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I was thinking about different ways of finding $\pi$ and stumbled upon what I'm sure is a very old method: dividing a circle of radius $r$ up into $n$ isosceles triangles each with radial side length $r$ and central angle $$\theta=\frac{360^\circ}{n}$$ Use $s$ for the side opposite to $\theta$.

An example for n=8.

Then we can approximate the circumference as $sn$. By the law of cosines: $$s=\sqrt{2r^2-2r^2 \cos{\theta}}=r\sqrt{2-2\cos{\left(\frac{360^\circ}{n}\right)}}$$

We know $\pi=\text{circumference}/\text{diameter} \approx \frac{sn}{2r}=\frac{n}{2}\sqrt{2-2\cos\left(\frac{360^\circ}{n}\right)}$. This becomes exact at the limit: $$\pi=\lim_{n \to \infty}\frac{n}{2}\sqrt{2-2\cos\left(\frac{360^\circ}{n}\right)}$$

Now for my question: How would you solve the opposite problem? To make my meaning more clear, above I used the definition of $\pi$ to determine a limit that gives its value. But if I had just been given the limit, what technique(s) could I have used to determine that it evaluates to $\pi$?

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  • $\begingroup$ You can approximate the opposite side length to be $ r sin \theta $ which is actually the height of the triangle. This gives a weaker limit $ \pi = \frac {180 sin \theta }{ \theta}$ , where $ \theta $ tends to zero. $\endgroup$ – N.S.JOHN Dec 12 '16 at 9:03
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Notice, $$\lim_{n\to \infty}\frac{n}{2}\sqrt{2-2\cos\left(\frac{2\pi}{n}\right)}$$ $$=\lim_{n\to \infty}\frac{n}{2}\sqrt{2\left(1-\cos\left(\frac{2\pi}{n}\right)\right)}$$ $$=\lim_{n\to \infty}\frac{n}{2}\sqrt{2\left(2\sin^2\left(\frac{\pi}{n}\right)\right)}$$ $$=\lim_{n\to \infty}\frac{2n}{2}\sin\left(\frac{\pi}{n}\right)$$ $$=\lim_{n\to \infty}\frac{\sin\left(\frac{\pi}{n}\right)}{\frac{1}{n}}$$ $$=\pi\lim_{n\to \infty}\frac{\sin\left(\frac{\pi}{n}\right)}{\left(\frac{\pi}{n}\right)}$$ Let $\frac{\pi}{n}=t\implies t\to 0\ as\ n\to \infty$ $$=\pi \lim_{t\to 0}\frac{\sin t}{t}$$ $$=\pi\times 1=\pi$$

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    $\begingroup$ Ah, the third step is the connection I was failing to make. Thanks! $\endgroup$ – hexaflexagonal Jul 21 '15 at 19:10
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Since there were already two solid answers that provided efficient approaches, I thought that it would be instructive to see a different way forward.

Here, we will use the expansion of the cosine as

$$\cos x=1-\frac12 x^2+O(x^4) \tag 1$$

Letting $x=\frac{2\pi}{n}$ in $(1)$ yields

$$2-2\cos \left(\frac{2\pi}{n}\right)=\frac{4\pi^2}{n^2} +O(n^{-4})$$

Finally, we have

$$\begin{align} \frac{n}{2}\sqrt{2-2\cos \left(\frac{2\pi}{n}\right)}&=\frac n2 \sqrt{\frac{4\pi^2}{n^2} +O(n^{-4})}\\\\ &=\frac n2 \frac{2\pi}{n}\left(1+O(n^{-2})\right)\\\\ &=\pi+O(n^{-3})\to \pi \end{align}$$

as expected!

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  • $\begingroup$ Love it! Yeah, I was definitely interested in seeing several approaches (and almost said as much in the original post). Thanks! $\endgroup$ – hexaflexagonal Jul 22 '15 at 2:18
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    $\begingroup$ @DivergentQueries Wow!! Thank you. And you're welcome! It was my pleasure. I am still home and recovering from surgery 3 weeks ago. Your comment just made my day. $\endgroup$ – Mark Viola Jul 22 '15 at 2:20
  • $\begingroup$ nice answer. i have edited a little $\endgroup$ – Bhaskara-III Sep 21 '16 at 13:16
  • $\begingroup$ @Mats Granvik Please refrain from accepting edits without checking first to see their effects. In this case, you accepted an edit that rendered an equation unreadable and changed language indiscriminantly (e.g., changed "letting" to "substituting," which had no positive impact, added the word "same," which made the sentence awkward). I have reversed out all of the edits and restored the original version. $\endgroup$ – Mark Viola Sep 21 '16 at 14:39
  • $\begingroup$ @Bhaskara-III Please refrain from making edits that either add nothing or actually degrade the quality of the post. In this case, you made edits that rendered an equation unreadable and changed language indiscriminantly (e.g., removed the "$$" from equation $(1)$ making it unreadable, changed "letting" to "substituting," which had no positive impact, added the word "same," which made the sentence awkward). I have reversed out all of the edits and restored the original version. $\endgroup$ – Mark Viola Sep 21 '16 at 14:41
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$$\sqrt{2-2\cos x}=2\left|\sin\frac{x}{2}\right|$$ hence everything boils down to: $$ \lim_{x\to 0}\frac{\sin x}{x}=1.$$

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Just recall double angle trig. identity: $\cos 2x=1-2\sin^2x$,

Starting from, $$\begin{align} \lim_{n\to \infty}\frac{n}{2}\sqrt{2-2\cos\left(2\pi/n\right)}\\=\lim_{n\to \infty}\frac{n}{2}\sqrt{2-2+4\sin^2\left(\pi/n\right)}\\ =\lim_{n\to \infty}\frac{n}{2}\cdot 2\sin\left(\pi/n\right)\\ =\pi\cdot \lim_{n\to \infty}\frac{\sin(\pi/n)}{\pi/n}\\ =\pi\cdot \lim_{t\to 0}\frac{\sin(t)}{t}\\ =\pi\cdot 1\\=\pi \end{align}$$

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