How do you evaluate $sin(((−5pi)/12))$ ?

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How do you evaluate $sin(((−5pi)/12))$ ?

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Answer 1 · 2016-04-05T03:13:10

$- (sqrt(2 + sqrt3)/2)$

Explanation:

Trig unit circle and property of complementary arcs -->
$sin ((-5pi)/12) = sin (pi/12 - (6pi)/12)) = sin (pi/12 - pi/2) =$
$= - cos (pi/12)$
Evaluate cos (pi/12) by applying the trig identity:
$cos 2a = 2cos^2 a - 1.$
$cos (pi/6) = sqrt3/2 = 2cos^2 (pi/12) - 1$
$2cos^2 (pi/12) = 1 + sqrt3/2 = (2 + sqrt3)/2$
$cos^2 (pi/12) = (2 + sqrt3)/4$
$cos (pi/12) = (sqrt(2 + sqrt3))/2$ --> since $cos (pi/12)$ is positive.
Therefor,
$sin ((-5pi)/12) = - cos (pi/12) = - (sqrt(2 + sqrt3)/2)$

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How do you evaluate $sin(((−5pi)/12))$ ?

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