M. Conroy
Dept. of Mathematics
University of Washington
May 2006
The Symmetry of the Curve x = cos(t + cos αt), y = sin(t + sin αt)
Let
fα (t) = cos(t + cos(αt)) and gα (t) = sin(t + sin(αt)).
We wish to establish the symmetry of the family of curves
x = fα (t), y = gα (t)
for integer values of α. The symmetry depends on the residue class of α modulo 4.
We will treat three cases.
0.1
The case α ≡ 0 mod 2
In this case, the curves have symmetry with respect to the origin. That is, if (a, b) is on the curve,
then so is (−a, −b). To show this, we show that
fα (t) = −fα (t + π) and gα (t) = −gα (t + π)
for all t.
Proof:
fα (t + π) =
=
=
=
=
cos(t + π + cos(α(t + π)))
cos(t + π + cos(αt + απ))
cos(t + π + cos(αt)) (since α is even )
− cos(t + cos(αt))
−fα (t).
The proof for gα (t) is identical.
0.2
The case α ≡ 1 mod 4
In this case, we have symmetry with respect to the line y = −x. That is, if the point (a, b) lies on
the curve, then the point (−b, −a) does as well. We will show the following:
fα −
π
π
+ t = −gα − − t
4
4
for all t.
Proof:
For all t, we have
π
π
π
+ t = cos − + t + cos α − + t
4
4
4
π
π
π
= cos − + t + cos − α cos αt − sin − α sin αt
4
4
4
√
√
!
2
2
π
cos αt +
sin αt
= cos − + t +
4
2
2
fα −
using the fact that α ≡ 1 mod 4.
Similarly,
π
π
π
gα − + t = sin − − t + sin α − − t
4
4
4
π
π
π
= sin − − t + sin(− α) cos(−αt) + cos(− α) sin(−αt)
4
4
4
√
√
!
2
2
π
cos αt −
sin αt
= sin − − t −
4
2
2
Since
cos(−
π
π
+ z) = sin(− − z)
4
4
for all z, we conclude that
fα −
π
π
+ t = −gα − − t
4
4
for all t.
Thus, we also have
gα
π
π
π
− + t = gα − − (−t) = −fα − − t
4
4
4
so the curve is symmetric with respect to the line y = −x.
0.3
The case α ≡ 3 mod 4
In this case we have symmetry with respect to the line y = x. That is, if the point (a, b) is on the
curve, then the point (b, a) is on the curve. We will show the following:
fα
and
π
π
+ t = gα
−t
4
4
π
π
+ t = fα
−t
gα
4
4
for all t. This will show that the point (f (t), g(t)) corresponding to t = π4 + ǫ and the point
corresponding to t = π4 − ǫ are symmetric with respect to the line y = x. Since every point on the
curve, i.e., each t value, can be expressed as t = π4 + ǫ for some ǫ, this shows that for every point
(a, b) on the curve, there is a symmetric point (b, a) on the curve as well, and hence the curve is
symmetric with respect to the line y = x.
Proof:
Suppose α ≡ 3 ( mod 4). Then for all t, we have
fα
π
+t
4
=
=
=
=
=
=
=
=
=
π
π
cos
+ t + cos α
+t
4
4
π
π
cos
+ t + cos − + αt
4
4
π
π
π
− sin −
sin αt
+ t + cos −
cos
4
4
4
π
π
π
cos
+ t + cos cos αt + sin sin αt
4
4
4
π
π
π
− t + sin cos αt − cos sin αt
sin
4
4
4 π
π
π
sin
sin(−αt)
− t + sin − cos(−αt) + cos −
4
4
4
π
π
− t + sin
− αt
sin
4
4
π
π
sin
− t + sin α
−t
4 4
π
−t
gα
4
This also shows that fα π4 − t = gα
with respect to the line y = x.
π
4
+ t . Hence, we can conclude that the curve is symmetric
0.4
Figures
α=2
1
0
-1
0
1
0
1
0
1
-1
α=4
1
0
-1
-1
α=6
1
0
-1
-1
α=8
1
0
-1
0
1
-1
α = −2
1
0
-1
0
1
0
1
-1
α = −4
1
0
-1
-1
α = −6
1
0
-1
0
1
-1
α=1
1
0
-1
0
1
0
1
-1
α=5
1
0
-1
-1
α=9
1
0
-1
0
1
-1
α = 13
1
0
-1
0
1
-1
α=3
1
0
-1
0
-1
1
α=7
1
0
-1
0
1
-1
α = 11
1
0
-1
0
-1
1