Answer
$x^{2}+(1-y)^{2}=1$,
See image:
Work Step by Step
Build a table of coordinates (x,y)
$x=f(t)=\sin t,\quad \qquad y=g(t)=1-\cos t$
Plot the points and join with a smooth curve.
Taking the initial t to be the first point in the table, track the direction in which the points "travel" as t increases.
From $ x=\sin t, 0\leq t\leq 2\pi\Rightarrow-1\leq x\leq 1, $
$x^{2}=\sin^{2}t$
From $y=1-\cos t,\quad 0\leq t\leq 2\pi\Rightarrow 0\leq y\leq 2.$
$(1-y)^{2}=\cos^{2}t$
Since $\sin^{2}t+\cos^{2}t=1$ substituting, we get
$x^{2}+(1-y)^{2}=1$,
a circle of radius 1 centered at (0,1)
