irrational trig functions
Phil Colville
comply with protocol (or rudeness even!).
========================================================
Here is my problem.
n is a natural number. The only values of n for which I know sin(2*pi/n) to
be rational are 1, 2, 4 and 12 (unless I've missed something obvious). Two
questions:
1. Are these the only values ?
2. Is the proof of 1. "elementary" ?
Many thanks for any help you can offer
Phil
(this arises out of a problem which I set my department - to prove that the
only regular polygon which can be "drawn on a square lattice" - i.e.
vertices at lattice points - is the square)
Franz Lemmermeyer
>
> THIS IS MY FIRST EVER POSTING TO A NEWSGROUP. Please forgive any failures to
> comply with protocol (or rudeness even!).
You really must be new -)
> Here is my problem.
>
> n is a natural number. The only values of n for which I know sin(2*pi/n) to
> be rational are 1, 2, 4 and 12 (unless I've missed something obvious). Two
> questions:
> 1. Are these the only values ?
> 2. Is the proof of 1. "elementary" ?
[...]
> (this arises out of a problem which I set my department - to prove that the
> only regular polygon which can be "drawn on a square lattice" - i.e.
> vertices at lattice points - is the square)
This last question reminds me of a problem that was - if I am
correct - posed by Littlewood in the 1920's and proved around 1950,
all in the American Mathematical Monthly. The article
L. Tan,
The group of rational points on the unit circle
Math. Mag. 69 (1996), no. 3, 163--171
contains relevant results and references, but I don't have
it here. The proofs are elementary (Tan uses unique
factorization in Z[i]).
franz
Robert Israel
"Phil Colville" <pcol...@tesco.net> writes:
> THIS IS MY FIRST EVER POSTING TO A NEWSGROUP. Please forgive any failures to
> comply with protocol (or rudeness even!).
>
> Here is my problem.
>
> n is a natural number. The only values of n for which I know sin(2*pi/n) to
> be rational are 1, 2, 4 and 12 (unless I've missed something obvious). Two
> questions:
> 1. Are these the only values ?
1. Yes. The only rational numbers r for which sin(r pi) is rational are
those for which 2 sin(r pi) is an integer. This is because 2 sin(r pi) is
an algebraic integer. To prove this, note that
2 cos(n t) = exp(int) + exp(-int) is a monic polynomial in 2 cos(t)
= exp(it) + exp(-it) with integer coefficients. This is easy to show by
induction, using
(exp(it) + exp(-it))^n = exp(int) + exp(-int) +
sum_{1<=k<n/2} (n choose k) (exp(i(n-2k)t) + exp(-i(n-2k)t))
+ (if n even) (n choose n/2)
If t = m Pi/n, then cos(n t) is an integer, so 2 cos(t) is an
algebraic integer. But the only algebraic integers that are
rational numbers are rational integers. So cos(t) is rational
iff 2 cos(t) is an integer. Since sin(t) = cos(Pi (1/2 - m/n)),
the same goes for sin(t).
> 2. Is the proof of 1. "elementary" ?
I would say so. It's not hard to prove that the only algebraic integers
that are rational numbers are rational integers.
Robert Israel isr...@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2