how can I solve this type of problem?

Question

Given the following function $$\psi: \begin{cases} \mathbb{R} \rightarrow \mathbb{R}^2 \\ t \mapsto (2\cos(t-\frac{\pi}{2}),\sin(2(t-\frac{\pi}{2}))) \end{cases}$$ show that $$M = \psi(\mathbb{R}) = \bigl\{ (x,y)\in \mathbb{R}^2: x^4 -4x^2+ 4y^2 = 0 \bigr\} $$

Answer 1

Step 1: Take $x(t)=2\cos(t-\pi/2)=2\sin t$ and $y(t)=\sin(2t-\pi)=-\sin 2t$.

Now verify that $x(t)^4-4x(t)^2+4y(t)^2=0$.

$$4(x(t)^2-y(t)^2)=4(4\sin^2t-\sin^22t)=4(4\sin^2t-4\sin^2t\cos^2t)=16\sin^4t$$

This gives $\psi(\mathbb R)\subseteq S=\{(x,y)\in\Bbb R^2:x^4-4x^2+4y^2=0\}$.

---

Step 2: Now we need to show $\psi$ is onto over $S$.

Let $(X,Y)\in S$, then $Y=\pm X\sqrt{1-X^2/4}$ and $X\in[-2,2]$. So $X/2\in[-1,1]$.

Take $t=\arcsin(X/2)\in\Bbb R$ and verify that $x(t)=X,y(t)=-X\sqrt{1-X^2/4}$.

Take $t=\pi-\arcsin(X/2)\in\Bbb R$ and verify that $x(t)=X,y(t)=X\sqrt{1-X^2/4}$.

Hint: Use the fact that $\cos(\arcsin(z))=\sqrt{1-z^2}$.