cos (theta) = - sqrt(6)/3 theta lies in the second quad
x = -sqrt(6), y = sqrt( 9 - 6) = sqrt (3) so.....sin(theta) = sqrt(3) / 3
(1) sin(2 theta) = 2 cos(theta)sin(theta) = 2[-sqrt(6)/3 ] [sqrt(3)/3] = (2/9)[-sqrt(6)*sqrt(3)] =
(-2/9)sqrt(18) = (-2/9)sqrt(9 * 2) = (-2*3)/ 9 * sqrt(2) = [ -2sqrt(2)]/ 3
(2) cos(2 theta) = cos^2 ( theta) - sin^2(theta) = [6/9] - [3/9] = 3/9 = 1/3
(3) sin (theta/2) = sqrt [ ( 1 - cos(theta)) / 2 ] = sqrt [ (1 - -(sqrt(6)/3) / 2 ] =
sqrt [ (1 + sqrt(6) / 3) / 2] = sqrt [ (3 + sqrt(6)) / 6]
(4) cos (theta/2) = sqrt[ [ ( 1 + cos(theta)) / 2] = sqrt [ (1 -sqrt(6)/3)/2] =
sqrt [ (3 - sqrt(6))/ 6)
