Trigonometry 7th Edition

$\sqrt{\dfrac{3+2\sqrt{2}}{6}}$
$\because B$ is in $QIII \hspace{30pt} \therefore \cos{B}$ is negative $\cos{B} = -\sqrt{1-\sin^2{B}} = -\sqrt{1-(-\dfrac{1}{3})^2} = -\dfrac{2\sqrt{2}}{3}$ $180 < B < 270$ $90 < \dfrac{B}{2} < 135$ Using the half angle formula $\sin{\dfrac{B}{2}} = \sqrt{\dfrac{1-\cos{B}}{2}}$ $\sin{\dfrac{B}{2}} = \sqrt{\dfrac{3+2\sqrt{2}}{6}}$