info prev up next book cdrom email home

Trigonometry Values Pi/11

Trigonometric functions of $n\pi/11$ for $n$ an integer cannot be expressed in terms of sums, products, and finite root extractions on real rational numbers because 11 is not a Fermat Prime. This also means that the Undecagon is not a Constructible Polygon.


However, exact expressions involving roots of complex numbers can still be derived using the trigonometric identity
$\sin(11\alpha)=\sin(12\alpha-\alpha)\cos\alpha-\cos(12\alpha)\sin\alpha$
$ = 2 \sin(6\alpha)\cos(6\alpha)\cos\alpha-[1-2\sin^2(6\alpha)]\sin\alpha.\quad$ (1)
Using the identities from Beyer (1987, p. 139),

\begin{displaymath}
\sin(6\alpha)=\sin\alpha \cos\alpha [32\cos^4\alpha-32\cos^2\alpha+6]
\end{displaymath} (2)


\begin{displaymath}
\cos(6\alpha)=32\cos^6\alpha-48 \cos^4\alpha+18\cos^2\alpha-1
\end{displaymath} (3)

gives

$\sin(11\alpha) = 2 \cos^2\alpha \sin\alpha(32 \cos^4\alpha-32\cos^2\alpha+6)$
$\quad \times(32\cos^6\alpha-48 \cos^4\alpha+18\cos^2\alpha-1)$
$\quad -\sin\alpha [1-2 \sin^2\alpha \cos^2\alpha(32\cos^4\alpha-32\cos^2\alpha+6)^2]$
$= \sin\alpha(11-220 \sin^2\alpha+1232 \sin^4\alpha -2816 \sin^6\alpha+2816\sin^8-1024 \sin^{10}\alpha).$ (4)
Now, let $\alpha\equiv\pi/11$ and $x \equiv \sin^2\alpha$, then


\begin{displaymath}
\sin\pi = 0 = 11-220x+1232x^2-2816x^3+2816x^4-1024x^5.
\end{displaymath} (5)

This equation is an irreducible Quintic Equation, so an analytic solution involving Finite Root extractions does not exist. The numerical Roots are $x = 0.07937$, 0.29229, 0.57115, 0.82743, 0.97974. So $\sin\alpha=0.2817$, $\sin(2\alpha)=0.5406$, $\sin(3\alpha)=0.7557$, $\sin(4\alpha)=0.9096$, $\sin(5\alpha)=0.9898$. From one of Newton's Identities,

$\sin\left({\pi\over 11}\right)\sin\left({2\pi\over 11}\right)\sin\left({3\pi\ov...
...ight)\sin\left({5\pi\over 11}\right)= \sqrt{11\over 1024} = {\sqrt{11}\over 32}$ (6)
$\cos\left({\pi\over 11}\right)\cos\left({2\pi\over 11}\right)\cos\left({3\pi\ov...
...ght)\cos\left({4\pi\over 11}\right)\cos\left({5\pi\over 11}\right)= {1\over 32}$ (7)
$\tan\left({\pi\over 11}\right)\tan\left({2\pi\over 11}\right)\tan\left({3\pi\ov...
...ight)\tan\left({4\pi\over 11}\right)\tan\left({5\pi\over 11}\right)= \sqrt{11}.$ (8)
The trigonometric functions of $\pi/11$ also obey the identity

\begin{displaymath}
\tan\left({3\pi\over 11}\right)+4\sin\left({2\pi\over 11}\right)=\sqrt{11}.
\end{displaymath} (9)

See also Undecagon


References

Beyer, W. H. ``Trigonometry.'' CRC Standard Mathematical Tables, 28th ed. Boca Raton, FL: CRC Press, 1987.




© 1996-9 Eric W. Weisstein
1999-05-26