Question

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

Answer 1

`- (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)`