What is the smallest regular $n+1$-gon that contains a regular $n$-gon with unit length sides? 
Given a regular $n$-gon with sides of unit length, what is the side length of the smallest regular $n+1$-gon containing it?

For $n=3$ a bit of calculus yields a square of side length
$$\cos\frac{\pi}{24}=\frac{1+\sqrt{3}}{2\sqrt{2}}\approx0.965925826289...$$
For $n=4$ a bit more calculus and a few more cases to check yield a regular pentagon of side length
$$\frac{\sin\tfrac{7\pi}{40}}{\sin\tfrac{3\pi}{10}}=\frac{\sqrt{5-\sqrt{5}+\sqrt{25-10\sqrt{5}}}}{\sqrt{5}}\approx0.936859701734...$$
What is the minimal side length for $n=5$? I hope someone, somewhere has already taken the time to publish a list values for small $n$. Any reference is welcome, though I'll be satisfied with any effective method to compute the exact values as well.
 A: If a regular n-gon
is inscribed in a circle
of radius $r$,,
the central angle
of a side is
$2\pi/n$,
so the distance of the side
from the center 
$r_i(n)$
satisfies
$r_i(n)
= r\cos(\pi/n)
$
and the length of the side is
$s_i
=2r\sin(\pi/n)
$.
If a regular n-gon
is circumscribed around a circle
or radius $r$,
the central angle
of a side is
$2\pi/n$,
so the distance of the corner
from the center 
$s(n)$
satisfies
$r_c(n)
= r/\cos(\pi/n)
$
and the length of the side is
$s_c
=2r\tan(\pi/n)
$.
If the sides of 
a regular n-gon
are of unit length,
the distance to the vertices
$d(n)$
satisfies
$\dfrac{1/2}{d(n)}
=\sin(\pi/n)
$
so
$d(n)
=\dfrac1{2\sin(\pi/n)}
$.
If this is
the distance to the side
of a regular m-gon
(I will set
$m = n+1$ later),
then,
if $s$ is the side of the m-gon
then
$\dfrac{s/2}{d(n)}
=\tan(\pi/m)
$
so
$s
=2d(n)\tan(\pi/m)
=2\dfrac1{2\sin(\pi/n)}\tan(\pi/m)
=\dfrac{\tan(\pi/m)}{\sin(\pi/n)}
$.
If $m = n+1$,
this is
$\dfrac{\tan(\pi/(n+1))}{\sin(\pi/n)}
$.
This, of course,
does not take into account
the possibility that
the outer m-gon
might be able to be smaller
if it is rotated,
so this is
an upper bound.
If we use the approximation
$\sin(x) \approx x$,
this is about
$\dfrac{n}{m}
$.
If the side of the m-gon
is the same of the
side of the n-gon,
then
$\tan(\pi/m)
=\sin(\pi/n)
$
so
$\dfrac{\pi}{m}
=\arctan(\sin(\pi/n))
$
or
$m
=\dfrac{\pi}{\arctan(\sin(\pi/n))}
$.
