Question

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

• A. $\frac{1}{\sqrt{10}}$
• B. $-\frac{1}{\sqrt{10}}$
• C. $-\frac{1}{\sqrt{5}}$
• D. $\frac{1}{5}$

Answer 1

The correct option is C $-\frac{1}{\sqrt{5}}$

Find the sign of required trigonometric functions in given quadrant

$\because \theta$ lies in third quadrant then cos⁡ θ is negative.

If $\pi <\theta <\frac{3\pi }{2}$ then $\frac{\pi }{2}<\frac{\theta }{2}<\frac{3\pi }{4}$

$\Rightarrow \cos \frac{\theta }{2}$ is negative

Use right angled triangle to get the value of required trigonometric functions

Given: $\sin \theta =-\frac{4}{5}$

$\cos \theta =-\sqrt{1-{\sin }^2\theta }(\because \cos \theta \text{is negative}$

$=-\sqrt{1-{(-\frac{4}{5})}^2}$

$=-\sqrt{1-\frac{16}{25}}$

$=-\frac{3}{5}$

Apply the trigonometric formula of submultiple angles

$\Rightarrow \cos \theta =2{\cos }^2\left(\frac{\theta }{2}\right)-1$

$\Rightarrow -\frac{3}{5}=2{\cos }^2\left(\frac{\theta }{2}\right)-1$

$\Rightarrow 2{\cos }^2\left(\frac{\theta }{2}\right)=1-\frac{3}{5}=\frac{2}{5}$

$\Rightarrow {\cos }^2\left(\frac{\theta }{2}\right)=\frac{1}{5}$

$\Rightarrow \cos \left(\frac{\theta }{2}\right)=\pm \frac{1}{\sqrt{5}}$

$\Rightarrow \cos \left(\frac{\theta }{2}\right)=-\frac{1}{\sqrt{5}}[\because \cos \frac{\theta }{2}\text{is negative}]$

Hence, the option $\left(C\right)$ is correct.