Question

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. $\frac{1}{\sqrt{10}}$

Answer 1

Answer: (C)

Given that, $sin\theta =-\frac{4}{5}$ and $\theta$ lies in the $II{I}^{rd}$ 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.