Question

If $\cos x=-\frac{1}{3}$ and $x$ is in III quadrant, then which among the following is/are correct ?
I. $\cos \frac{x}{2}=-\frac{\sqrt{3}}{3}$
II. $\sin \frac{x}{2}=\frac{\sqrt{6}}{3}$
III. $\tan \frac{x}{2}=-\sqrt{2}$

• A. I is correct
• B. II is correct
• C. III is correct
• D. All are correct

Answer 1

Answer: (D)

$\cos x=-\frac{1}{3}$
Given that, $x$ is in third quadrant.
i.e., $\pi <x<\frac{3\pi }{2}$
( $\because$ in third quadrant $\theta$ lies between $\pi$ and $3\pi /2$ )
From the formula, $\cos x=2{\cos }^2\frac{x}{2}-1$
$\Rightarrow 2{\cos }^2\frac{x}{2}=1+\cos x=1-\frac{1}{3}$
$\Rightarrow 2{\cos }^2\frac{x}{2}=\frac{2}{3}$
$\Rightarrow {\cos }^2\frac{x}{2}=\frac{1}{3}$
$\Rightarrow \cos \frac{x}{2}=\pm \frac{1}{\sqrt{3}}$
Now,$\pi <x<\frac{3\pi }{2}\Rightarrow \frac{\pi }{2}<\frac{x}{2}<\frac{3\pi }{4}$
i.e., $\frac{x}{2}$ lies in second quadrant, therefore $\cos \frac{x}{2}$ will be negative.
$\Rightarrow \cos \frac{x}{2}=-\frac{1}{\sqrt{3}}=-\frac{\sqrt{3}}{3}$
Now, ${\sin }^2\frac{x}{2}=1-{\cos }^2\frac{x}{2}$
${\sin }^2\frac{x}{2}=1-{\left(-\frac{1}{\sqrt{3}}\right)}^2$
$=1-\frac{1}{3}=\frac{2}{3}$
$\sin \frac{x}{2}=\pm \sqrt{\frac{2}{3}}$
$\Rightarrow \sin \frac{x}{2}=\sqrt{\frac{2}{3}}=\frac{\sqrt{6}}{3}$
$(an\frac{x}{2}\text{ lies in second quadrant) }$
$\Rightarrow \tan \frac{x}{2}=\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}=\frac{\sqrt{\frac{2}{3}}}{\frac{1}{3}}$
$=-\sqrt{2}$