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May 31st, 2018, 01:50 PM | #1 |
Newbie Joined: May 2018 From: Brazil Posts: 2 Thanks: 0 | Transformations
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May 31st, 2018, 03:26 PM | #2 |
Global Moderator Joined: Dec 2006 Posts: 20,307 Thanks: 1976 |
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 07:22 PM. |
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May 31st, 2018, 05:19 PM | #3 |
Newbie Joined: May 2018 From: Brazil Posts: 2 Thanks: 0 |
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