# 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.

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.

• Wow! very nice. Actually few seconds ago, for some reason, I was graphing $(\frac{x-y}4)^{2/5}+(\frac{x+y}4)^{2/5}=1$. Oh, that terrible $4$. Indeed $\to(+1)$ since I can't upvote more than once :(
– ACB
Commented Dec 27, 2021 at 5:17
• @ACB. Thank you ! Glad to help. Commented Dec 27, 2021 at 5:31
• May I ask whether you used some special trick in catching the similarity in $(x+y)^2$ and the 10th degree polynomial, and similarly for the other? Even the expansion of $(c+s)^{10}$ seems hard to be simplified to the desired form.
– ACB
Commented Dec 27, 2021 at 17:44
• @ACB. By analogy with what you did, I thought that we had to consider (x+y)2 and (x−y)2. Using the last parametrization, it is not so hard ! Take WA and type $$\text{FullSimplify[(Cos[t]*(5 - 4*Cos[t]^4) + Sin[t]*(5 - 4*Sin[t]^4))^2]}$$ Commented Dec 28, 2021 at 3:05