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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.

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    $\begingroup$ It is $\cos(t)=1-y$. $\endgroup$
    – Jon
    Jun 7, 2017 at 14:32
  • $\begingroup$ 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 $-1$. $\endgroup$
    – Michael Burr
    Jun 7, 2017 at 14:36

3 Answers 3

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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?).

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we have $$x=\sin(t)$$ and $$1-y=\cos(t)$$ squaring both sides we get $$(1-y)^2+x^2=1$$

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Hint:

from $ y=1-\cos t$ we have $\cos t=(1-y)$ so ... yes, you have a mistake !

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  • $\begingroup$ It's worth mentioning $$\sin^2(t) + \cos^2(t) = 1$$ $\endgroup$
    – Zubin Mukerjee
    Jun 7, 2017 at 14:33
  • $\begingroup$ This was just noted in the OP :) $\endgroup$
    – Emilio Novati
    Jun 7, 2017 at 14:36

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