Transformations

u17159BR · May 31st, 2018, 12:50 PM

Prove:

skipjack · May 31st, 2018, 02:26 PM

Using $\sin4x + \sin2x = 2\sin3x\cos x \text{ and } \cos4x + \cos2x = 2\cos3x\cos x$,

LHS = $\displaystyle \frac{(2\cos x - 1)\sin3x}{(2\cos x - 1)\cos3x} = \tan3x$.

u17159BR · May 31st, 2018, 04:19 PM

Thanks

May 31st, 2018, 12:50 PM	#1
u17159BR Newbie Joined: May 2018 From: Brazil Posts: 2 Thanks: 0	Transformations Prove:
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May 31st, 2018, 02:26 PM	#2
skipjack Global Moderator Joined: Dec 2006 Posts: 19,702 Thanks: 1804	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
u17159BR Newbie Joined: May 2018 From: Brazil Posts: 2 Thanks: 0	Thanks
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