How do you find the sine and cosine from the tangent? I'm given the problem:

If $\cot(\theta) = 1.5$ and $\theta$ is in quadrant 3, what is the value of $\sin(\theta)$?

I looked at all the related answers I could find on here, but I haven't been able to piece together the answer I need from them. 
I know that $\sin^2\theta + \cos^2\theta = 1$, $ \cot^2\theta + 1 = \csc^2\theta $, and $\csc^2\theta = \frac{1}{\sin^2\theta}$
Substituting 3.25 for $\cot^2\theta + 1$ and $\frac{1}{\sin^2\theta}$ for $\csc^2\theta$ I get:
$3.25 = \frac{1}{\sin^2\theta}$ 
then
$\sin\theta = -\sqrt{\frac{1}{3.25}}$
This doesn't seem correct though. Can anyone help please?
edit: Sorry, meant to make that answer negative.
 A: Indeed, we know that
$$1.5 = \cot\theta = \frac{\cos\theta}{\sin\theta}$$
hence
$$1.5\sin\theta = \cos\theta.$$
Squaring both sides we have
$$2.25\sin^2\theta = \cos^2\theta$$
and since $\cos^2\theta = 1-\sin^2\theta$ we have
$$\begin{align*}
2.25\sin^2\theta &= 1-\sin^2\theta\\
2.25\sin^2\theta + \sin^2\theta &= 1\\
3.25\sin^2\theta &= 1.
\end{align*}$$
From this, you can figure out the value of $\sin^2\theta$. Taking square roots will tell you something about the absolute value of $\sin\theta$.
Now... why did they tell you $\theta$ was in the third quadrant?
A: If $\cot\theta=1.5$, then $\tan\theta=\frac23$. This means that if $\theta$ were in the first quadrant, it would be one of the angles of a right triangle whose legs measure $2$ and $3$ and whose hypotenuse measures $\sqrt{2^2+3^2}=\sqrt{13}$. Specifically, it would be the angle opposite the side of length $2$. Sketch the triangle, and you’ll see that in that case we’d have $$\sin\theta=\frac2{\sqrt{13}}\;.$$
But $\theta$ is in the third quadrant, not the first; what effect does this have on its sine?
A: Consider the point $P=(-3,-2)$, and the (reflex) angle $\theta$ formed by the positive $x$-axis and the line from the origin to $P$. Can you see why this is the $\theta$ in the problem? Can you work out the other functions from this diagram?
