sin 2a=sin(a+a)=sin a*cos a + cos a*sin a = 2sin a*cos a
We have the value for cos a, which is negative. The value for sin a is negative and it could be found from the fundamental formula of trigonometry.
sin a= sqrt[1-(cos a)^2]
sin 2a = 2sin a*cos a
sin 2a = 2(2sqrt2/3)(-1/3)
sin 2a = -4sqrt2/9
cosa =-1/3 a belongs to 9pi, 3pi/2). To find sin2a.
Sin2a = 2sinA*cosa = 2 sqrt(1-cos^2a)cosA
= 2sqrt(1-(-1/3^2)(-1/3) = 2[-sqrt(1-1/9)](-1/3) = (2sqrt8)/9 = 4sqrt2/9 is positive sign is valid as 2a is 1st or second quadrant is positive.