can you show me the steps on how to?

Question

Find the approximate value of pi by circumscribing a regulare heagon (n=6) about a unite circle and then doubling the number of sides four times until you reach n=96.



note: for the regular hexagon, S(6)=(2sqrt(3))/3
note: this is based on the forumlas s(2n)=(sqrt(2-sqrt(4-(s(n))^2))) and s(2n)=(2sqrt(4+(s(n))^2)-4)/s(n) for the side of a regular 2n-gon inscribed in and circumscribed about a unit circle, respectively.


this was a hint given but i still don't know: find s(n) in each case, then multiply by n and divide by 2. Starting with n=12, the expressions will get increasingly messy, so it is enough to find the decimal value in each case.

PLEASE HELP ASAP..thank you so much

Answer 1

Consider the circumscribed hexagon as a collection of 6 triangles with a common vertex at the center of the circle.

All the triangles are equilateral triangles with the same side, and since the circle is circumscribed, the circle is tangent at the center of the base of each triangle, so the altitude of the triangles is the radius of the circle.

What is the ratio of the altitude of an equilateral triangle to the side? Well, you know:

half the central angle (60/2 = 360/(6 x 2) degrees)
the tangent of that angle
and that the tangent equals (base/2)/altitude

So, assuming a radius of 1, the length of the side of the hexagon = 2 x tan(60/2) = 2 x tan(180/6)

Then the circumference is approximately the sum of the 6 sides of the hexagon. But we know the circumference is exactly 2 x pi x r, so the sum gives us an approximation for pi.

Now suppose we double the number of sides to 12. We now have 12 isosceles triangles with the altitudes still being radii. The approximation to the circumference is going to be 12 times the length of the base. The formula is still the same, but the central angle is now only 15 degrees.

We can generalize:

S(n) = number of sides x length of each side
S(n) = n x 2 x tan(180/n)

so:
S(2n) = 2 x n x 2 x tan(180/(2n))

Using the half-angle formulas, it is possible to compute the sine, cosine, and tangent of the new half angle, etc.

http://mathworld.wolfram.com/Half-AngleFormulas.html

(My guess, not having worked through the algebra, is that if you do the substitutions for the half angle, you get the hint)

Repeat for 12 sides, 24 sides, 48 sides, and 96 sides.