trigonometry - 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.

Answer

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?