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July 2nd, 2014, 07:27 AM   #1
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find sin27 and find 63

find sin27
find 63
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July 2nd, 2014, 07:44 AM   #2
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Quote:
Originally Posted by mared View Post
find sin27
Code:
find sin 27
find s in 27
find s in twentyseven
                     ^
Found it! Sorry, I couldn't leave this pun.
Thanks from soroban

Last edited by skipjack; July 3rd, 2014 at 12:32 AM.
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July 2nd, 2014, 07:45 AM   #3
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$\displaystyle \sin 27^\circ = \sin(45^\circ-18^\circ)$


answers.yahoo.com/question/index?qid=20090405224513AAwtYD9

Last edited by skipjack; July 3rd, 2014 at 12:31 AM.
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July 2nd, 2014, 08:09 AM   #4
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$\displaystyle (\cos (\theta) + i \sin (\theta))^5 = \cos (5\theta) + i \sin (5\theta)$

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 2nd, 2014 at 11:28 PM.
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