Trigonometry Values Pi/11

Trigonometric functions of $n\pi/11$ for 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

, 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.

$\sin(11\alpha)=\sin(12\alpha-\alpha)\cos\alpha-\cos(12\alpha)\sin\alpha$
$= 2 \sin(6\alpha)\cos(6\alpha)\cos\alpha-[1-2\sin^2(6\alpha)]\sin\alpha.\quad$	(1)

$\sin(11\alpha) = 2 \cos^2\alpha \sin\alpha(32 \cos^4\alpha-32\cos^2\alpha+6)$
$\quad \times(32\cos^6\alpha-48 \cos^4\alpha+18\cos^2\alpha-1)$
$\quad -\sin\alpha [1-2 \sin^2\alpha \cos^2\alpha(32\cos^4\alpha-32\cos^2\alpha+6)^2]$
$= \sin\alpha(11-220 \sin^2\alpha+1232 \sin^4\alpha -2816 \sin^6\alpha+2816\sin^8-1024 \sin^{10}\alpha).$	(4)

$\sin\left({\pi\over 11}\right)\sin\left({2\pi\over 11}\right)\sin\left({3\pi\ov... ...ight)\sin\left({5\pi\over 11}\right)= \sqrt{11\over 1024} = {\sqrt{11}\over 32}$	(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)