Trigonometric elimination possibly related to hypocycloids

This is the question, which has been previously asked on Math.SE.

Eliminate $\theta$ from the system of equations. $$x\sin\theta-y\cos\theta=-\sin4\theta$$ $$x\cos\theta+y\sin\theta=\frac52-\frac32\cos4\theta$$

I encountered this problem while browsing through some trigonometric elimination problems. At the first glance, I thought that this is definitely from the famliy of problems such as:

Eliminate $\theta$ from $$x\sin\theta-y\cos\theta=\cos2\theta$$ $$x\cos\theta+y\sin\theta=2\sin2\theta$$

where the eliminant (usually) gives evolute of a hypocycloid etc.

For instance, the elimination of the second problem is $(x-y)^{2/3}+(x+y)^{2/3}=2$ which is the envelope of normals to the astroid $x^{2/3}+y^{2/3}=1$.

So I first decided to play with geogebra graphing tool.

Solving for $x$ and $y$ we get,

$$x=\frac52\cos\theta-\frac12\cos\theta\cos4\theta-\cos3\theta$$ $$y=\frac52\sin\theta-\frac12\sin\theta\cos4\theta-\sin3\theta$$

The plot looks like this:

I then tried some other functions and I noted that the locus of point $A$ (in the figure) lies on (approximately),

$$\color{blue}{[(x-y)^{1/3}+(x+y)^{1/3}]^2=4}$$ $$\color{#F80}{[(y-x)^{1/3}+(x+y)^{1/3}]^2=4}$$

Thus, the curve, $$\left(\left((x-y)^{1/3}+(x+y)^{1/3}\right)^2-4\right)\cdot\left(\left((y-x)^{1/3}+(y+x)^{1/3}\right)^2-4\right)=0$$ seems to be doing a very good job, but the matter is it is not bounded.

With this clue, could you please help me to end this solution?

Solution 1:

If you solve the equations for $(x,y)$ and play with some trigonometric identities, there is another (more symmetric but equivalent) parametrization $$x=\frac{1}{4} (10 \cos (t)-5 \cos (3 t)-\cos (5 t))$$ $$y=\frac{1}{4} (10 \sin (t)+5 \sin (3 t)-\sin (5 t))$$

Using the multiple angle formulae, this reduces to $$x=\cos(t)\,(5-4\cos^4(t))$$ $$y=\sin(t)\,(5-4\sin^4(t))$$

This makes $$(x+y)^2=(\cos (t)+\sin (t))^{10} \qquad \text{and} \qquad (x-y)^2=(\cos (t)-\sin (t))^{10}$$

I am sure that, from here, you can finish.