Given that,

\( \Large \sin \theta = -\frac{4}{5} \) and \( \theta \) lies in the third quadrant

=> \( \Large \cos \theta = -\sqrt{1-\frac{16}{25}} = -\frac{3}{5} \)

Now, \( \Large \cos \frac{ \theta }{2} = \pm \sqrt{\frac{1+\cos \theta }{2}} = \sqrt{\frac{1-\frac{3}{5}}{2}} = \pm \sqrt{\frac{1}{5}} \)

But we take \( \Large \cos \frac{ \theta }{2} = -\frac{1}{\sqrt{5}} \); Since, if \( \Large \theta \) lies in IIIrd quadrant, then \( \Large \frac{ \theta }{2} \) will be in IInd quadrant.

Hence, \( \Large \cos \frac{ \theta }{2} = - \frac{1}{\sqrt{5}} \)