Proof using trigonometry that circle circumference is $2 \pi R$ Using trigonometry, I would like to prove that the circumference of a circle is $2\pi$ times its radius. Can someone help please?
 A: The question as it stands is not well posed as you are asking the wrong question.  In order to prove that the circumference formula $C = 2\pi r$ holds for all circle of radius $r$, we first have to understand what $\pi$ is.  
First, let's discuss what $\pi$ is not.  In grade school, $\pi$ is typically defined as a number which is about $3.14159\dots$  In reality, $\pi$ is more than just a number.  It is defined as the proportion of a circle's circumference to its diameter.  First, we need to ask ourselves: Why is that no matter what diameter of a circle we have, the ratio of the circumference to the diameter are always the same? The answer to this question can be found here, here, or at any of these links via a google search.  Once we know that the circumference of a circle and its diameter are always proportional, then we can ask: What is the proportionality constant?  This amounts to precisely evaluating digits of pi, which is a computation problem.
With these two question behind us, your original question now answers itself.  We have $C = 2 \pi r$, since $\pi = \frac{C}{d} = \frac{C}{2r}$ by definition.
A: HINT:
If we divide an $n$ sided equilateral polygon into $n$ triangles (which are generated by the lines to centre from the vertices), the angle at the centre $\frac{2\pi}n$
Let each side be $x$ and the distance from the centre to any vertex be $r$
The rest two angles are same $=\theta$(say)
So, $2\theta=\pi-\frac{2\pi}n\implies \theta =\frac\pi2-\frac\pi n$
Now, using Sine Law of triangle, $$\frac r{\sin \left(\frac\pi2-\frac\pi n\right) }=\frac x{\sin \frac{2\pi}n}$$
$$\implies x=\frac{\sin \frac{2\pi}n}{\cos \frac\pi n }=2r\sin\frac\pi n  $$
Now, the circumference $=n\cdot 2r\sin\frac\pi n $ 
Now, the polygon will become a circle if $n\to\infty$
