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 - firstname.lastname@example.org