Introduction: Finding the Value of Π

I will presnt a way to prove π = 3.14159 ............ using geometry and trigonometry................

it was originally made by David Coulson, Teacher / writer in maths and science

Step 1: Basic Diagram

The circumference of the circle is bigger than the perimeter of the inner hexagon,and is smaller than the perimeter of the outer hexagon...........................

Step 2: Inner Hexagon

The perimeter of the inner hexagon is exactly 6 radii..................

The perimeter of the outer hexagon is harder to work out.................

Step 3: Outer Hexagon

The length of each side is 2R Tan 30 2R Tan 30

So, the perimeter of the outer hexagon is

6 x 2R Tan 30 = 12R Tan 30

Step 4: Value of Pi

The circle is sandwiched between the inner hexagon and the outer hexagon.

6R < Circumference < 12RTan 30

c= 2 π R

therefore,

3 < π < 3.4

Step 5: More Detailed

we can improve on the estimate by doubling the number of sides.

Step 6: More Detailed Vale of Π

12 sin15 < π < 12 tan15

therefore,

3.106 < π < 3.21

Step 7: FORMULA

In general, for an n-sided polygon inside the circle and another one outside of the circle,

n x Sin (180/n) < p < n x Tan (180/n)

As the number of sides increase,

the value of p is squeezed more tightly by the upper and lower limits.

Step 8: Final Value

SEE the chart above based on previous formula.........

the max and min valve of π is same when no.of sides = 6,114

and thus the value of π is 3.141593

Step 9: Solving Problem

You can use geometry to get the COS of 60 degrees.........that is 1/2 or 0.5 ....and then we can use the half angle formulas to generate all the SINs and TANs you need after that.

Step 10: Value Found to Be 3.14159

So starting from 60 degrees, we can calculate the Sin, Cos and Tan of increasingly small angles..............

from the image you can see the value of π to be

3.14159

Runner Up in the
Pi Day Challenge 2016

Participated in the
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}