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Q.
If $sin\theta=\frac{-4}{5}$ and $\theta$ lies in third quadrant, then the value of $cos \frac{\theta}{2}$ is
Trigonometric Functions
Solution:
Given that, $sin\theta=-\frac{4}{5}$ and $\theta$ lies in the $III^{rd}$ quadrant.
$\Rightarrow cos\theta=-\sqrt{1-\frac{16}{25}}=-\frac{3}{5}$
Now, $cos \frac{\theta}{2}= \pm \sqrt{\frac{1+cos\,\theta}{2}}$
$= \pm\sqrt{\frac{1-\frac{3}{5}}{2}}$
$=\pm \sqrt{\frac{1}{5}}$
But we take $cos \frac{\theta}{2}=-\frac{1}{\sqrt{5}}$. Since, if $\theta$ lies in $III$ quadrant, then $\frac{\theta}{2}$ will be in $II$ quadrant.