How do you evaluate $- sin (\frac{13 π}{12})$ $-sin((13pi)/12)$ ?

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How do you evaluate $- sin (\frac{13 π}{12})$ $-sin((13pi)/12)$ ?

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Answer 1 · 2016-04-07T02:44:19

$\frac{\sqrt{2 - \sqrt{3}}}{2}$ $sqrt(2 - sqrt3)/2$

Explanation:

Trig unit circle and property of supplementary arc-->
$- sin (\frac{13 π}{12}) = - sin (\frac{π}{12} + π) = sin (\frac{π}{12})$ $- sin ((13pi)/12) = - sin (pi/12 + pi) = sin (pi/12)$
Evaluate $sin (\frac{π}{12})$ $sin (pi/12)$ by applying the trig identity:
$cos 2 a = 1 - 2 {sin}^{2} a$ $cos 2a = 1 - 2sin^2 a$
$cos (\frac{π}{6}) = \frac{\sqrt{3}}{2} = 1 - 2 {sin}^{2} (\frac{π}{12})$ $cos (pi/6) = sqrt3/2 = 1 - 2sin^2 (pi/12)$
$2 {sin}^{2} (\frac{π}{12}) = 1 - \frac{\sqrt{3}}{2} = \frac{2 - \sqrt{3}}{2}$ $2sin^2 (pi/12) = 1 - sqrt3/2 = (2 - sqrt3)/2$
${sin}^{2} (\frac{π}{12}) = \frac{2 - \sqrt{3}}{4}$ $sin^2 (pi/12) = (2 - sqrt3)/4$
$sin (\frac{π}{12}) = \pm \frac{\sqrt{2 - \sqrt{3}}}{2}$ $sin (pi/12) = +- sqrt(2 - sqrt3)/2$ -->
Since $sin (\frac{π}{12})$ $sin (pi/12)$ is positive, therefor:
$- sin (\frac{13 π}{12}) = sin (\frac{π}{12}) = \frac{\sqrt{2 - \sqrt{3}}}{2}$ $- sin ((13pi)/12) = sin (pi/12) = sqrt(2 - sqrt3)/2$

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How do you evaluate $- sin (\frac{13 π}{12})$ $-sin((13pi)/12)$ ?

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