Eliminate the parameter from a lissajous figure

[Guest]

Hi, I have to eliminate the parameter 't' from the following parametric equations for a Lissajous figure:
x=sin(2t)
y=cos(t)

I have to end up with y as a function of x.
I've tried... Somehow, ending up with y=cos(ArcSin(x)/2) doesn't seem right. If someone could just explain to me how I should go about doing it I would be grateful.

 

[galactus]

There's a reason these are mostly in parametric form.

Here's a graph of your bowditch curve(another name for a lissajous)

We can use trig to get it in terms of x:

\(\displaystyle \L\\x=sin(2t)\)

\(\displaystyle \L\\x=2sin(t)cos(t)\)

\(\displaystyle \L\\x^{2}=4sin^{2}(t)cos^{2}(t)\)

\(\displaystyle \L\\x^{2}=4(1-cos^{2}(t))cos^{2}(t)\)

\(\displaystyle \L\\x^{2}=4(1-y^{2})y^{2}\)

\(\displaystyle \L\\4y^{4}-4y^{2}+x^{2}=0\)

\(\displaystyle \L\\y^{2}=\frac{4\pm\sqrt{16-16x^{2}}}{8}\)

\(\displaystyle \L\\=\frac{1\pm\sqrt{1-x^{2}}}{2}\)

\(\displaystyle \L\\y=\pm\sqrt{\frac{1\pm\sqrt{1-x^{2}}}{2}}\)

See why parametrics are used?. This way, you'd have 4 functions to plot in order to arrive at your lissajous.

 

[Guest]

Thank you very much for the solution.