so we know that θ is between π and 3π/2, which is another of saying that θ is in the III Quadrant, and obviously its cosine will be negative.
we also know that cos(θ/2) is negative, well, that rules out the I and IV Quadrants, so most likely θ/2 is on the II Quadrant, since is smaller than θ anyway, and on the II Quadrant as we know, the sine or "y" value is positive.
[tex]\stackrel{\textit{Half-Angle Identities}}{ sin\left(\cfrac{\theta}{2}\right)=\pm \sqrt{\cfrac{1-cos(\theta)}{2}} \qquad\qquad cos\left(\cfrac{\theta}{2}\right)=\pm \sqrt{\cfrac{1+cos(\theta)}{2}}} \\\\[-0.35em] ~\dotfill\\\\ cos\left(\cfrac{\theta}{2}\right)=\pm \sqrt{\cfrac{1+\left(-\frac{1}{9} \right)}{2}}\implies cos\left(\cfrac{\theta}{2}\right)=\pm \sqrt{\cfrac{1-\left(\frac{1}{9} \right)}{2}}[/tex]
[tex]cos\left(\cfrac{\theta}{2}\right)=-\sqrt{\cfrac{~~ \frac{8}{9} ~~}{2}}\implies cos\left(\cfrac{\theta}{2}\right)=-\sqrt{\cfrac{4}{9}}\implies cos\left(\cfrac{\theta}{2}\right)=-\cfrac{2}{3} \\\\[-0.35em] ~\dotfill[/tex]
[tex]sin\left(\cfrac{\theta}{2}\right)=\pm \sqrt{\cfrac{1-\left( -\frac{1}{9} \right)}{2}}\implies sin\left(\cfrac{\theta}{2}\right)=+ \sqrt{\cfrac{~~\frac{10}{9}~~}{2}}\implies sin\left(\cfrac{\theta}{2}\right)=+\cfrac{\sqrt{5}}{3} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \stackrel{~~~a_1~\hfill a_2~~~}{\left( -\cfrac{2}{3}~~,~~ \cfrac{\sqrt{5}}{3}\right)}~\hfill[/tex]
