If #cosa=-1/3#, #a# must be between #90# and #180# degrees. In other words, a in #2#nd quadrant in unit circle. In this quadrant #sina>0#. So, if #cosa=-1/3#, #sina=(2sqrt(2))/3#
After using #sin2a=2sina*cosa# identity, I found #sin2a=-(4sqrt(2))/9#
Explanation:
1) I found #sina# after being given #cosu=-1/3# and #90 < a < 180# conditions.