# Converting cardioid from parametric form into Cartesian form

I'm trying to find the cartesian equation of the curve defined by the parametric equations:

$$x=2\cos(t) - \cos(2t), \qquad y=2\sin(t) - \sin(2t)$$

I feel stumped. How can I go about this?

## 4 Answers

Another way:

we have $x=2\cos t-\cos2t\ \ \ \ (1)$

$\implies x=2\cos t-(2\cos^2t-1)\iff x-1=2\cos t(1-\cos t)$

and $y=2\sin t-\sin2t\ \ \ \ (2)$

$\implies y=2\sin t -2\sin t\cos t=2\sin t(1-\cos t)$

On division, $\displaystyle\frac yx=\tan t$

$\displaystyle\implies \cos2t=\frac{1-\tan^2t}{1+\tan^2t}=\frac{x^2-y^2}{x^2+y^2}$

and $\displaystyle\cos t=\frac1{\sec t}=\pm\frac1{\sqrt{1+\tan^2t}}=\frac{\pm x}{\sqrt{x^2+y^2}}$

Put these values in $(1)$ and square to remove $\pm$

HINT:

Squaring & adding we get $$x^2+y^2=(2\cos t-\cos2t)^2+(2\sin t-\sin2t)^2$$

$$=4(\cos^2t+\sin^2t)+(\cos^22t+\sin^22t)-4(\cos2t\cos t+\sin2t\sin t)$$ $$=4+1-4(\cos2t\cos t+\sin2t\sin t)$$

$$=5-4\cos(2t-t) (\text{ using }\cos(A-B)=\cos A\cos B+\sin A\sin B)$$

$$\implies x^2+y^2=5-4\cos t$$

Put the value of $\cos t$ in $x=2\cos t-\cos2t=2\cos t-(2\cos^2t-1)$

• I don't understand this: x^2+y^2=4+1−4(cos2tcost+sin2tsint) x^2+y^2=1 - how is the sums formula tied in? Thanks – nio Sep 14 '13 at 17:22
• @nio, please find the edited version – lab bhattacharjee Sep 14 '13 at 18:20

The Maple command eliminate({x = 2*cos(t)-cos(2*t), y = 2*sin(t)-sin(2*t)}, t) outputs, in particular, $$\left\{ \sqrt {2\,\sqrt {3-2\,x}+2\,x}\sqrt {3-2\,x}+\sqrt {2\,\sqrt {3-2\,x}+2\,x}-2\,y \right\}$$ and $$-\sqrt {-2\,\sqrt {3-2\,x}+2\,x}\sqrt {3-2\,x}+\sqrt {-2\,\sqrt {3-2\, x}+2\,x}-2\,y .$$ Up to Maple syntax (see ?eliminate for info), these expressions equal zero. I think the squarings can be produced with Maple too.

PS. Also see Wiki concerning the elimination of $t$ with complex numbers.

The Cartesian equation, in x and y, equivalent to the two given pair of parametric equations is the following order-6 determinant, set equal to zero: $\begin{vmatrix} 2&{-2}&{x-1}&0&0&0\\ 0&2&{-2}&{x-1}&0&0\\ 0&0&2&{-2}&{x-1}&0\\ 0&0&0&2&{-2}&{x-1}\\ 4&{-8}&0&8&{y^2-1}&0\\ 0&4&{-8}&0&8&{y^2-1} \end{vmatrix}$

This determinant is Sylvester's dialytic eliminant with the coëfficients of the Cartesian forms of both of the two given parametric equations plugged in. Donn S. Miller dsm5442@gmail.com