# Convert advanced parametric equation to regular/cartesian

can anybody help me to convert following parametric equation in a form Y =Y(X): $$x = cos(t) \sqrt{(2 - cos^2(3t))} \\ y = sin(t) \sqrt{(2 - cos^2(3t))}$$

I've tried also with Wolfram Alpha and it seem not to work:

Reduce[x == Cos[t] Sqrt[2 - Cos[3 t]^2], {t}]


Actually what I need is a function in a form Y = Y(X) to get this kind a result: http://postimg.org/image/6n8f839yp/

Best! Vito

• Anyone guys? Using half angle substitution and other trigonometry formula I still cannot do that... – Vito Oct 9 '15 at 12:19

I would say

$x^2+y^2=(\cos^2t+\sin^2t) (2 - \cos^2(3t)) = 2 - \cos^2(3t)=1+\sin^2(3t)$

$\Rightarrow$equations in polar coordinates:

$r = \sqrt{1+ \sin^2(3t)}=\cdots=\sqrt{1+ \sin^2t(3 \cos^2t-sin^2t)^2}$

and transformation from polar into Cartesian coordinates:

$\displaystyle \,\cos t = \frac{x}{r},\,\, \sin t= \frac{y}{r},\,\,r=\sqrt{x^2+y^2}$

$\displaystyle \Rightarrow r^2=1+\frac{y^2}{r^2}\left(\frac{3x^2}{r^2}-\frac{y^2}{r^2}\right)^2\Rightarrow \cdots \Rightarrow r^8 - r^6 - y^2 (3 x^2 - y^2)^2 = 0$

$\Rightarrow \,\,$Cartesian:

$(x^2+y^2)^4 - (x^2+y^2)^3 -y^2 (3 x^2 - y^2)^2=0$

$x^8 + 4x^6 y^2 - x^6 + 6x^4 y^4 - 12x^4 y^2 + 4x^2 y^6 + 3x^2 y^4 + y^8 - 2y^6 = 0$

See: Wolfram

• Thanks, that's very nice! Just few days before I figured out the translation, the problem was I've never tried the (x^2 + y^2). Using cos(3t) = ... and (y^2/x^2) = ... finally I get (x^2+y^2)^4 - 2(x^2+y^2)^3 + x^2(x^2-3y^2)^2 = 0 which if you extend is the same result as you got. (: – Vito Oct 20 '15 at 13:38