How do you evaluate $sin (\frac{19 π}{12}) cos (- \frac{π}{6}) - cos (\frac{19 π}{12}) sin (- \frac{π}{6})$ $sin((19pi)/12)cos(-pi/6) - cos((19pi)/12)sin(-pi/6)$ ?

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How do you evaluate $sin (\frac{19 π}{12}) cos (- \frac{π}{6}) - cos (\frac{19 π}{12}) sin (- \frac{π}{6})$ $sin((19pi)/12)cos(-pi/6) - cos((19pi)/12)sin(-pi/6)$ ?

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Answer 1 · 2016-03-24T22:06:28

I found: $- \frac{\sqrt{2}}{2}$ $-sqrt(2)/2$

Explanation:

I think we can use the identity:
$sin (x - y) = sin (x) cos (y) - sin (y) cos (x)$ $sin(x-y)=sin(x)cos(y)-sin(y)cos(x)$
where in our case:
$x = \frac{19}{12} π$ $x=19/12pi$
$y = - \frac{π}{6}$ $y=-pi/6$
so we get:
$sin (\frac{19}{12} π) cos (- \frac{π}{6}) - sin (- \frac{π}{6}) cos (\frac{19}{12} π) = sin (\frac{19}{12} π + \frac{π}{6}) =$ $sin(19/12pi)cos(-pi/6)-sin(-pi/6)cos(19/12pi)=sin(19/12pi+pi/6)=$
$= sin (\frac{19 + 2}{12} π) = sin (\frac{21}{12} π) = sin (\frac{7}{4} π) = - \frac{\sqrt{2}}{2}$ $=sin((19+2)/12pi)=sin(21/12pi)=sin(7/4pi)=-sqrt(2)/2$

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How do you evaluate $sin (\frac{19 π}{12}) cos (- \frac{π}{6}) - cos (\frac{19 π}{12}) sin (- \frac{π}{6})$ $sin((19pi)/12)cos(-pi/6) - cos((19pi)/12)sin(-pi/6)$ ?

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How do you evaluate $sin (\frac{19 π}{12}) cos (- \frac{π}{6}) - cos (\frac{19 π}{12}) sin (- \frac{π}{6})$ $sin((19pi)/12)cos(-pi/6) - cos((19pi)/12)sin(-pi/6)$ ?