The double angle formula is
# cos 2x = 2 cos ^2 x - 1 #
Solving for #cos x # yields the half angle formula,
# \cos x = \pm sqrt{ 1/2 ( cos 2 x + 1) } #
So we know
# cos( theta/2) = pm sqrt{ 1/2 (cos theta + 1) } ## = pm sqrt{ 1/2 (24/25 + 1) } = pm sqrt {49/50} #
The question is slightly ambiguous on this point, but we're obviously talking about #theta# a positive angle in the fourth quadrant, meaning its half angle between #135^circ# and #180^circ# is in the second quadrant, so has a negative cosine.
We could be talking about the "same" angle but say it's between #-90^circ# and #0^circ# and then the half angle would be in the fourth quadrant with a positive cosine. That's why there's a #pm# in the formula.
In this problem we conclude
# cos(theta/2) = - sqrt {49/50} #
That's a radical we can simplify a bit, let's say
#cos(theta/2) =-sqrt{{2(49)}/100} = - 7/10 sqrt{2} #
