Answer
$$\sin\theta=-\frac{\sqrt{33}}{6}$$
Work Step by Step
$$\cos(-\theta)=\frac{\sqrt 3}{6}\hspace{1cm}\cot(\theta)\lt0$$
First, $\cos\theta=\cos(-\theta)=\frac{\sqrt 3}{6}$
Using this Pythagorean Identity,
$$\cos^2\theta+\sin^2\theta=1$$
we have,
$$\sin^2\theta=1-\cos^2\theta$$
$$\sin^2\theta=1-(\frac{\sqrt 3}{6})^2=1-\frac{3}{36}=\frac{33}{36}$$
Therefore,
$$\sin\theta=\pm\frac{\sqrt{33}}{6}$$
We also know that $$\cot\theta\lt0$$
$$\frac{\cos\theta}{\sin\theta}\lt0\hspace{1cm}\text{(Quotient Identities)}$$
That means the signs of $\cos\theta$ and $\sin\theta$ must be opposite from each other.
$\cos\theta=\frac{\sqrt 3}{6}\gt0$. So, $\sin\theta\lt0$.
That means, $$\sin\theta=-\frac{\sqrt{33}}{6}$$
