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May 31st, 2018, 12:50 PM   #1
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Transformations

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May 31st, 2018, 02:26 PM   #2
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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.
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May 31st, 2018, 04:19 PM   #3
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