Circumradius of a regular heptagon My graphing software says that the value of circumradius of a regular heptagon, of side unity, upto 5 decimal places, is 1.15238. Just as the circumradius of a regular pentagon of length unity can be expressed as 1/(2 sin36°), can it be expressed in terms of trigonometric ratios? See figure
 A: Let $R$ be the circumradius, $r$ the inradius and $a$ the side-length of
a regular $n$-gon. Then
$$\frac{a}{R}=2\sin\frac\pi n$$
and
$$\frac{a}{r}=2\tan\frac\pi n.$$
The reason? There's a right-angled triangle with hypotenuse $R$,
and side-lengths $r$ and $a/2$ adjacent and opposite to an angle $\pi/n$.
A: The answer of Lord Shark clearly is correct.
So, its use even could be extended a bit.
Let $a_d$ be the $d$-th short chord of the regular $\{n\}$-gon.
(Esp. $a_1$ being its edge, $a_2$ just connects every second vertex, etc.)
Then the above formula clearly extends into
$$\frac{a_d}{R}=2\sin(\frac{d\pi}{n})$$
(as this would be also the edge ratio of the regular $\{n/d\}$-gon star.)
And thus the ratio of those short chord lengths - in units of the edge size - becomes
$$\frac{a_d}{a_1}=\frac{\sin(\frac{d\pi}{n})}{\sin(\frac{\pi}{n})}$$
As special benefit therefrom you even could derive the nice following formula 
$$\frac{a_2}{a_1}=\frac{\sin(\frac{2\pi}{n})}{\sin(\frac{\pi}{n})}=\frac{2\sin(\frac{\pi}{n})\cos(\frac{\pi}{n})}{\sin(\frac{\pi}{n})}=2\cos(\frac{\pi}{n})$$
--- rk
