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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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    What triangle? And I want the equation in terms of y = f(x) yes2012-02-19

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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.

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    Using the two equations to eliminate $t$ and arrive at a relation between $x$ and $y$ would require the usage of $Arcsin$ one-to-many relation instead of $arcsin$ one-to-one function. You need to choose values for x in the range $\{-\pi,\pi\}$ that satisfy the $Arcsin$ relation. That should do the trick2015-09-09