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 August 6th, 2018, 08:11 AM #1 Newbie   Joined: Aug 2018 From: Czech Republic Posts: 2 Thanks: 0 sqrt(2)*cos(2x)=sin(4x) Hi, I've been struggling with this equation. I tried to expand cos(2x), but I just can't get to the result. Could someone please help me with this one? Thanks. August 6th, 2018, 08:39 AM #2 Global Moderator   Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,945 Thanks: 1136 Math Focus: Elementary mathematics and beyond it's not an identity. Did you type it correctly? August 6th, 2018, 08:54 AM   #3
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Quote:
 Originally Posted by ducanhle Hi, I've been struggling with this equation. I tried to expand cos(2x), but I just can't get to the result. Could someone please help me with this one? Thanks.
There are five solutions to the equation on $\displaystyle [ 0, ~2 \pi )$ which, I suspect, can only be found numerically.

Is the problem somehow 2 sin(2x) cos(2x) = sin(4x) ? This is an identity.

-Dan August 6th, 2018, 09:25 AM #4 Math Team   Joined: Jul 2011 From: Texas Posts: 2,947 Thanks: 1555 $\sqrt{2}\cos(2x) = \sin(4x)$ $\sqrt{2}\cos(2x) = 2\sin(2x)\cos(2x)$ $\sqrt{2}\cos(2x) - 2\sin(2x)\cos(2x) = 0$ $\cos(2x){\large[}\sqrt{2} - 2\sin(2x){\large]} = 0$ for $x \in [0, 2\pi )$ ... $\displaystyle\cos(2x) = 0 \implies x \in \left\{\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4} \right\}$ $\displaystyle\sin(2x) = \frac{\sqrt{2}}{2} \implies x \in \left\{ \frac{\pi}{8}, \frac{3\pi}{8}, \frac{9\pi}{8}, \frac{11\pi}{8} \right\}$ Thanks from greg1313, topsquark and ducanhle Last edited by skipjack; August 6th, 2018 at 01:40 PM. Tags sqrt2cos2xsin4x Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post DecoratorFawn82 Algebra 2 October 1st, 2017 06:03 PM Fredrick75 Elementary Math 8 March 26th, 2015 12:25 AM unistu Algebra 3 January 24th, 2015 02:21 PM ungeheuer Calculus 1 August 30th, 2013 04:34 AM nikkor180 Real Analysis 3 June 25th, 2009 03:18 AM

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