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Rational Trigonometry
There are an infinite number of points on the unit circle whose coordinates are
each rational numbers. Indeed, every Pythagorean triple gives one!

23
265
,
264
265


7
25
,
24
25

 203 ,
445
396
445

3, 4
5
5
 275 ,
373
252
373


153
185
,
104

80
185
89
,


39
89
35
37

,

12
37
357
365


,
399
401
76
365
,

40

401
Rational Trigonometry
Consider the tetrahedron with vertices (1, 1, 1), (1, %1, %1), (%1, 1, %1),
(%1, %1, 1). What is the cosine of the angle formed by rays from the origin to two
of these vertices?
The cosine is % 1 , and the sine is
3
2
2
3
.
Rational Trigonometry
A rational angle is an angle whose degree measure is a rational number. Equivalently, it's an angle whose radian measure is a rational number times &.
It's not too hard to see that the sine and cosine of a rational angle are algebraic
numbers.
Toward this end, one may use the angle sum formulas for sine and cosine to build
up expressions for cos(3 a), cos(4 a), cos(5 a), etc. that are polynomials in cos(a).
cos(x + y) = cos(x) cos(y) % sin(x) sin(y)
sin(x + y) = sin(x) cos(y) + cos(x) sin(y)
cos(3 a) = cos(2 a + a)
= cos(2 a) cos(a) % sin(2 a) sin(a)
= 2 cos2(a) % 1 cos(a) % (2 sin(a) cos(a)) sin(a)
= 2 cos3(a) % cos(a) % 2 sin2(a) cos(a)
= 2 cos3(a) % cos(a) % 2 1 % cos2(a) cos(a)
= 4 cos3(a) % 3 cos(a)
Rational Trigonometry
This process builds the Chebyshev polynomials of the first kind.
cos(2 a)
cos(3 a)
cos(4 a)
cos(5 a)
cos(6 a)
cos(7 a)
cos(8 a)
cos(9 a)
cos(10 a)
=
=
=
=
=
=
=
=
=
2 cos2(a) $ 1
4 cos3(a) $ 3 cos(a)
8 cos4(a) $ 8 cos2(a) + 1
16 cos5(a) $ 20 cos3(a) + 5 cos(a)
32 cos6(a) $ 48 cos4(a) + 18 cos2(a) $ 1
64 cos7(a) $ 112 cos5(a) + 56 cos3(a) $ 7 cos(a)
128 cos8(a) $ 256 cos6(a) + 160 cos4(a) $ 32 cos2(a) + 1
256 cos9(a) $ 576 cos7(a) + 432 cos5(a) $ 120 cos3(a) + 9 cos(a)
512 cos10(a) $ 1280 cos8(a) + 1120 cos6(a) $ 400 cos4(a) + 50 cos2(a) $ 1
Rational Trigonometry
Letting a =
m
n
&, we see that cos(n a) = ± 1, so cos(a) is a root of the appropriate
Chebyshev polynomial plus or minus 1.
Example: With n = 3, we get cos(& , 3) is a root of 4 x3 % 3 x + 1, so the cosine of
& , 3 is algebraic. (But we knew that!)
This reasoning shows cos m & is algebraic for all rational numbers
n
m
.
n
1 % cos2 m & is also algebraic, so sin m & is algebraic too.
Finally,
n
n
Rational Trigonometry
Which rational angles have a super-easy-to-remember sine and cosine?
Specifically: Which rational angles - have sin2(-) and cos2(-) both rational?
Here are a few such rational angles in the first quadrant: 0, & , & , & , & .
6
Are there more?

7 2
24 2
 +
 =1
25
25
%
1 2
 +
3
2
10
2
+
2
2
3
7
2
10
2
=1
2
=1
4
3
2
Rational Trigonometry
LEMMA 1. Let b0 be any rational number with %2 ≤ b0 ≤ 2. Define a sequence recursively by bk+1 = bk2 % 2. Then:
(i) Each bk is rational.
(ii) Each bk satisfies %2 ≤ bk ≤ 2.
(iii) If the same value is ever attained twice, say bm = bm+k, then
bm ∈ {%2, %1, 0, 1, 2}.
(iv) If bm ∈ {%2, %1, 0, 1, 2} and m > 0, then bm%1 ∈ {%2, %1, 0, 1, 2}.
Thus if the sequence b0, b1, b2, … ever repeats a term, then b0 ∈ {%2, %1, 0, 1, 2}.
Rational Trigonometry
(iii) If the same value is ever attained twice, say bm = bm+k, then
bm ∈ {%2, %1, 0, 1, 2}.
Proof of (iii)
Let f (x) = x2 % 2, and for any natural number n, let f n(x) = f (f (f (4 f (x) 4 ))) be the nfold composition of f . This is a monic polynomial of degree 2 n, with integer
coefficients.
Note that f k(bm) = f k(f m(b0)) = f m+k(b0) = bm+k = bm.
This means bm is a rational root of f k(x) % x.
This means that bm is an integer! (Gauss’ Lemma: Every rational root of a monic
polynomial with integer coefficients must be an integer.)
Rational Trigonometry
LEMMA 2.
Let - ∈ 6, and let b0 = 2 cos(-). Then bk = 2 cos2k - for all k.
Proof:
This is a simple inductive argument that makes use of the double angle formula for
cosine.
bk = bk%12 % 2
2
bk = 2 cos2k%1 - % 2
bk = 4 cos22k%1 - % 2
bk = 2 2 cos22k%1 - % 1
bk = 2 cos2k -
THEOREM
Suppose - is a rational angle, and that cos2(-) is rational. Then cos(-) and sin(-)
both belong to the set 0, ± 1, ± 1 , ±
2
Proof:
Write - =
m
n
2
2
,±
3
2
.
&, and we suppose that cos2(-) is rational (which implies sin2(-) is ratio-
nal). It suffices to show cos(-) belongs to the set 0, ± 1 , ±
2
2
2
,±
3
2
, ± 1.
Let b0 = 2 cos(2 -). By the double angle formula, b0 = 4 cos2(-) % 2, which is rational,
and clearly %2 ≤ b0 ≤ 2, so this defines the sequence b0, b1, b2,… of lemma 1.
k
By lemma 2, bk = 2 cos m 2 2 &. The key is that the only possible remainders upon
n
division of an integer by n are 0, 1, 2, …, n % 1. So after n + 1 terms in the
sequence, there must be two terms that are equal. But lemma 1 then tells us that
b0 ∈ {%2, %1, 0, 1, 2}.
This gives cos m 2 & ∈ %1, % 1 , 0, 1 , 1. Or (using the double angle formula yet
n
again):
2
2
2 cos2 m & % 1 ∈ %1, % 1 , 0, 1 , 1
n
2
2
A few short calculations should allow you to reach the conclusion.
Rational Trigonometry
COROLLARY
Suppose - is a rational angle, and that cos(-) is rational. Then cos(-) belongs to
the set 0, ± 1, ± 1 .
2
Proof:
SInce cos2(-) is also rational, the first theorem tells us that
cos(-) ∈ 0, ± 1 , ±
0, ± 1, ±
1
.
2
2
2
2
,±
3
2
, ± 1. The only rational members of this set are
Further Reading
Bergen, Jeffery, “Values of trigonometric functions,” Math Horizons, Feb. 2009, pp
17-19.
Jahnel, Jörg, “When is the (co)sine of a rational angle equal to a rational number?,”
arXiv:1006.2938 [math.HO].
Nivan, Ivan., Irrational Numbers, MAA, 1956.