Question #fa554

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Answer 1 · 2016-07-12T02:43:26

$sin (\frac{π}{3} + \frac{7}{6} π) + sin (\frac{π}{3}) + sin (\frac{7}{6} π)$ $sin(pi/3+7/6pi)+ sin(pi/3)+ sin(7/6pi)$

$= sin (\frac{2}{6} π + \frac{7}{6} π) + sin (\frac{π}{3}) + sin (π + \frac{π}{6})$ $=sin(2/6pi+7/6pi)+ sin(pi/3)+ sin(pi+pi/6)$

$= sin (\frac{9}{6} π) + sin (\frac{π}{3}) - sin (\frac{π}{6})$ $=sin(9/6pi)+ sin(pi/3)-sin(pi/6)$

$= sin (\frac{3}{2} π) + sin (\frac{π}{3}) - sin (\frac{π}{6})$ $=sin(3/2pi)+ sin(pi/3)-sin(pi/6)$

$= sin (π + \frac{π}{2}) + sin (\frac{π}{3}) - sin (\frac{π}{6})$ $=sin(pi+pi/2)+ sin(pi/3)-sin(pi/6)$

$= - sin (\frac{π}{2}) + sin (\frac{π}{3}) - sin (\frac{π}{6})$ $=-sin(pi/2)+ sin(pi/3)-sin(pi/6)$

$= - 1 + \frac{\sqrt{3}}{2} - \frac{1}{2}$ $=-1+sqrt3/2-1/2$

$= - \frac{3}{2} + \frac{\sqrt{3}}{2}$ $=-3/2+sqrt3/2$

$= \frac{1}{2} (\sqrt{3} - 3)$ $=1/2(sqrt3-3)$