Answer:
cos(θ/2) = -1/√17
Step-by-step explanation:
The given angle is in quadrant III, so its half-angle will be in quadrant II. The cosine in that quadrant is negative, so a relevant half-angle formula is ...
[tex]\cos{\dfrac{\theta}{2}}=-\sqrt{\dfrac{1+\cos{\theta}}{2}}[/tex]
The cosine can be found from ...
[tex]\cos{\theta}=-\sqrt{1-\sin^2{\theta}}=-\sqrt{\dfrac{225}{289}}=-\dfrac{15}{17}[/tex]
Using this value in the half-angle formula, you find ...
[tex]\cos{\dfrac{\theta}{2}}=-\sqrt{\left(1-\dfrac{15}{17}\right)\cdot\dfrac{1}{2}}=\boxed{\dfrac{-1}{\sqrt{17}}}[/tex]
You have the right idea and the right magnitude, but the wrong sign.
_____
You can also get there from your two correct answers:
cos = sin/tan = (4/√17)/(-4) = -1/√17
