July 2nd, 2014, 07:27 AM  #1 
Senior Member Joined: Nov 2013 Posts: 434 Thanks: 8  find sin27 and find 63
find sin27 find 63 
July 2nd, 2014, 07:44 AM  #2 
Senior Member Joined: Mar 2012 From: Belgium Posts: 654 Thanks: 11  Code: find sin 27 find s in 27 find s in twentyseven ^ Last edited by skipjack; July 3rd, 2014 at 12:32 AM. 
July 2nd, 2014, 07:45 AM  #3 
Senior Member Joined: Apr 2014 From: Europa Posts: 571 Thanks: 176 
$\displaystyle \sin 27^\circ = \sin(45^\circ18^\circ)$ answers.yahoo.com/question/index?qid=20090405224513AAwtYD9 Last edited by skipjack; July 3rd, 2014 at 12:31 AM. 
July 2nd, 2014, 08:09 AM  #4 
Senior Member Joined: Mar 2012 From: Belgium Posts: 654 Thanks: 11 
$\displaystyle (\cos (\theta) + i \sin (\theta))^5 = \cos (5\theta) + i \sin (5\theta)$ therefore $\displaystyle \cos (5\theta)= \cos^5 (\theta)10\cos (\theta)^3\sin (\theta)^2 + 5\cos (\theta)\sin (\theta)^4$ $\displaystyle \cos (5\theta)= \cos (\theta)^510\cos (\theta)^3+ 10 \cos(\theta)^5 + 5\cos (\theta) 10\cos (\theta)^3 + 5\cos (\theta)^5$ $\displaystyle \cos (5\theta)= 16\cos (\theta)^520\cos (\theta)^3+ 5\cos (\theta)$ $\displaystyle \theta = 27^\circ$ and we substitute $\displaystyle t=\cos (\theta)$ $\displaystyle \frac{\sqrt{2}}{2}= 16t^520t^3+ 5t$ Using Wolfram (I won't bother solving that...) I get that $\displaystyle t = \frac{1}{4} \left( \sqrt{\frac{5}{2}}  \frac {1}{\sqrt{2}} + \sqrt{5+\sqrt{5}} \right )$ and thus $\displaystyle \cos(27^\circ) = \sin(63^\circ) = \frac{1}{4} \left( \sqrt{\frac{5}{2}}  \frac {1}{\sqrt{2}} + \sqrt{5+\sqrt{5}} \right )$ using $\displaystyle \sin(27^\circ) = \sqrt {1\cos(27^\circ)^2 }$ one could find the value of $\displaystyle \sin(27^\circ)$ too. Last edited by skipjack; July 2nd, 2014 at 11:28 PM. 

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