Finding polynomial order of parametric curve with trig functions

79 Views Asked by At

I was given a parametric curve defined by: \begin{align} x(t) &= \cos(t) \\ y(t) &= \cos(2^kt) \end{align} And asked:

For any natural number k, eliminating the parameter gives a polynomial of what order?

When I eliminate the parameter, I end up with something that looks like $$y = \cos(2^k\arccos(x)),$$ which seems correct. But this isn’t a polynomial and doesn’t have an order. Am I making a mistake? Did I misunderstand the question or the definition of a polynomial?

1

There are 1 best solutions below

0
On

A heavy hint:
The comment of @Macavity has it all; let me expand on what they say.

First, look at the simplest case, $k=1$, when the equations are $x=\cos t$, $y=\cos 2t=\cos^2t-\sin^2t=2\cos^2t-1=2x^2-1$. So there you are for the first case.

And $k=2$ asks about $x=\cos t$, $y=\cos4t=\cos\bigl(2(2t)\bigr)=2\cos^2(2t)-1=2(2\cos^2t-1)^2=8\cos^4t-8\cos^2t+2-1\\=8x^4 -8x^2+1$.

Now you see the inductive process.