I have been trying to calculate pi using the formula.

$$n \sin\bigg(\frac{180}{n}\bigg) \cos\bigg(\frac{180}{n}\bigg)$$

This formula is a simplified regular polygon area formula, without being multiplied by the radius squared. The basis for it calculating pi is that as a circle is effectively a regular polygon with an infinite number of sides, they also have an area of pi when having a radius of 1. As you get larger and larger numbers of sides the area of the regular polygon matches more and more of pi. The problem I'm having is that all of the $\sin/\cos$ functions on the internet use radians, I can't do that if I want to calculate pi sadly. But, I had an idea, the top half of the unit circle can be graphed with the following, $Y = \sqrt{1 - x^2}.$ I could then calculate the intercept between this and the $\sin/\cos$ line with the slope of the angle. My problem so far has been getting the slopes to make coherent angles to me. For instance, when I do this with, $Y = x$, I get the same answer as $\sin(45^\circ)$. However, if I want to do $22.5$ degrees, it ends up just shy of $Y = .4X$, not $Y = .5X$, as I initially assumed. So, ultimately I'm wondering if there is a way to find the angle of a line against the $X$-Axis without using trig functions if you know the slope. Also, it doesn’t have to work above 45 degrees as it is unlikely I will ever use these $\sin/\cos$ functions for any other purpose.

  • $\begingroup$ What exactly about $n\cdot\sin(\frac{n}{180})\cdot\cos(\frac{n}{180})$ will help you calculate $\pi$? $\endgroup$ – Rhys Hughes May 7 '18 at 0:34
  • $\begingroup$ it is the simplified formula to calculate the area of a regular polygon with a radius of 1. The non-simplified one is multiplied by r^2. N is the number of sides. $\endgroup$ – Nick B May 7 '18 at 0:40
  • $\begingroup$ Could you include some more of your work here so we can get a better idea of exactly what you're doing? It certainly seems like an interesting problem $\endgroup$ – Robert Howard May 7 '18 at 1:59
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    $\begingroup$ You complicate things by trying to use degrees here. The radian measure is just nicely adapted to taking limits of trig functions as you move the argument to zero. $\endgroup$ – Allawonder May 7 '18 at 2:27
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    $\begingroup$ The problem is radians are in terms of pi, sort of a problem when trying to calculate it. $\endgroup$ – Nick B May 7 '18 at 10:20

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