Cartesian equation for $\pi \le t \le 2\pi$

Question

The graph for: $x=\cos{t}, y=\sin{2t} $

Looks like a figure 8. I've found the Cartesian equation for the line by doing: $$ t = \arccos{x} \rightarrow y = \sin{2\arccos{x}} = 2x \cdot \sin{\arccos{x}} = 2x\sqrt{1-x^2}$$

A later part of the question asks for the equation for the part of the line where $\pi \le t \le 2\pi$

Wouldn't this be the same equation?

The question I solved asked to prove the cartesian equation for the line where $0 \le t \le 2\pi$ was $ 2x\sqrt{1-x^2}$

So would I rework my solution to this replacing $t= \arccos{x}$ with $2\pi-t = \arccos{x}$

getting: $$ t = 2\pi - \arccos{x} \rightarrow\\ y = \sin{(2\cdot(2\pi - \arccos{x}))} \\=\sin(4\pi-2\arccos{x})\\= \sin{4\pi}\cos{2\arccos{x}} + \cos{4\pi}\sin{2\arccos{x}}\\=\sin{2\arccos{x}}$$ ... = same answer.

How would I solve for the new limit for t?

Answer 1

You could solve the problem in this way:

\begin{equation} \begin{cases} x=\cos t\\y=\sin2t \end{cases} \begin{cases} x^2=\cos^2t\\y^2=4\sin^2t\cos^2t=4(1-\cos^2t)\cos^2t=4x^2(1-x^2) \end{cases} \end{equation}

So we have $y^2=4x^2(1-x^2)\longrightarrow y=\pm2|x|\sqrt{1-x^2}$ for $-1\le x\le1$.