# Solving Parametric Equation: Multiple coefficients of trigonomic functions

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.

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Not entirely sure what you want, but if you're looking for $y=f(x)$, you can generally find $\cos ( \arcsin ( \theta ))$ by using the pythagorean theorem (draw the relevant right triangle if you're confused). – Yun William Yu Feb 19 '12 at 16:00
What triangle? And I want the equation in terms of y = f(x) yes – Kian Feb 19 '12 at 18:01

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.
There are two reasons you're not getting the same graph: (1) you typed it in wrong; make sure you put $(x/16)$ in parentheses, or else exponentiation takes precedence, and (2), even if you do that, arcsin is only the inverse of sin from $-1$ to $1$, so you'll only get a small portion of the graph you want. – Yun William Yu Feb 19 '12 at 22:07