A) \[\frac{1}{\sqrt{5}}\]
B) \[-\frac{1}{\sqrt{5}}\]
C) \[\sqrt{\frac{2}{5}}\]
D) \[-\sqrt{\frac{2}{5}}\]
Correct Answer: B
Solution :
Given that \[\sin \theta =-\frac{4}{5}\]and \[\theta \]lies in the III quadrant. \[\Rightarrow \cos \theta =\sqrt{1-\frac{16}{25}}=\pm \frac{3}{5}\] \[\cos \frac{\theta }{2}=\pm \sqrt{\frac{1+\cos \theta }{2}}=\sqrt{\frac{1-3/5}{2}}=\pm \sqrt{\frac{1}{5}}\] But \[\cos \frac{\theta }{2}=-\frac{1}{\sqrt{5}}.\]since \[\frac{\theta }{2}\]will be in II quadrant. Hence\[\cos \frac{\theta }{2}=-\frac{1}{\sqrt{5}}\].
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