If $\sin \theta=-\frac{4}{5}$ and $\theta$ 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) $\frac{1}{\sqrt{10}}$
$\sin \theta=\frac{-4}{5}$
where $\theta$ lies in third quadrant.
since $\cos ^{2} \theta=1-\sin ^{2} \theta$
$=1-\left(\frac{-4}{5}\right)^{2}$
$=1-\frac{16}{25}$
$\cos ^{2} \theta=\frac{9}{25}$
i. e. $\cos \theta=\pm \frac{3}{5}$
since $\pi<\theta<\frac{3 \pi}{2}$
i. e. $\cos \theta=-\frac{3}{5}$
$\Rightarrow \frac{\pi}{2}<\frac{\theta}{2}<\frac{3 \pi}{4}$
lies in II quadrant using identity,
$\cos \frac{\theta}{2}=\pm \sqrt{\frac{1+\cos \theta}{2}}$
$=\pm \sqrt{\frac{1-\frac{3}{5}}{2}}$
$=\pm \sqrt{\frac{5-\frac{3}{5}}{2}}$
$=\pm \sqrt{\frac{2}{5} \times \frac{1}{2}}$
$=\pm \frac{1}{\sqrt{5}}$
since $\frac{\theta}{2}$ lies II quadrant $=\cos \frac{\theta}{2}=-\frac{1}{\sqrt{5}}$
Hence, the correct answer is option C.
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