# Eliminate the parameter

Eliminate the parameter to find a Cartesian equation. \begin{align}x&=\sin(t)\\ y&=1-\cos(t)\\ 0\leq t& \leq \pi \end{align}

So I need to get into the form of $x^{2}+y^{2} =1$

$\cos$ needs to be isolated so: \begin{align}y&=1-\cos(t)\\ y-1&=-\cos(t)\end{align} I think this is where I made a mistake $$y+1=\cos(t)$$ and and have the incorrect equation $$x^{2}+(y+1)^{2} =1$$ just not sure the proper equation and where I went wrong.

• It is $\cos(t)=1-y$. – Jon Jun 7 '17 at 14:32
• To get rid of the negative sign on $\cos$, you need to multiply through by $-1$, remember to distribute the $-1$ on the left hand side to both the $y$ and the $-1$. – Michael Burr Jun 7 '17 at 14:36

I think this is where I made a mistake $$y+1=\cos(t)$$
Careful! If $y = 1-\cos(t)$, then $\cos t = 1-y$; so: $$\sin^2t +\cos^2t = 1 \implies x^2+\left( 1-y \right)^2= 1$$ Since $\left( 1-y \right)^2=\left( y-1 \right)^2$, this is also: $$x^2+\left( y-1\right)^2= 1$$ This is a Cartesian equation of a circle with radius $1$ and center in $(0,1)$, but note that the parameter $t$ goes from $0$ to $\pi$ so it's only half of the circle (which half?).
we have $$x=\sin(t)$$ and $$1-y=\cos(t)$$ squaring both sides we get $$(1-y)^2+x^2=1$$
from $y=1-\cos t$ we have $\cos t=(1-y)$ so ... yes, you have a mistake !
• It's worth mentioning $$\sin^2(t) + \cos^2(t) = 1$$ – Zubin Mukerjee Jun 7 '17 at 14:33