Answer
$$\sin\theta=\frac{\sqrt{102}}{12}$$
$$\cos\theta=-\frac{\sqrt{42}}{12}$$
Work Step by Step
$$\cos2\theta=-\frac{5}{12} \hspace{1.5cm}90^\circ\lt\theta\lt180^\circ$$
$$\sin\theta=?\hspace{2cm}\cos\theta=?$$
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, $$\cos^2\theta=\frac{\cos2\theta+1}{2}=\frac{-\frac{5}{12}+1}{2}=\frac{\frac{7}{12}}{2}=\frac{7}{24}$$
$$\cos\theta=-\frac{\sqrt7}{\sqrt{24}}=-\frac{\sqrt7}{2\sqrt6}=-\frac{\sqrt{42}}{12}\hspace{1.5cm}\cos\theta\lt0$$
Now, $\cos2\theta$ can also be written this way
$$\cos2\theta=1-2\sin^2\theta$$
Thus,
$$\sin^2\theta=\frac{1-\cos2\theta}{2}=\frac{1-\Big(-\frac{5}{12}\Big)}{2}=\frac{1+\frac{5}{12}}{2}=\frac{\frac{17}{12}}{2}=\frac{17}{24}$$
$$\sin\theta=\frac{\sqrt{17}}{\sqrt{24}}=\frac{\sqrt{17}}{2\sqrt{6}}=\frac{\sqrt{102}}{12}\hspace{1.5cm}\sin\theta\gt0$$