My Math Forum find sin27 and find 63

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 July 2nd, 2014, 08:27 AM #1 Senior Member   Joined: Nov 2013 Posts: 434 Thanks: 8 find sin27 and find 63 find sin27 find 63
July 2nd, 2014, 08:44 AM   #2
Senior Member

Joined: Mar 2012
From: Belgium

Posts: 654
Thanks: 11

Quote:
 Originally Posted by mared find sin27
Code:
find sin 27
find s in 27
find s in twentyseven
^
Found it! Sorry, I couldn't leave this pun.

Last edited by skipjack; July 3rd, 2014 at 01:32 AM.

 July 2nd, 2014, 08:45 AM #3 Senior Member     Joined: Apr 2014 From: Europa Posts: 571 Thanks: 176 $\displaystyle \sin 27^\circ = \sin(45^\circ-18^\circ)$ answers.yahoo.com/question/index?qid=20090405224513AAwtYD9 Last edited by skipjack; July 3rd, 2014 at 01:31 AM.
 July 2nd, 2014, 09:09 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 $\displaystyle \cos (5\theta)= \cos^5 (\theta)-10\cos (\theta)^3\sin (\theta)^2 + 5\cos (\theta)\sin (\theta)^4$ $\displaystyle \cos (5\theta)= \cos (\theta)^5-10\cos (\theta)^3+ 10 \cos(\theta)^5 + 5\cos (\theta) -10\cos (\theta)^3 + 5\cos (\theta)^5$ $\displaystyle \cos (5\theta)= 16\cos (\theta)^5-20\cos (\theta)^3+ 5\cos (\theta)$ $\displaystyle \theta = 27^\circ$ and we substitute $\displaystyle t=\cos (\theta)$ $\displaystyle -\frac{\sqrt{2}}{2}= 16t^5-20t^3+ 5t$ 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 3rd, 2014 at 12:28 AM.

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