θ lies in quadrant III, which means
π < θ < 3π/2 ⇒ π/2 < θ/2 < 3π/4
which is to say that θ/2 lies in quadrant II. For quadrant III, we have sin(θ) < 0 and cos(θ) < 0. For quadrant II, we have sin(θ/2) > 0 and cos(θ/2) < 0.
Recall the half-angle identity for sine:
sin²(θ/2) = (1 - cos(θ))/2
Then in this case,
sin(θ/2) = - √((1 - cos(θ))/2)
Also recall the Pythagorean identity,
1 + tan²(θ) = sec²(θ)
It follows that
sec(θ) = - √(1 + tan²(θ)) = -√73/3
⇒ cos(θ) = -3/√73
⇒ sin(θ/2) = - √((1 + 3/√73)/2) = √(1/2 + 3/(2√73))