Cartesian equation for $((\sin\theta)^3+\cos\theta,(\cos\theta)^3+\sin\theta)$

Question

Can the parametric equations $x=(\sin\theta)^3+\cos\theta$ and $y=(\cos\theta)^3+\sin\theta$ be expressed as an equation with $x$ and $y$ only? If so, how can it be done?

Current progress

By letting $t=\sin\theta\cos\theta$, I managed to get rid of the trig functions. The equations can be converted into

\begin{align} (x+y)^2&=(1+2t)(2-t)^2\\ (x-y)^2&=(1-2t)t^2 \end{align}

and then I'm stumped. How can I continue?

Answer 1

To find a polynomial $P(x,y)$ such that:

$$ P\left(\sin^3\theta+\cos\theta,\,\cos^3\theta+\sin\theta\right) = 0 \quad \quad \forall\,\theta \in [0,2\pi) $$

there are at least two automated approaches.

---

Thanks to the identity:

$$ \sin^2\theta + \cos^2\theta = 1 $$

it follows that:

$$ \begin{cases} x = \sin^3\theta + \cos\theta \\ y = \cos^3\theta + \sin\theta \\ \end{cases} \quad \quad \Rightarrow \quad \quad \begin{cases} x - u^3 - v = 0 \\ y - v^3 - u = 0 \\ u^2 + v^2 - 1 = 0 \\ \end{cases} $$

so we can use Wolfram|Alpha to select the first polynomial of a Gröbner basis:

First[GroebnerBasis[{x - u^3 - v, y - v^3 - u, u^2 + v^2 - 1}, {v, u, y, x}]]

from which:

$$ \small \boxed{x^6+3 x^4 y^2-4 x^4-18 x^3 y+3 x^2 y^4+19 x^2 y^2+23 x^2-18 x y^3-20 x y+y^6-4 y^4+23 y^2-9 = 0}. $$

---

Thanks to identities:

$$ z \equiv e^{\text{i}\,\theta}, \quad \quad \sin\theta = \frac{z-z^{-1}}{2\,\text{i}}, \quad \quad \cos\theta = \frac{z+z^{-1}}{2} $$

it follows that:

$$ \begin{cases} x = \sin^3\theta + \cos\theta \\ y = \cos^3\theta + \sin\theta \\ \end{cases} \quad \quad \Rightarrow \quad \quad \begin{cases} 8\,x\,z^3 - \text{i}\,z^6 - (4-3\,\text{i})\,z^4 - (4+3\,\text{i})\,z^2 + \text{i} = 0 \\ 8\,y\,z^3 - z^6 - (3-4\,\text{i})\,z^4 - (3+4\,\text{i})\,z^2 - 1 = 0 \\ \end{cases} $$

so we can use Wolfram|Alpha to compute the factored resultant with respect to $z$:

Factor[Resultant[8 x z^3 - I z^6 - (4 - 3 I) z^4 - (4 + 3 I) z^2 + I, 
                 8 y z^3 - z^6 - (3 - 4 I) z^4 - (3 + 4 I) z^2 - 1, z]]

from which:

$$ \small \boxed{x^6+3 x^4 y^2-4 x^4-18 x^3 y+3 x^2 y^4+19 x^2 y^2+23 x^2-18 x y^3-20 x y+y^6-4 y^4+23 y^2-9 = 0}. $$

---

Although both are well suited to a computer application, the second can be more easily applied autonomously by calculating the determinant of a matrix, for example of the Sylvester matrix.

Answer 2

An experimental answer, in need of explanations :

I used Geogebra for sketching the initial curve (green curve above). It is made of 2 symmetric arcs, one obtained for $\theta \in [0,\pi)$, the other one for $\theta \in [\pi,2 \pi)$ (which doesn't come as a surprize because changing $\theta$ into $\theta + \pi$ gives two opposite points), these two parts being symmetrical with respect to line bissector $y=x$.

I had the idea to compare the first arc with the curve $(C_q)$ (featured in blue above) whose equation is :

$$ |x+1|^q+|y-1|^q=2^q$$

(generalized disks for distance $d_q$ with radius $2$).

I attempted first $q=2$, then progressively diminished the exponent till reaching an almost perfect matching with $q=1.775$.

Similar matching, of course, for the other arc, now with curve $(C'_q)$ (featured in red) with equation

$$ |x-1|^q+|y+1|^q=2^q$$

How can this almost perfect matching be explained ?

Remark : the two little "sparrowtails" deserve a specific analysis.

Answer 3

The Maple function eliminate() eliminates the parameter immediately:

eliminate([(x + y)^2 = (1 + 2 * t) * (2 - t)^2, (x - y)^2 = (1 - 2 * t) * t^2], t);

$$\boxed{(x^2 + y^2)^3 - 4 x^4 - 18 x^3 y + 19 x^2 y^2 - 18 x y^3 - 4 y^4 + 23 x^2 - 20 x y + 23 y^2 - 9 = 0}$$

In terms of the coordinates $x, y$, we have: $$t = \frac{4 x^2 - 9 x y + 4 y^2 + 1}{3 x^2 + 3 y^2 - 7} .$$