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# If $\sin \theta=-\large\frac{4}{5}$ and $\theta$ lies in third quadrant then the value of $\cos\large\frac{\theta}{2}$ is

$\begin{array}{1 1}(A)\;\large\frac{1}{5}&(B)\;\large\frac{-1}{\sqrt {10}}\\(C)\;\large\frac{-1}{\sqrt 5}&(D)\;\large\frac{1}{\sqrt{10}}\end{array}$

Toolbox:
• $\cos \large\frac{\theta}{2}$$=\pm \sqrt{\large\frac{1+\cos \theta}{2}} Given : \sin \theta=-\large\frac{4}{5} in III quadrant \cos \theta=\large\frac{-3}{5} in III quadrant \cos\large\frac{\theta}{2}$$=-\sqrt{\large\frac{1+\cos \theta}{2}}$
$\Rightarrow -\sqrt{\large\frac{1+\Large\frac{-3}{5}}{2}}$
$\Rightarrow -\sqrt{\large\frac{2}{10}}$
$\Rightarrow -\large\frac{1}{\sqrt 5}$
Hence (C) is the correct answer.