This is a fine technique, although there's a crucial step missing in how to calculate the cosine or sine of the angles (assuming that like Archimedes you don't have a calculator). One could use an infinite series, for example:
sin(x) = x - x^3/3! + x^5/5! - x^7/7! + ...
but this is cumbersome, and requires a knowledge of calculus that I doubt the ancients had. So I wonder if they didn't use another technique, such as this: given the knowledge that cos(pi/4) = sqrt(2)/2, and the relationship cos(x) = sqrt[(cos(2x)-1)/2], you can play around using basic trig identities and you find that you get a neat recursive formula:
 = \frac {\sqrt 2} 2 \\<br />
\sin \(\frac {\pi} 8 \) = \frac 12 \sqrt {2 - \sqrt 2} \\<br />
\sin \(\frac {\pi} {16}\) = \frac 12 \sqrt {2 - \sqrt {2+ \sqrt 2}}\\<br />
\sin \(\frac {\pi} {32}\) = \frac 12 \sqrt {2 - \sqrt {2+ \sqrt {2 + \sqrt 2}}}<br />
)
etc.
The length of the side of an n-sided polygon is 2* sin(pi/(n)), so if we choose polygons where each has twice the number of sides as its predecessor, the accuracy in determining pi is limited only by one's accuracy in calculating square roots. The formula for estimating the length of half the the perimeter (that is, pi) is:
 = N \times \sin(\pi/N).<br />
)
We choose the value for N to be 1, 2, 4, 8, 16, etc., and what we find is that the approximation for pi converges pretty quickly to a reasonable level of accuracy:
 \\<br />
4 & 2.828 \\<br />
8 & 3.061 \\<br />
16 & 3.121 \\<br />
32 & 3.136 \\<br />
64 & 3.140 \\<br />
128 & 3.1413 \\<br />
256 & 3.14151 \\<br />
512 & 3.14157<br />
\end{matrix}<br />
)
etc.
Linear Mode
