Answer
$$\sin\theta=\frac{\sqrt6}{6}$$
$$\cos\theta=-\frac{\sqrt{30}}{6}$$
Work Step by Step
$$\cos2\theta=\frac{2}{3} \hspace{1.5cm}90^\circ\lt\theta\lt180^\circ$$
$$\sin\theta=?\hspace{2cm}\cos\theta=?$$
1) First, we decode the signs of the $\sin\theta$ and $\cos\theta$
From the information given, we know that $90^\circ\lt\theta\lt180^\circ$. In the trigonometric circle, this is the realm of quadrant II. So $\theta$ must terminate in quadrant II.
As $\theta$ terminates in quadrant II, it can, therefore, be stated that $\sin\theta\gt0$ and $\cos\theta\lt0$.
2) Using only the Double-Angle Identities for $\cos2\theta$, we can find out the value of $\sin\theta$ and $\cos\theta$ as follows.
$\cos2\theta$ can be written as
$$\cos2\theta=2\cos^2\theta-1$$
Thus, $$\cos^2\theta=\frac{\cos2\theta+1}{2}=\frac{\frac{2}{3}+1}{2}=\frac{\frac{5}{3}}{2}=\frac{5}{6}$$
$$\cos\theta=-\frac{\sqrt5}{\sqrt6}=-\frac{\sqrt{30}}{6}\hspace{1.5cm}\cos\theta\lt0$$
Now, $\cos2\theta$ can also be written this way
$$\cos2\theta=1-2\sin^2\theta$$
Thus,
$$\sin^2\theta=\frac{1-\cos2\theta}{2}=\frac{1-\frac{2}{3}}{2}=\frac{\frac{1}{3}}{2}=\frac{1}{6}$$
$$\sin\theta=\frac{1}{\sqrt{6}}=\frac{\sqrt{6}}{6}\hspace{1.5cm}\sin\theta\gt0$$
The last four exercises show that with just the value of $\cos2\theta$ and the position of angle $\theta$ in the trigonometric circle, $\sin\theta$ and $\cos\theta$ can be readily calculated. This is due to the fact that $\cos2\theta$ can be flexibly rewritten in 3 different ways, with both $\sin\theta$ and $\cos\theta$ or with just one element out of two.
$\sin2\theta$, nevertheless, does not possess such capacity, since there is only one way to write $\sin2\theta$ in terms of $\sin\theta$ and $\cos2\theta$:
$$\sin2\theta=2\sin\theta\cos\theta$$
And as you can see, it involves both $\sin\theta$ and $\cos\theta$. Therefore, when $\sin2\theta$ and the position of angle $\theta$ are given and the requirements ask for $\sin\theta$ and $\cos\theta$, there must be another information given as only $\sin2\theta$ is not enough to figure out any of $\sin\theta$ or $\cos\theta$.
