If I have $2\pi/n$, where $n$ is the number of sides of a polygon, does the answer give me a length of sides necessary for me to draw the polygon with those $n$ sides as a regular polygon inside the circle?
I know it works or I believe it does. But analytically, without anything but the drawing, how can I know this is true? I am having trouble visualizing it. I want to know why it works.
What you can do is divide the angle at the center of the circle by $n$, which gives angles of $\frac {2 \pi}n$ radians at the center. Now draw radii from the center to the circle at this spacing and make the points where the radii hit the circle. As all the central angles are the same, the lengths of the sides between the radii are all the same. The angles between the sides of your polygon are twice the base angle of the isoceles triangles formed by the radii, so the angles are all the same. This proves this gives a regular polygon.
The lengths of the side will scale with the radius of the circle. For a regular polygon of $n$ sides in a circle of radius $r$, the side will be $s=2r \sin \frac {\pi}n$, which you can prove by looking at the isoceles triangles.