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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.

Kian
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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

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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.

Yun William Yu
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  • How can you explain my answer above? – Kian Feb 19 '12 at 19:39
  • 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
  • Note that the graph of the parametric cannot be a single function of x because it does not pass the vertical line test. – Yun William Yu Feb 19 '12 at 22:07
  • So how do I fix your second point? – Kian Feb 19 '12 at 22:08
  • It's not possible for a single function y=f(x) to have the same graph as your parametric, because the graph of the parametric does not pass the vertical line test. You'll need to graph multiple functions on the same plot to achieve the same effect. – Yun William Yu Feb 19 '12 at 22:13
  • Ahh right, I understand so it would be like $$y=\sqrt(f(x)) \\ y = -\sqrt(f(x))$$ – Kian Feb 19 '12 at 22:21
  • Something along those lines, although it'd be far more complicated. – Yun William Yu Feb 19 '12 at 22:28
  • Yes of course. Thanks for all the help! I'd upvote you if I could – Kian Feb 19 '12 at 22:29
  • No problem. :-) – Yun William Yu Feb 19 '12 at 22:30
  • 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 trick – Deepak Gupta Sep 09 '15 at 09:21