How would I express the following parametric pair in Cartesian form? $\bigl(4\sin(4t), 3\sin(3t)\bigr)$

Question

Since I know that $\sin(4t) = 4\cos^3t\sin t-4\cos t\sin^3t$ and $\sin(3t) = 3\sin t-4\sin^3 t$ (or at least I think I do), I think I'm halfway there.

Answer 1

$$\dfrac{x}{4} = \sin4t \hspace{1cm}\text{and}\hspace{1cm}\dfrac{y}{3} = \sin 3t\\ \implies \sin12t = 3\left(\dfrac{x}{4}\right)-4\left(\dfrac{x}{4}\right)^3$$ using $\sin(3\theta) = 3\sin\theta-4\sin^3\theta$ and $$\sin 12t = \pm4\left(\dfrac{y}{3}\right)\dfrac{\sqrt{3-y^2}}{3}\left(1-2\left(\dfrac{y^2}{9}\right)\right)$$ using $\sin(4\theta) = 4\sin\theta\cos\theta(1-2\sin^2\theta)$

Now, equate the square of the two expressions and get the desired equation (need to square to avoid sign ambiguity of square root in the second equation).