Answer
$x^{2}+y^{2}=1, \ y\geq 0$
Work Step by Step
$\left\{\begin{array}{l}
x=\cos(\pi-t)\\
y=\sin(\pi-t)
\end{array}\right\}\Rightarrow\quad\left[\begin{array}{lll}
t & x & y\\
0 & -1 & 0\\
\pi/6 & -1/2 & \sqrt{3}/2\\
\pi/4 & \sqrt{2}/2 & \sqrt{2}/2\\
\pi/2 & 1 & 0\\
3\pi/4 & \sqrt{2}/2 & \sqrt{2}/2\\
\pi & -1 & 0
\end{array}\right]\Rightarrow\quad y\geq 0$
Square both parametric equations: $\left\{\begin{array}{l}
x^{2}=\cos^{2}(\pi-t)\\
y^{2}=\sin^{2}(\pi-t)
\end{array}\right.$
Add the two equations:
$ \fbox{$ x^{2}+y^{2}=1 $}, y\geq 0$
To graph, create a table using several values for $t$, and then calculate $(x(t), y(t)),$ plotting the points as you go. Join with a smooth curve, noting the direction in which $t$ increases.
