So, last year I got obsessed with the idea of finding a way to calculate π that wasn't already done. After reading some history, the Greek idea of measuring polygons inscribed within circles and cutting their vertices to increase their side counts and get closer to calculating π intrigued me. From this I decided to use perimeter as the base, but I had to figure out how that would actually work. I went on for weeks trying to figure out an answer, but kept coming short and basically wasting my time by looking in the completely wrong places. Then I thought about the beautiful circle that we all know and love: the unit circle and its perfect radius of 1 unit. Within a week, I had answered the question and since then have improved upon it.
My first answer to the problem was $$\lim_{n\to ∞} (2^n)\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+...\sqrt2}}}}=π$$ and a similar answer of $$\lim_{n\to ∞} 3(2^{n-1})\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2+...\sqrt3}}}}=π$$ when n=the # of radicals in the problem. Both answers heavily relied on the half-angle identity and the simpler exact values of higher degree cosine functions. As said, this has been simplified greatly since then so that it is shorter in length and possibly more versatile.
Now I think it's about time I actually explain the math that led to the solution. If you inscribe a regular polygon of r=1 within a circle, you will notice that adding more sides gradually makes the polygon look more like the circle it is in. The points found at the degrees of a circle can be linked to create this polygon (ex. Drawing a line between consecutive multiples of 120 degrees makes a triangle and 90 degrees makes a square). The perimeter(P) of a regular polygon is found by multiplying side length(s) by the total number of sides(t). We can easily find the number of sides, so the side length is next. That becomes easy as well when you use the distance formula: $\sqrt{(x_2-x_1)^2+(x_2-x_1)^2}$. The point at 0 degrees will be our anchor point and will be used every time, and our other point will be whatever point we choose within the unit circle. We can then say $x_2=cos(0)=1$, $y_2=sin(0)=0$, $x_1=cosx$, and $y_1=sinx$. We plug the values in to get $\sqrt{(1-cosx)^2+(sinx)^2}$.
I quite stupidly didn't realize for some reason at the time that I could simplify that easily, so we'll get to it later. Now that we have s, we just multiply it by t to get P. We can then correlate P with circumference(C) by saying $P≈C=2pr=2p(1)=2p$ and single out the π with $P/2≈π$. So now we have $(t/2)\sqrt{(1-cosx)^2+(sinx)^2}≈π$. Using a 90 degree angle, you solve to get $2\sqrt2$. 45 degrees gives $4\sqrt{2-\sqrt2}$, 22.5 degrees gives $8\sqrt{2-\sqrt{2+\sqrt2}}$, and so on. An interesting detail about the increments of $90/2^n$ degrees is that its values directly correlate with Francois Viete's infinite product that was published in his 1593 work Variorum de rebus mathematicis responsorum. Go ahead and use a calculator to check that if you want.
After checking some more angles to show that they followed the same pattern, I realized that since $\sqrt{2+2cosx}/2=cos(x/2)$ and t correlates to the angle used, then the whole thing can simplify to $$f(x)=(180/x)\sqrt{2-2cosx}$$ for degrees and $$f(x)=(π/x)\sqrt{2-2cosx}$$ for radians. I could have simplified this much longer ago if I had realized that I overlooked simplifying the original side length. We now have a function though! Great! I'm sure some may notice that π in the radian function and will likely question it, but rest assured that it is only there to cancel out the π found in x (ex. π/(π/3)=3). This function can be simplified even further though. Since $\sqrt{2-2cosx}/2=sin(x/2)$, we can simplify it down to $(π/x)2sin(x/2)=((π/x)2sinx)/2$ (We divide by 2 because we halve t when we double the angle), so we are left with simply $$f(x)=(π/x)sinx$$ While the first function type goes to negative π when approaching 0 from the left and positive π when from the right, this simpler function goes to postive π from either direction. Simple trig for a simple answer. That is all that I have worked out and I'm not sure whether or not more can come from it. Since I believe this is the final result, I simply wanted to give what I've learned and ask one simple question. Has anyone posted, published, or given this answer in any way before this post? I was unable to find anything from any of my searching, but my answer is so simple that I am curious whether or not someone else has or can find such information. I also apologize in advance for if I made a mistake in some way, such as if this is the wrong place to submit this type of post.
