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?
Best Answer
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.