Transformations

u17159BR · May 31st, 2018, 01:50 PM

Prove:

skipjack · May 31st, 2018, 03:26 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, 05:19 PM

Thanks

May 31st, 2018, 01:50 PM	#1
u17159BR Newbie Joined: May 2018 From: Brazil Posts: 2 Thanks: 0	Transformations Prove:
$u17159BR is offline$

May 31st, 2018, 03:26 PM	#2
skipjack Global Moderator Joined: Dec 2006 Posts: 20,003 Thanks: 1862	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.
$skipjack is offline$

May 31st, 2018, 05:19 PM	#3
u17159BR Newbie Joined: May 2018 From: Brazil Posts: 2 Thanks: 0	Thanks
$u17159BR is offline$

Thread Tools
$Show Printable Version$ Show Printable Version $Email this Page$ Email this Page
Display Modes
$Linear Mode$ Linear Mode $Hybrid Mode$ Switch to Hybrid Mode $Threaded Mode$ Switch to Threaded Mode

Similar Threads
Thread	Thread Starter	Forum	Replies	Last Post
Transformations	Adrian	Algebra	20	March 6th, 2011 06:12 PM
exp - log transformations	reto11	Applied Math	1	October 18th, 2010 11:08 PM
Transformations	julian21	Algebra	4	July 17th, 2010 02:34 PM
TRANSFORMATIONS PLEASE HELP	maigowai	Algebra	1	July 9th, 2010 01:49 AM