The double angle formula is
cos 2x = 2 cos ^2 x - 1 cos2x=2cos2x−1
Solving for cos x cosx yields the half angle formula,
\cos x = \pm sqrt{ 1/2 ( cos 2 x + 1) } cosx=±√12(cos2x+1)
So we know
cos( theta/2) = pm sqrt{ 1/2 (cos theta + 1) } cos(θ2)=±√12(cosθ+1) = pm sqrt{ 1/2 (24/25 + 1) } = pm sqrt {49/50} =±√12(2425+1)=±√4950
The question is slightly ambiguous on this point, but we're obviously talking about thetaθ a positive angle in the fourth quadrant, meaning its half angle between 135^circ135∘ and 180^circ180∘ 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−90∘ and 0^circ0∘ 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} cos(θ2)=−√4950
That's a radical we can simplify a bit, let's say
cos(theta/2) =-sqrt{{2(49)}/100} = - 7/10 sqrt{2} cos(θ2)=−√2(49)100=−710√2
