Rational Sines of Rational Multiples of PI
For which integer number of degrees is the sine a rational number?
Excluding the trivial cases sin(0 deg) = 0 and sin(90 deg) = 1,
and the "negative" case sin(150 deg) = sin(-30 deg) = -1/2, our
question is really whether 30 degrees is the only positive integer
number of degrees k less than 90 such that sin(k deg) is rational.
There are several ways of showing that it is.
One interesting approach is to consider the polynomials whose
roots are sin(pi*d/n) for d=1,2,..,2n-1. The coefficients of
the first several such polynomials are shown below:
n x x^3 x^5 x^7 x^9 x^11 x^13
--- --- --- ---- ----- ---- ----- -----
1 1
2 -4 4
3 9 -24 16
4 -16 80 -128 64
5 25 -200 560 -640 256
6 -36 420 -1792 3456 -3072 1024
7 49 -784 4704 -13440 19712 -14336 4096
For example, the 11 quantities sin(d*pi/6) for d=1,2,..,11 are
the roots of
1024x^11 - 3072x^9 + 3456x^7 - 1792x^5 + 420x^3 - 36x = 0
In general, the coefficient of x^(2n-1) is 2^(2n-2), and the
coefficient of x is +-n^2. Therefore, in the special case of
integer degree angles, we consider the values of sin(d*pi/180)
where d is an integer number of degrees. These numbers are the
roots of a polynomial of degree 359; the coefficient of x^359
is 2^358 and the coefficient of x is 180^2. Of course this
polynomial has the root x=0, so we can factor that out.
It follows from "Gauss's Lemma" that any rational roots of the
remaining polynomial must have a numerator that divides 180^2
and a denominator that divides 2^358. Thus, the denominator
must be a power of 2, so we can divide out the powers of 2
from 180^2, meaning that the numerator of the reduced fraction
must be one of the fifteen divisors of (3^4)(5^2) = 2025.
By inspection we can check the integer angles from 1 to 89,
and we find that only d=30 has a rational root, namely 1/2.
The above isn't the most efficient demonstration, but it leads to
the interesting family of polynomials whose roots are sines of
rational multiples of pi with a given denominator. Notice that
the sums of the coefficients of these polynomials are alternately
0 and 1. On the other hand, if we sum the absolute values of the
coefficients they have the sums
1 = (1)^2
8 = 2 (2)^2
49 = (7)^2
244 = 2 (12)^2
1681 = (41)^2
9800 = 2 (70)^2
57121 (239)^2
These numbers satisfy the recurrence s[n] = 6s[n-1] - s[n-2] + 2,
corresponding to the Pell equation x^2 - 8y^2 = 1. Also, the values
of s[n](s[n]+1)/2 are all the square triangular numbers.
Incidentally, if we consider just the absolute values in the table
of coefficients above, and begin with a leading column of 1/2's, this
array of absolute values can be generated by the recurrence
A[n,k] = 4 A[n-1,k-1] + 2 A[n-1,k] - A[n-2,k]
For example, in the 7th row we have
13440 = 4(1792) + 2(3456) - 640
Also, we can simplify the array by including the null root
sin(0*pi/n) and then consider the polynomials whose roots are
TWICE the sines. The result is given by dividing the coefficient
of x^j by the power 2^j. This gives the following array of
coefficients for the polynomials whose roots are the 2n values
of x = 2sin(k*pi/n) with k = 0,1,2,..,2n-1
n x^2 x^4 x^6 x^8 x^10 x^12 x^14 x^16 x^18
--- ----- ----- ----- ----- ----- ----- ----- ----- -----
1 1
2 4 -1
3 9 -6 1
4 16 -20 8 -1
5 25 -50 35 -10 1
6 36 -105 112 -54 12 1
7 49 -196 294 -210 77 -14 -1
8 64 -336 672 -660 352 -104 16 -1
9 81 -540 1386 -1782 1287 -546 135 -18 1
These polynomials are closely related to the Chebyshev polynomials.
Let's let S_n(x) denote the nth polynomial in this table. The
absolute values of the coefficients in this table satisfy the
recurrence
A[n,k] = A[n-1,k-1] + 2 A[n-1,k] - A[n-2,k]
For example, in the 9th row we have
1287 = (660) + 2(352) - (77)
Naturally, if n divides m then S_n(x) divides S_m(x). Also,
notice that for even integers n the set of values sin(k*pi/n)
for k=0,1,..,2n-1 is precisely the same set of values of
cos(k*pi/n) for k=0,1,..,2n-1, although in a different order.
