For a better visual than the title of this question, its Problem #5 on this page: https://youtu.be/FlcDPm78D2M?t=12m31s
I've done this problem two times, and I keep getting something wrong. When I graph I'm getting sin theta = -sqrt(15)/3 , and when I use the formula sin (2theta) = 2 sin theta x cos theta with the sin I got and the cos the problem provides, I get the wrong answer (according to the book it should be -2xsqrt(2)/3 ) (also I'm not the one getting the wrong answer, it was CPhill, and I'm pretty sure it was because I gave him wrong numbers in the first place, but I really don't know how I'm doing this wrong. I got -1 for cos (2theta), which the book says should be 1/3. I should be getting the right answers, because I'm doing it right, but I'm not getting the right answers, so apparently I'm not doing it right at all.
+128731 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)
