Question

If $\sin \alpha =-\frac{3}{5}$, where $\pi <\alpha <\frac{3\pi }{2}$, then $\cos \frac{\alpha }{2}$

• A. $-\frac{1}{\sqrt{10}}$
• B. $\frac{1}{\sqrt{10}}$
• C. $\frac{3}{\sqrt{10}}$
• D. $-\frac{3}{\sqrt{10}}$

Answer 1

The correct option is A

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

Explanation for the correct option.

Given that, $\sin \alpha =-\frac{3}{5}$and $\pi <\alpha <\frac{3\pi }{2}$.

$\Rightarrow \frac{\pi }{2}<\frac{\alpha }{2}<\frac{3\pi }{4}$
So, $\cos \frac{\alpha }{2}$ is negative as it lies in II quadrant.
$\sin \alpha =-\frac{3}{5}$$\Rightarrow \cos \alpha =-\frac{4}{5}$
Also, $\cos 2A=2{\cos }^{2}A-1$
$\begin{array}{rcl}\cos \alpha & =& 2{\cos }^2\left(\frac{\alpha }{2}\right)-1\\ -\frac{4}{5}& =& 2{\cos }^2\left(\frac{\alpha }{2}\right)-1\\ 2{\cos }^2\left(\frac{\alpha }{2}\right)& =& 1-\frac{4}{5}\\ 2{\cos }^2\left(\frac{\alpha }{2}\right)& =& \frac{1}{5}\\ {\cos }^2\left(\frac{\alpha }{2}\right)& =& \frac{1}{10}\\ \cos \left(\frac{\alpha }{2}\right)& =& -\frac{1}{\sqrt{10}}\end{array}$

Hence, option A is correct.