Therefore, if we let C_n(x) denote the polynomial whose roots
are 2 cos(k*pi/n) for k=0,1,..,2n-1 we have the equality
C_{2j}(x) = S_{2j}(x)
However, if n is odd, the values of sin(k*pi/n) and cos(k*pi/n)
are shifted "half a step" from each other, so the sine and cosine
polynomials are not the same. Of course, it's clear that C_n(x)
must always divide S_{2n}(x), and likewise S_n(x) must divide
C_{2n}(x). The actual coefficients for the polynomials C_n(x)
are listed below
n x^0 x^2 x^4 x^6 x^8 x^10 x^12 x^14 x^16 x^18
--- ----- ----- ----- ----- ----- ----- ----- ----- ----- -----
1 -4 1
2 0 4 -1
3 -4 9 -6 1
4 0 16 -20 8 -1
5 -4 25 -50 35 -10 1
6 0 36 -105 112 -54 12 1
7 -4 49 -196 294 -210 77 -14 -1
8 0 64 -336 672 -660 352 -104 16 -1
9 -4 81 -540 1386 -1782 1287 -546 135 -18 1
Obviously, when evaluating either sin(k*pi/n) or cos(k*pi/n) over
an entire cycle from k=0 to k=2n-1, every value occurs precisely
twice (because, for example, sin(30)=sin(150)), except for the
peak values, which occur in sin(k*pi/n) when n is even, and in
cos(k*pi/n) for all n. Consequently, we know S_n(x) must be of
the form (x-2)(x+2)p(x)^2 when n is even, and of the form p(x)^2
when n is odd. On the other hand, C_n(x) must always be of the
form (x-2)(x+2)p(x)^2.
To confirm this, here are the factorizations for the first few
polynomials S_n(x):
n S_n(x)
--- -------------------------------------------------------
1 (x)^2
2 (x-2) (x+2) (x)^2
3 (x)^2 (x^2 - 3)^2
4 (x-2) (x+2) (x)^2 (x^2 - 2)^2
5 (x)^2 (x^4 - 5x^2 + 5)^2
6 (x-2) (x+2) (x)^2 (x-1)^2 (x+1)^2 (x^2 - 3)^2
7 (x)^2 (x^6 - 7x^4 + 14x^2 - 7)^2
8 (x-2) (x+2) (x)^2 (x^2 - 2)^2 (x^4 - 4x^2 + 2)^2
9 (x)^2 (x^2 - 3)^2 (x^6 - 6x^4 + 9x^2 - 3)^2
and here are the factorizations of the first few polynomials C_n(x):
n C_n(x)
--- -------------------------------------------------------------
1 (x-2) (x+2)
2 (x-2) (x+2) (x)^2
3 (x-2) (x+2) (x-1)^2 (x+1)^2
4 (x-2) (x+2) (x)^2 (x^2 - 2)^2
5 (x-2) (x+2) (x^2 + x - 1)^2 (x^2 - x - 1)^2
6 (x-2) (x+2) (x)^2 (x-1)^2 (x+1)^2 (x^2 - 3)^2
7 (x-2) (x+2) (x^3 - x^2 - 2x + 1)^2 (x^3 + x^2 - 2x - 1)^2
8 (x-2) (x+2) (x)^2 (x^2 - 2)^2 (x^4 - 4x^2 + 2)^2
9 (x-2) (x+2) (x-1)^2 (x+1)^2 (x^3 - 3x - 1)^2 (x^3 - 3x + 1)^2
These tables confirm all the required divisibility properties for
these polynomials. The number of "new" roots of each successive
polynomial S_n(x) and C_n(x) is just phi(n), i.e., the number of
positive integers less than and coprime to n. For example, the new
roots appearing at n=180 are just the 48 roots (which occur in
pairs with opposite signs) of the polynomial
x^48 - 48x^46 + 1080x^44 - 15136x^42 + 148092x^40
- 1074528x^38 + 5995185x^36 - 26320356x^34
+ 92286216x^32 - 260824576x^30 + 597177831x^28
- 1109376324x^26 + 1669616130x^24 - 2026629360x^22
+ 1969138215x^20 - 1514843020x^18 + 908975295x^16
- 417023856x^14 + 142445393x^12 - 34943820x^10
+ 5851386x^8 - 619152x^6 + 36216x^4 - 864x^2 + 1
The roots of this equation are twice the sines of any integer number
of degrees coprime to 180, i.e., 1, 7, 11, 13, and so on. Of course,
this polynomial appears squared in S_180(x), along with the least
common multiple of all the S_d(x) for every integer d that divides
180.
In any case, our polynomials in 2sin(k*pi/180) show that these non-
zero roots can be rational only if 2sin(k*pi/180) is an integer
dividing (180)^2, but since the sine is always between -1 and +1, the
only possible rational values of 2sin(k*pi/180) are -2, -1, 0, +1,
and +2, and so the only possible rational values of sin(k*pi/180)
are -1, -1/2, 0, +1/2, and +1.
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