Given: $\sin x=\frac{\sqrt{5}}{3}$ and $\frac{\pi}{2}<x<\pi$ i.e, $x$ lies in the Quadrant II .

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

Now, since $\sin x=\frac{\sqrt{5}}{3}$

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

$\cos x=\pm \sqrt{1-\left(\frac{\sqrt{5}}{3}\right)^{2}}$

$\cos x=\pm \sqrt{1-\frac{5}{9}}$

$\cos x=\pm \sqrt{\frac{4}{9}}=\pm \frac{2}{3}$

since $\cos x$ is negative in II quadrant, hence $\cos x=-\frac{2}{3}$

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

Formula used:

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

Now, $\sin \frac{x}{2}=\pm \sqrt{\frac{1-\left(\frac{-2}{3}\right)}{2}}=\pm \sqrt{\frac{\frac{5}{3}}{2}}=\pm \sqrt{\frac{5}{6}}$

Since sinx is positive in II quadrant, hence $\sin ^{\frac{x}{2}}=\sqrt{\frac{5}{6}}$

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

Formula used:

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

now, $\cos \frac{x}{2}=\pm \sqrt{\frac{1+\left(\frac{-2}{3}\right)}{2}}=\pm \sqrt{\frac{\frac{1}{3}}{2}}=\pm \sqrt{\frac{1}{6}}$

since $\cos x$ is negative in II quadrant, hence $\cos \frac{x}{2}=-\frac{1}{\sqrt{6}}$

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

Formula used:

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

hence, $\tan \frac{x}{2}=\frac{\sin \frac{x}{2}}{\cos \frac{2}{2}}=\frac{\sqrt{\frac{5}{6}}}{-\frac{1}{\sqrt{6}}}=\frac{\sqrt{5}}{\sqrt{6}} \times \frac{\sqrt{6}}{-1}=-\sqrt{5}$

Here, tan x is negative in II quadrant.