Question

If $\sin \theta =-\frac{4}{5}$ and θ lies in third quadrant, then the value of $\cos \frac{\theta }{2}$ is
(a) $\frac{1}{5}$

(b) $-\frac{1}{\sqrt{10}}$

(c) $-\frac{1}{\sqrt{5}}$

(d) $\frac{1}{\sqrt{10}}$

Answer 1

$$\begin{gathered} \sin \theta =\frac{-4}{5} \\ \mathrm{where}\mathrm{\theta }\mathrm{lies}\mathrm{in}\mathrm{third}\mathrm{quadrant}. \\ \mathrm{since}{\cos }^2\theta =1-{\sin }^2\theta \\ =1-{\left(\frac{-4}{5}\right)}^2 \\ =1-\frac{16}{25} \\ {\cos }^2\theta =\frac{9}{25} \\ \mathrm{i}.\mathrm{e}.\cos \theta =\pm \frac{3}{5} \\ \mathrm{since}\mathrm{\pi }<\theta <\frac{3\mathrm{\pi }}{2}\mathrm{i}.\mathrm{e}.\cos \theta =-\frac{3}{5} \\ \Rightarrow \frac{\mathrm{\pi }}{2}<\frac{\theta }{2}<\frac{3\mathrm{\pi }}{4} \\ \mathrm{lies}\mathrm{in}\mathrm{II}\mathrm{quadrant} \\ \mathrm{using}\mathrm{identity}, \\ \cos \frac{\theta }{2}=\pm \sqrt{\frac{1+\cos \theta }{2}} \\ =\pm \sqrt{\frac{1-{\displaystyle \frac{3}{5}}}{2}} \\ =\pm \sqrt{\frac{5-{\displaystyle \frac{3}{5}}}{2}} \\ =\pm \sqrt{\frac{2}{5}\times \frac{1}{2}} \\ =\pm \frac{1}{\sqrt{5}} \\ \mathrm{since}\frac{\theta }{2}\mathrm{lies}\mathrm{II}\mathrm{quadrant}=\cos \frac{\theta }{2}=-\frac{1}{\sqrt{5}} \\ \mathrm{Hence},\mathrm{the}\mathrm{correct}\mathrm{answer}\mathrm{is}\mathrm{option}\mathrm{C}. \end{gathered}$$