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 July 12th, 2018, 10:25 AM #1 Newbie   Joined: Jul 2018 From: High Wycombe Posts: 2 Thanks: 0 Urgent help needed! Hi guys, I was just wondering how I should proceed to answer this question. I've tried the double-angle cosine formula (advised to try by a friend), but still can't figure out the answer. Any help would be much appreciated! http://tinypic.com/r/29bhop1/9 Last edited by skipjack; July 12th, 2018 at 02:20 PM. Reason: Forgot to add a URL
 July 12th, 2018, 11:01 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,205 Thanks: 512 Math Focus: Calculus/ODEs That link doesn't work for me...try attaching the image perhaps.
July 12th, 2018, 11:48 AM   #3
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Here's the image:
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Last edited by greg1313; July 12th, 2018 at 02:41 PM.

 July 12th, 2018, 01:03 PM #4 Senior Member     Joined: Sep 2015 From: USA Posts: 2,165 Thanks: 1139 Well you've got an isosceles triangle with a angle of $\dfrac \pi 4$ between the two equal sides. because $\dfrac \pi 4 = \dfrac {2\pi}{8}$ so $\dfrac s 2 = r \sin\left(\dfrac 1 2 \dfrac \pi 4\right)$ $s = 2 r \sin\left(\dfrac \pi 8\right)$ $s^2 = 4r^2 \sin^2\left(\dfrac \pi 8\right)$ $s^2 = 4r^2 \cdot \dfrac{1}{4} \left(2-\sqrt{2}\right)$ $s^2 = r^2 \left(2-\sqrt{2}\right)$ of course the magic here is $\sin^2\left(\dfrac \pi 8\right) = \dfrac{1}{4} \left(2-\sqrt{2}\right)$ I leave that to you to show Thanks from topsquark
 July 12th, 2018, 01:08 PM #5 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,881 Thanks: 1088 Math Focus: Elementary mathematics and beyond Construct a perpendicular from where $r$ meets $s$ to $r$ on the opposite side and use the Pythagorean theorem.
 July 12th, 2018, 01:11 PM #6 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,205 Thanks: 512 Math Focus: Calculus/ODEs What happened to the OP?
July 12th, 2018, 01:57 PM   #7
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Quote:
 Originally Posted by MarkFL What happened to the OP?
eaten... by snakes

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