Call #sin ((-17pi)/12) = sin t#
#sin ((-17pi)/12) = sin t = sin ((-5pi)/12 - 2pi) = sin ((-5pi)/12) = - sin ((5pi)/12)#
#cos 2t = cos ((10pi)/12) = cos ((5pi)/6) = - sqrt3/2#
Apply the trig identity: #cos 2a = 2cos^2 a - 1#
#cos ((5pi)/6) = cos 2t = - sqrt3/2 = 1 - 2sin^2 t#
#2sin^2 t = 1 + sqrt3/2 = (2 + sqrt3)/2#
#sin^2 t = (2 + sqrt3)/4#
#sin ((-17pi)/12) = +- sqrt(2 + sqrt3)/2#
Since the arc (-17pi/12) is located in Quadrant III, its sin is positive, then #sin ((-17pi)/12) = sqrt(2 + sqrt3)/2#
Call #sin (pi/12) = sin u#
#cos 2u = cos (pi/6) = sqrt3/2#
#cos 2u = sqrt3/2 = 1 - 2sin^2 u#
#sin^2 u = (2 - sqrt3)/4#
#sin (pi/12) = sin u = +- sqrt(2 - sqrt3)/2#
Since #sin pi/12 > 0#, then
#sin (pi/12) = sqrt(2 - sqrt3)/2# (2)
From (1) and (2), we get: #sin((-17pi)/12) + sin (pi/12) =#
#sqrt(2 + sqrt3)/2 + sqrt(2 - sqrt3)/2 = 1#
