Find $\sin \frac{x}{2}, \cos \frac{x}{2}$ and $\tan \frac{x}{2}$ for $\cos x=-\frac{1}{3}, x$ in quadrant III
Question.

Find $\sin \frac{x}{2}, \cos \frac{x}{2}$ and $\tan \frac{x}{2}$ for $\cos x=-\frac{1}{3}, x$ in quadrant III

solution:

Here, $x$ is in quadrant III.

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

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

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

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

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

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

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

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

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

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

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

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

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

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 $-\sqrt{2}$.
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