Theorem 5.8c: The area of an {n/k} star inscribed in a circle of radius r is

nr2sin(180/n)cos(180k/n)sec(180(k-1)/n)

proof 4: Notice that in the geometric proofs we have been taking n copies of figures which are obtained by doubling the triangle in the following figure.

There are several ways to compute twice the area of this triangle. One way would be to use Theorem 5.1a which would give us an area of

nsrk-1sin(a)

where a is the angle at Pk-1 which is seen to be 180 - (90 - 180k/n) - 180/n = 90 + 180(k-1)/n. Substituting for s from Theorem 5.7c and rk-1 from Theorem 4.3 we get an area of

= nrsin(180/n)sec(180(k-1)/n)rcos(180k/n)sec(180(k-1)/n)sin(90 + 180(k-1)/n)

= nrsin(180/n)sec(180(k-1)/n)rcos(180k/n)sec(180(k-1)/n) cos(180(k-1)/n)

= nrsin(180/n)sec(180(k-1)/n)rcos(180k/n)

= nr2 sin(180/n)sec(180(k-1)/n)rcos(180k/n)

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Proof 1

Proof 2

Proof 3

Proof 5

Proof 6