Answer
$$\sin\theta=\frac{\sqrt5}{5}$$
Work Step by Step
$$\cos2\theta=\frac{3}{5}\hspace{1cm}\theta\hspace{0.2cm}\text{terminates in quadrant I}\hspace{1cm}\sin\theta=?$$
From the original half-angle identity for sines:
$$\sin\frac{\theta}{2}=\pm\sqrt{\frac{1-\cos\theta}{2}}$$
we can rewrite to use to find $\sin\theta$
$$\sin\theta=\pm\sqrt{\frac{1-\cos2\theta}{2}}$$
Also, whether to choose positive or negative square root now depends on the sign of $\sin\theta$.
$\theta$ terminates in quadrant I, where sines are positive. Therefore, $\sin\theta\gt0$.
As a result, we need to choose positive square root:
$$\sin\theta=\sqrt{\frac{1-\cos2\theta}{2}}$$
Now we can start calculating $\sin\theta$.
$$\sin\theta=\sqrt{\frac{1-\frac{3}{5}}{2}}$$
$$\sin\theta=\sqrt{\frac{\frac{2}{5}}{2}}$$
$$\sin\theta=\sqrt{\frac{2}{5\times2}}=\sqrt{\frac{1}{5}}$$
$$\sin\theta=\frac{1}{\sqrt5}=\frac{\sqrt5}{5}$$
Therefore, $$\sin\theta=\frac{\sqrt5}{5}$$
