Hello fellow StackExchange members!
Some time ago, I was thinking about the proofs concerning the circumference of a circle and had the following idea. Having insufficient knowledge in trigonometry and calculus at that time, I couldn't get myself to "do the math". But after taking classes, I now have the ability (and time due to quarantine!) to share my proof with you. I do not know if this counts as a proof since we have a predefined definition of $π$, but here is my way of thinking:
In a regular polygon of $n$ sides and with a distance $r$ between its vertices and its center, we can use trigonometry to find the length of one side. ($n∈\mathbb{Z}^+$ and $r∈\mathbb{R}^+$)
First of all, we have to find the central angle $α$ between any two neighboring "radii" forming an isosceles triangle when looked at with the side connecting the two vertices. To find this, we can divide the central angle of $2π$ to the number of sides:
$α=\frac{2π}{n}$
Next, we can use the law of cosines to calculate the length of the third side of the isosceles triangle, which also happens to be the side length of the regular polygon. Let's call the side length $s$.
$s=\sqrt{r^2+r^2-2⋅r⋅r⋅\cos(α)}=\sqrt{2r^2-2⋅r^2⋅\cos(\frac{2π}{n})}\\=r\sqrt{2-2\cos(\frac{2π}{n})}$
To find the total perimeter ($P$) of the regular polygon, we have to multiply the number of sides by the side lenght:
$P=n⋅r\sqrt{2-2\cos(\frac{2π}{n})}$
Now, when $n$ approaches infinity, our polygon should approach to being a circle; so if we take the limit of this expression with $n$ approaching infinity, we should be able to approximate the circumference of the circle with a radius of $r$. (I urge you to plot this function into a graphic calculator with $x$ instead of $n$ as a variable. As $x$ fets larger, the function approaches the value of $2πr$. I was really happy to see this for the first time.)
Before the limit, let's simplify our expression. I was careful with the domains and ranges of the trigonometric functions and rewriting the expression in the following way should not cause any problems while taking the limit at infinity.
$n⋅r\sqrt{2-2\cos(\frac{2π}{n})}=n⋅r\sqrt{2(1-\cos(2⋅\frac{π}{n}))}\\=n⋅r\sqrt{2[1-(1-2\sin^2(\frac{π}{n}))]}=n⋅r\sqrt{4\sin^2(\frac{π}{n})}=2⋅n⋅r⋅\sin(\frac{π}{n})$
(That simplification was quite magical for my simple mind!)
Let's now take the limit:
$\lim_{n \to \infty} n⋅r\sqrt{2-2\cos(\frac{2π}{n})}=\lim_{n \to \infty}2⋅n⋅r⋅\sin(\frac{π}{n})=2r⋅\lim_{n \to \infty}n⋅\sin(\frac{π}{n})$
Now this was where I got stumped for a while; until I learned L'Hospital's Rule of course! We can rewrite the expression as follows:
$2r⋅\lim_{n \to \infty}n⋅\sin(\frac{π}{n})=2r⋅\lim_{n \to \infty}\frac{\sin(\frac{π}{n})}{\frac{1}{n}}$
We now see that the limit gives us $\frac{0}{0}$ as $n$ approaches infinity so let's use L'Hospital's Rule:
$2r⋅\lim_{n \to \infty}\frac{\sin(\frac{π}{n})}{\frac{1}{n}}=2r⋅\lim_{n \to \infty}\frac{\frac{d}{dn}(\sin(\frac{π}{n}))}{\frac{d}{dn}(\frac{1}{n})}\\=2r⋅\lim_{n \to \infty}\frac{\cos(\frac{π}{n})⋅(-\frac{π}{n^2})}{-\frac{1}{n^2}}=2r⋅\lim_{n \to \infty}π⋅\cos(\frac{π}{n})\\=2πr⋅\lim_{n \to \infty}\cos(\frac{π}{n})$
Since the limit of the fraction $\frac{π}{n}$ approaches $0$, the value of $\cos(\frac{π}{n})$ approaches $1$:
$2πr⋅\lim_{n \to \infty}\cos(\frac{π}{n})=2πr⋅1=2πr$
And this is how I ended up with the circumference of a circle. Thank you very much for sticking around until the end!
Is this a mathematically correct way of proving the circumference of a circle?