# Alternative cartesian equation for epicycloid (quatrefoil)

In the problem:

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

it is stated in a previous answer that the resultant is

$$x^{10}+5 x^8 y^2+10 x^6 x^4 + 10 x^4 y^6+5 x^2 y^8+y^{10}-705 x^8+12180 x^6 y^2 -24230 x^4 y^4+12180 x^2 y^6-705 y^8+122560 x^6-112320 x^4 y^2 -112320 x^2 y^4 +122560 y^6+599040 x^4-1361920 x^2 y^2 +599040 y^4+327680 x^2+327680 y^2-1048576=0$$

(generated via Mathematica)

which ultimately turned out to be

$$\boldsymbol{(x+y)^{2/5}+(x-y)^{2/5}=2}$$

that was accomplished by the parametrization, $$x=\cos(\theta)\,(5-4\cos^4(\theta)),$$ $$y=\sin(\theta)\,(5-4\sin^4(\theta)).$$

That represents the cartesian equation of the evolute of an astroid.(?)

With the victory over that problem (thanks to @Claude Leibovici), I was attracted by another similar problem: Eliminate $\theta$ from $4x=5\cos\theta -\cos 5\theta$ and $4y=5\sin\theta -\sin 5\theta$, where the answer is the equation of an epicycloid.

In that question, one can find that $$x=\cos^3(\theta)\,(5-4\cos^2(\theta))$$ $$y=\sin^3(\theta)\,(5-4\sin^2(\theta))$$

which seems almost similar to the aforementioned approach.

Also as mentioned in one answer, the eliminant is $$-81 - 45 x^2 + 365 x^4 - 15 x^6 - 480 x^8 + 256 x^{10} - 45 y^2 - 2395 x^2 y^2 - 45 x^4 y^2 - 1920 x^6 y^2 + 1280 x^8 y^2 + 365 y^4 - 45 x^2 y^4 - 2880 x^4 y^4 + 2560 x^6 y^4 - 15 y^6 - 1920 x^2 y^6 + 2560 x^4 y^6 - 480 y^8 + 1280 x^2 y^8 + 256 y^{10}=0$$

(generated via Wolfram|Alpha)

So because of the similarity, there may be a neat solution as before.

I tried plugging some possible equations in Wolfram|Alpha and Desmos graphing calculator, seeking for a hint. Yet, I haven't found the exit.

Question. Can there be a nice equation for this curve?

If it helps, here's a table showing the coefficients of each term of each polynomial (for the ease of comparison).

$$(x+y)^{2/5}+(x-y)^{2/5}=2$$ unknown
$$x^{10},y^{10}$$ 1 256
$$x^8,y^8$$ -705 -480
$$x^6,y^6$$ 122560 -15
$$x^4,y^4$$ 599040 365
$$x^2,y^2$$ 327680 -45
$$x^8y^2,x^2y^8$$ 5 1280
$$x^6y^4,x^4y^6$$ 10 2560
$$x^6y^2,x^2y^6$$ 12108 -1920
$$x^4y^2,x^2y^4$$ -112320 -45
$$x^2y^2$$ -1361920 -2395
$$x^4y^4$$ -24230 -2880
constant -1048576 -81

Curves:

The red curve is the hypocycloid and the other is the epicycloid.

The symmetry of these curves is probably implying that they have well formed equations too.

If we have $$x=\cos^3(\theta)\,(5-4\cos^2(\theta))$$ $$y=\sin^3(\theta)\,(5-4\sin^2(\theta))$$ then, using $$S=(x^2+y^2)$$ $$S=\frac{1}{8} (13-5 \cos (4 \theta))\implies \theta_\pm=\pm\frac 14 \cos ^{-1}\left(\frac{13-8S}{5} \right)$$