Given the two equations $x(t) = acos(4t)$ and $y(t) = bsin(4t)$, find $\frac{dy}{dx}$ and the Cartesian representation of the curve
I have already easily solved the first bit for $\frac{dy}{dx}$, the second part is a bit trickier. My first approach was to square both equations and write it in the $x^2 + y^2 = r^2$ form, but the constants are different so I thought that'd make $r^2$ a bit tricky to figure out. Then I decided I'd substitute one into the other, so I did as shown:
$$x(t) = acos(4t) \implies cos^{-1}(\frac{x}{a}) = 4t$$ $$\because y(t) = bsin(4t) \implies y(t) = bsin(cos^{-1}(\frac{x}{a}))$$ I wanted to simplify it further, I thought I'd approach this the same way a sine cancels out an inverse sine if it's in its argument and the result simplifies to the inverse sine's argument $$sin(sin^{-1}(x)) = x$$ but I didn't know how I'd do that with an inverse cosine being the argument of a sine. Can this be simplified further? Is there an easier (more direct) way to find the Cartesian representation than just trying to guess?