Find $\sin \frac{\theta}{2}$ when $\sin \theta = \frac{3}{5}$, and $0^{\circ} < \theta < 90^{\circ}$

Question

I need to find $\sin \frac{\theta}{2}$ when $\sin \theta = \frac{3}{5}$, and $0^{\circ} < \theta < 90^{\circ}$

The answer I'm currently getting is $\sqrt{\frac{1}{10}}$, but the answer must be either $\frac{\sqrt{10}}{10}$, $\frac{3 \sqrt{10}}{10}$, or $\frac{1}{3}$.

My Process:

since, $\sin \frac{\theta}{2} = \pm \sqrt{\frac{1 - cos \theta}{2}}$

$\sqrt{ \frac{1 - \frac{4}{5}}{2}} \cdot \sqrt{\frac{5}{5}} = \sqrt{\frac{5 - 4}{10}} = \sqrt{\frac{1}{10}}$

What am I doing wrong?

Answer 1

...well
$$ \frac{\sqrt{10}}{10} = \frac{\sqrt{10}}{\sqrt{10} \times \sqrt{10}} \color{red}{=} \frac 1{\sqrt{10}}$$

Answer 2

I had forgotten that, $\sqrt{\frac{1}{10}} = \frac{\sqrt{1}}{\sqrt{10}} = \frac{1}{\sqrt{10}} \cdot (\frac{\sqrt{10}}{\sqrt{10}}) = \frac{\sqrt{10}}{10}$

So, there is my answer.