Recall the half-angle identity for cosine,
cos²(θ/2) = (1 + cos(θ))/2
Since θ lies in quadrant I, we also have θ/2 in quadrant I, since
0 < θ < π/2 ⇒ 0 < θ/2 < π/4
Then for this θ, we have
cos(θ/2) = + √((1 + cos(θ))/2)
Also recall the Pythagorean identity,
cos²(θ) + sin²(θ) = 1
Multiplying through both sides of this identity by cos²(θ) gives another form of it,
1 + tan²(θ) = sec²(θ)
Because θ belongs to quadrant I, we know cos(θ) > 0, so we also have sec(θ) = 1/cos(θ) > 0.
It follows that
sec(θ) = + √(1 + tan²(θ)) = √89/8
⇒ cos(θ) = 8/√89
and so
cos(θ/2) = + √((1 + 8/√89)/2) = √(1/2 + 4/√89)