Trigonometric functions of
for
an integer cannot be expressed in terms of sums, products, and finite root
extractions on real rational numbers because 7 is not a Fermat Prime. This also means that the Heptagon
is not a Constructible Polygon.
However, exact expressions involving roots of complex numbers can still be derived using the trigonometric identity
![\begin{displaymath}
\sin(n\alpha)=2\sin[(n-1)\alpha]\cos\alpha-\sin[(n-2)\alpha].
\end{displaymath}](t_1921.gif) |
(1) |
The case
gives
Rewrite this using the identity
,
|
|
|
(3) |
Now, let
and
, then
![\begin{displaymath}
\sin(\pi) = 0 = x^3-{\textstyle{7\over 4}} x^2+{\textstyle{7\over 8}} x - {\textstyle{7\over 64}},
\end{displaymath}](t_1935.gif) |
(4) |
which is a Cubic Equation in
. The Roots are numerically found to be
,
,
. But
, so these Roots correspond to
,
,
. By Newton's
Relation
![\begin{displaymath}
\prod_i r_i = -a_0,
\end{displaymath}](t_1943.gif) |
(5) |
we have
![\begin{displaymath}
x_1 x_2 x_3 = {\textstyle{7\over 64}},
\end{displaymath}](t_1944.gif) |
(6) |
or
![\begin{displaymath}
\sin\left({\pi\over 7}\right)\sin\left({2\pi\over 7}\right)\...
... 7}\right)= \sqrt{7\over 64} = {\textstyle{1\over 8}}\sqrt{7}.
\end{displaymath}](t_1945.gif) |
(7) |
Similarly,
![\begin{displaymath}
\cos\left({\pi\over 7}\right)\cos\left({2\pi\over 7}\right)\cos\left({3\pi\over 7}\right)= {1\over 8}.
\end{displaymath}](t_1946.gif) |
(8) |
The constants of the Cubic Equation are given by
The Discriminant is then
so there are three distinct Real Roots. Finding the first one,
![\begin{displaymath}
x = {\root 3 \of {R+\sqrt{D}}}+{\root 3 \of {R-\sqrt{D}}}-{\textstyle{1\over 3}} a_2.
\end{displaymath}](t_1953.gif) |
(12) |
Writing
![\begin{displaymath}
\sqrt{D} = 3^{-3/2} {\textstyle{7\over 128}} i,
\end{displaymath}](t_1954.gif) |
(13) |
plugging in from above, and anticipating that the solution we have picked corresponds to
,
See also Heptagon
© 1996-9 Eric W. Weisstein
1999-05-26