st, 2018, 04:19 PM  #3 
Newbie Joined: May 2018 From: Brazil Posts: 2 Thanks: 0 
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