How do you find the value of #sin(-(11pi)/12)#?

1 Answer
Aug 22, 2016

#- sqrt(2 - sqrt3)/2#

Explanation:

Trig table and unit circle -->
#sin ((-11pi)/12) = sin (pi/12 - (12pi)/12) = sin (pi/12 - pi) = - sin (pi/12)#
Evaluate #sin (pi/12)# by using trig identity:
#2sin^2 a = 1 - cos 2a#
#2sin^2 (pi/12) = 1 - cos (pi/6) = 1 - sqrt3/2 = (2 - sqrt3)/2#
#sin^2 (pi/12) = (2 - sqrt3)/4#
#sin (t/12) = +- sqrt(2 - sqrt3)/2#
Because sin (pi/12) is positive, take the positive answer.
Finally,
#sin ((-11pi)/12) = - sin (pi/12) = - sqrt(2 - sqrt3)/2#