# 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?

• How do you define $\pi$? May 17 '13 at 19:01
• In many contexts, $\pi$ is defined as the ratio of circumference to diameter in a circle. That's actually the most usual definition for maybe more than two centuries now. I see a couple of questions here: (1) How does one know that that ratio is the same for all circles?, and (2) How does one know that that is the same as the ratio of the area of a circle to the area of the square on its radius? (Note that I am careful here to refer to the square on, rather than of the radius---a distinction that is too often neglected.) May 17 '13 at 19:04
• Following up on my last comment: Nowadays people often take the word "radius" to refer to a number: the distance from the center to the curve. In older usage, a radius is any line segment having one endpoint at the center and the other on the curve. A square on the radius is a square one of whose sides is the radius. Those whose reading is restricted only to the curriculum often never find this out, I suspect. I think in Latin "radius" means a spoke of a wheel (but I could be wrong about that). May 17 '13 at 19:09

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.

• I should also mention that Michael Hardy mentioned that the circumference of a circle and its length are often confused, and I admit that I use the two terms interchangeably here. What I give up in precision, I make up for in getting my point across. May 18 '13 at 20:04

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 $n\to\infty$

• What is $\pi$ if not the ratio of a circle's circumference to the length of its diameter? May 17 '13 at 19:21
• The very last step is extremely problematic since lengths don't behave very nicely with respect to taking limits. The length function is only upper semi-continuous. May 17 '13 at 19:57
• @IttayWeiss, how about maa.org/joma/volume7/aktumen/Polygon.html or the Step n: Using an n-sided polygon of ugrad.math.ubc.ca/coursedoc/math101/notes/integration/… May 18 '13 at 4:54
• @labbhattacharjee I'm not sure what these links are supposed to relate to. Can you perhaps clarify? May 18 '13 at 5:13
• @labbhattacharjee you keep ignoring the fact that length is not continuous. The fact that the polygons approach the circle does not imply that the lengths of the polygons approach the length of the circle. May 18 '13 at 7:31