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July 12th, 2018, 09:25 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 doubleangle 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 01:20 PM. Reason: Forgot to add a URL 
July 12th, 2018, 10:01 AM  #2 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,184 Thanks: 481 Math Focus: Calculus/ODEs 
That link doesn't work for me...try attaching the image perhaps.

July 12th, 2018, 10:48 AM  #3 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,823 Thanks: 1049 Math Focus: Elementary mathematics and beyond 
Here's the image:
Last edited by greg1313; July 12th, 2018 at 01:41 PM. 
July 12th, 2018, 12:03 PM  #4 
Senior Member Joined: Sep 2015 From: USA Posts: 2,011 Thanks: 1044 
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 
July 12th, 2018, 12:08 PM  #5 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,823 Thanks: 1049 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, 12:11 PM  #6 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,184 Thanks: 481 Math Focus: Calculus/ODEs 
What happened to the OP?

July 12th, 2018, 12:57 PM  #7 
Senior Member Joined: Sep 2015 From: USA Posts: 2,011 Thanks: 1044  

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