#sin ((7pi)/12) = sin (pi/12 + (6pi)/12) = sin (pi/12 + pi/2) = cos (pi/12)#
Note. #sin (x + pi/2) = cos x# (complementary arcs).
Find #cos ((pi)/12)#. Call #cos ((pi)/12) = cos x#
#cos 2x = cos ((pi)/6) = sqrt3/2#.
Apply the trig identity: #cos 2x = 2cos^2 x - 1#. We get:
#sqrt3/2 = 2cos^2 x - 1#
#2cos^2 x = 1 + sqrt3/2 = (2 + sqrt3)/2#
#cos^2 x = (2 + sqrt3)/4#
#cos x = cos ((pi)/12) = +- sqrt(2 + sqrt3)/2#.
Since the arc #(pi/12)# is in Quadrant I, then its cos is positive. Finally,
#sin ((7pi)/12) = cos((pi)/12) = sqrt(2 + sqrt3)/2#.
Check by calculator. #sin ((7pi)/12) = sin 105^@ = 0.97#
#sqrt(2 + sqrt3)/2 = 1.93/2 = 0.97# .OK
