निम्नलिखित प्रत्येक प्रश्न में $\sin \frac{x}{2}, \cos \frac{x}{2}$ तथा $\tan \frac{x}{2},$ ज्ञात कीजिए

$\cos x=-\frac{1}{3}, x$ तृतीय चतुर्थांश में है।

Here, $x$ is in quadrant $III$.

i.e., $\pi < x < \frac{3 \pi}{2}$

$\Rightarrow \frac{\pi}{2}<\frac{x}{2}<\frac{3 \pi}{4}$

Therefore, $\cos \frac{x}{2}$ and $\tan \frac{x}{2}$ are negative, where $\sin \frac{x}{2}$ as is positive.

It is given that $\cos x=-\frac{1}{3}$

$\cos x=1-2 \sin ^{2} \frac{x}{2}$

$\Rightarrow \sin ^{2} \frac{x}{2}=\frac{1-\cos x}{2}$

$\Rightarrow \sin ^{2} \frac{x}{2}=\frac{1-\left(-\frac{1}{3}\right)}{2}=\frac{\left(1+\frac{1}{3}\right)}{2}=\frac{4 / 3}{2}=\frac{2}{3}$

$\Rightarrow \sin \frac{x}{2}=\frac{\sqrt{2}}{\sqrt{3}} \quad\left[\because \sin \frac{x}{2} \text { is positive }\right]$

$\therefore \sin \frac{x}{2}=\frac{\sqrt{2}}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{6}}{3}$

Now $\cos x=2 \cos ^{2} \frac{x}{2}-1$

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

$\Rightarrow \cos \frac{x}{2}=-\frac{1}{\sqrt{3}} \quad\left[\because \cos \frac{x}{2} \text { is negative }\right]$

$\therefore \cos \frac{x}{2}=-\frac{1}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{-\sqrt{3}}{3}$

$\tan \frac{x}{2}=\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}=\frac{\left(\frac{\sqrt{2}}{\sqrt{3}}\right)}{\left(\frac{-1}{\sqrt{3}}\right)}=-\sqrt{2}$

Thus, the respective values of $\sin \frac{x}{2}, \cos \frac{x}{2}$ and $\tan \frac{x}{2}$ are  $\frac{\sqrt{6}}{3}, \frac{-\sqrt{3}}{3},$ and $-\sqrt{2}.$