I was trying to find the area of a circle the ancient way. For example, here is the area of an octagon inscribed in a circle.(adsbygoogle = window.adsbygoogle || []).push({});

You formula is the same regardless of how many sides your figure has: (S/2)r^{2}sin(2π/S).

And so, the area of a circle must be lim_{S-->∞}(S/2)r^{2}sin(2π/S).

I can expand that out using the Taylor series.

(S/2)r^{2}[(2π/S) - (2π/S)^{3}/3! + (2π/3)^{5}/5! - (2π/3)^{7}/7! + ... - .... + ...... ]

I can factor a (2π/S) out of the sum.

r^{2}([STRIKE]S[/STRIKE]/[STRIKE]2[/STRIKE])([STRIKE]2[/STRIKE]π/[STRIKE]S[/STRIKE]) [ 1 - (2π/S)^{2}/3! + (2π/3)^{5}/4! - (2π/3)^{6}/7! + ... - .... + ...... ].

As you can see, the sum now lookslike the infinite series for cosine. Let me pull out a [1 + 1/3 + 1/5 + 1/7 + ....]sort of

r^{2}π cos(2π/S) [1 + 1/3 + 1/5 + 1/7 + .......].

cos(2π/S)--->1 as s--->∞.

This is where I'm stuck. I won't get πr^{2}unless [1 + 1/3 + 1/5 + 1/7 + .......] is equal to 1, which it obviously isn't.

Some help, please?

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# Area of circle using incribed polygons

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