Question

If $\sin\theta = \frac{- 4}{5}$ and $\theta$ lies in the third quadrant, then $\cos\frac{\theta}{2} =$

• A. $\frac{1}{\sqrt{5}}$
• B. $- \frac{1}{\sqrt{5}}$
• C. $\sqrt{\frac{2}{5}}$
• D. $- \sqrt{\frac{2}{5}}$

Answer 1

Answer: (B)

$- \frac{1}{\sqrt{5}}$

Answer 2

Given that $\sin\theta = - \frac{4}{5}$and $\theta$lies in the III quadrant.

$\Rightarrow \cos\theta = \sqrt{1 - \frac{16}{25}} = \pm \frac{3}{5}$

$\cos\frac{\theta}{2} = \pm \sqrt{\frac{1 + \cos\theta}{2}} = \sqrt{\frac{1 - 3/5}{2}} = \pm \sqrt{\frac{1}{5}}$

But $\cos\frac{\theta}{2} = - \frac{1}{\sqrt{5}}.$ since $\frac{\theta}{2}$will be in II quadrant.

Hence $\cos\frac{\theta}{2} = - \frac{1}{\sqrt{5}}$.