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Solution:

Given: $\sin x=\frac{-1}{2}$ and $x$ lies in Quadrant IV.

To Find: i) $\sin \frac{x}{2}$ ii) $\cos \frac{x}{2}$ iii) $\tan \frac{x}{2}$

Now, since $\sin x=\frac{-1}{2}$

We know that $\cos x=\pm \sqrt{1-\sin ^{2} x}$

$\cos x=\pm \sqrt{1-\left(\frac{-1}{2}\right)^{2}}$

$\cos x=\pm \sqrt{1-\frac{1}{4}}$

$\cos x=\pm \sqrt{\frac{3}{4}}=\pm \frac{\sqrt{3}}{2}$

since $\cos x$ is positive in IV quadrant, hence $\cos x=\frac{\sqrt{3}}{2}$

i) $\sin \frac{x}{2}$

Formula used:

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

Now, $\sin \frac{x}{2}=\pm \sqrt{\frac{1-\left(\frac{\sqrt{3}}{2}\right)}{2}}=\pm \sqrt{\frac{\frac{2-\sqrt{3}}{2}}{2}}=\pm \sqrt{\frac{2-\sqrt{3}}{4}}=\pm \frac{\sqrt{2-\sqrt{3}}}{2}$

Since $\sin x$ is negative in IV quadrant, hence $\sin \frac{x}{2}=-\frac{\sqrt{2-\sqrt{3}}}{2}$

ii) $\cos \frac{x}{2}$

Formula used:

$\cos \frac{x}{2}=\pm \sqrt{\frac{1+\cos x}{2}}$

now, $\cos \frac{x}{2}=\pm \sqrt{\frac{1+\left(\frac{\sqrt{3}}{2}\right)}{2}}=\pm \sqrt{\frac{\frac{2+\sqrt{3}}{2}}{2}}=\pm \frac{\sqrt{2+\sqrt{3}}}{2}$

since cos $x$ is positive in IV quadrant, hence $\cos \frac{x}{2}=\frac{\sqrt{2+\sqrt{3}}}{2}$

iii) $\tan \frac{x}{2}$

Formula used:

$\tan x=\frac{\sin x}{\cos x}$