How to write $x=2\cos(3t) y=3\sin(2t)$ in rectangular coordinates?

Question

How would I write the following in terms of $x$ and $y$? I think I use the inverse $\cos$ or $\sin$?

$$x=2\cos(3t)\,, \quad y=3\sin(2t)$$

Answer 1

Hint:

$$ x=2\cos(3t) = 2\cos(2t+t)= 2\cos (2t)\, \cos t - 2 \sin (2t) \,\sin t $$

$$ \implies x = 2\cos (2t)\, \cos t - 2 \sin (2t) \,\sin t\longrightarrow (*). $$

Added: To finish the problem you need to find $\cos(2t),\sin t$, and $\cos t$ in $(*)$. For $\cos (2t)$ we have

$$ \sin(2t)=\frac{y}{3}\implies \cos (2t) = \sqrt{1-\sin^2(2t)}= \sqrt{1-\frac{y^2}{9}}. $$

For $\cos t$ and $\sin t$, use the identities

$$ \cos(2t)=2\cos^2 t-1,\quad \cos(2t)=1-2\sin^2 t. $$

Answer 2

This is not an answer (with apologies to Magritte)...

The point is that there is no functional relationship between $x $ and $y$.

Answer 3

We have $\cos(3t)=\frac{x}{2}$. Using the identity $\cos 2u=2\cos^2(u)-1$, we get $$\cos(6t)=2\left(\frac{x}{2}\right)^2-1.\tag{1}$$

We also have $\sin(2t)=\frac{y}{3}$. Using the not so well-known identity $\sin(3u)=3\sin u-4\sin^3 u$, we get $$\sin(6t)=3\left(\frac{y}{3}\right)-4\left(\frac{y}{3}\right)^3.\tag{2}$$ Now use (1) and (2), and the fact that $\cos^2(6t)+\sin^2(6t)=1$ to eliminate $t$. We get the remarkably ugly equation $$\left( 2\left(\frac{x}{2}\right)^2-1 \right)^2+\left( 3\left(\frac{y}{3}\right)-4\left(\frac{y}{3}\right)^3 \right)^2=1.$$