<s>[INST] Here's a problem I'm having trouble with and would love some guidance. Problem: In an 'n'  sided polygon inscribed in a circle, prove that the length of one side of the polygon is 2*R*sin((π/n) ), where R is the radius of the circle.

This problem has several variables, so I thought I'd break it down. 

Step 1: I'd draw an inscribed diagram, which has an n-sided polygon with each vertex connected to the center of the circle. Since n is the number of sides, there must be n triangles. 

Step 2:  Identify 1 triangle to focus on. Assume the center of the circle is O, the midpoint of the chord is M, and the vertices of the polygon are A and B.

Step 3:  In this triangle OAB, angle AOB is 2 * (π / n). Multiply the angle by 2 because angles on the same side of a circumscribing polygon add up to an interior angle.

Step 4: In this triangle, OM = R * cos(π/n). The reason I say that is angle MOB = π/2 (since M is the midpoint of the chord)  Therefore, triangle MOB is a right triangle.    The side Mob is therefore R*cos(π / n).

R is the Hypotenuse

Step 5: However, until now I still don't have 2 * R * sin ( π / n). This is likely where I go wrong.

You see, OM = R * cos(π/n) determines one of the sides of a triangle. I need to determine another side.

OM was only half the chord. In order to determine AB, it would have two OM's because AB consists of two Adds of OM.
OM is from vertex A to M as well as from vertex B to M. If I double OM I'd get AM and BM (or the entire chord). Therefore, 2 * ( R * cos (π/n)) would give me the AB, right?[/INST]