Use trig identities:
#sin a = 2sin (a/2)cos (a/2)#.
#cos a = 2cos^2 (a/2) - 1#
First, find cos a.
#cos^2 a = 1 - sin^2 a = 1 - 36/49 = 13/49#
#cos a = - sqrt13/7# (because a is in Q. 3).
Next, find #cos (a/2)#:
#2cos^2 (a/2) = 1 + cos a = 1 - sqrt13/7 = (7 - sqrt13)/7#.
#cos^2 (a/2) = (7 - sqrt13)/14#
#cos (a/2) = +- sqrt(7 - sqrt13)/sqrt14#
Since a is in Q.3, then, #a/2# is in Q. 2, so, #cos (a/2)# is negative.
Last, find #sin (a/2)#.
#sin (a/2) = sin a/(2cos (a/2))#
#sin (a/2) = (-6/7)((-sqrt14)/(2sqrt(7 - sqrt13))) =#
#sin (a/2) = (6sqrt14)/( 14sqrt(7 - sqrt13)#
Check by calculator:
#sin a = - 6/7# --> #a = 180 - (- 59^@) = 239^@# (Quadrant 3)
#a/2 = 119^@50# --> #sin (a/2) = 0.87#
#(6sqrt14) = 22.45#, and #14sqrt(7 - sqrt13) = 25.79#
#sin (a/2) = 22.45/25.79 = 0.87#. Proved.
