How can I solve:
$
x = 16 \sin^3(t) \\
y = 13\cos(t) - 5\cos(2t) - 2\cos(3t) - \cos(4t)
$
I've derived $t = arcsin(\frac{x^\frac{1}{3}}{16^\frac{1}{3}})$ from the first equation but I am still unsure as to whether or not this is correct.
I believe I need to substitute the $t = arcsin(\frac{x^\frac{1}{3}}{16^\frac{1}{3}})$ into the y= ... equation, however when I do this, it does not produce the same graph as the parametric:
$y= 13cos(arcsin(\frac{x^\frac{1}{3}}{16^\frac{1}{3}})) - 5\cos(2arcsin(\frac{x^\frac{1}{3}}{16^\frac{1}{3}})) - 2\cos(3arcsin(\frac{x^\frac{1}{3}}{16^\frac{1}{3}})) - \cos(4arcsin(\frac{x^\frac{1}{3}}{16^\frac{1}{3}}))$
The above produces this graph, whereas the original parametric produces this graph.
Hint:
Let $ t= \arcsin (\theta)$. Then $\sin (t) = \theta$. Recall that $ \sin^2 (t) + \cos^2 (t) = 1 $ (which can be conveniently visualised as a right triangle with hypotenuse of length $1$).
Combining those equivations should allow you to solve for $\cos (t) = \cos (\arcsin(\theta))$. That should be enough to find $y=f(x)$, once you also apply the relevant double and triple angle identities for cosine.