).
For each 
= 2, 3, 4, ...,
we inscribe and circumscribe regular polygons having
sides.
Recall, too, that the perimeters
and
satisfy
.
In order to generate an iterative formula for the perimeters, we use a bit of geometry:
If we denote the central
angle in our -gon
by 2
, then
is the side opposite the
angle in a right triangle with
hypoteneuse 1. Hence,
= sin.
Next we rotate our picture and concentrate on
.
This time our right triangle has
as the side opposite
the angle , and
as the
adjacent side. Hence, =
tan.
Archimedes' iterative formulas for
and
will now follow from two trig identities.
Neal Carothers - carother@bgnet.bgsu.edu