[tex]\bf (\stackrel{\stackrel{r}{\downarrow }}{-2}~~,~~\stackrel{\stackrel{\theta }{\downarrow }}{2\pi })\qquad \begin{cases} x=&rcos(\theta )\\ &(-2)cos(2\pi )\\ &(-2)(1)\\ &-2 \\\cline{1-2} y=&rsin(\theta )\\ &(-2)sin(2\pi )\\ &(-2)(0)\\ &0 \end{cases}\qquad \implies \qquad (\stackrel{x}{-2}~,~\stackrel{y}{0})[/tex]
Answer:
y = rsen (θ)
So for r = -2 and θ = 2π we have
x = -2cos (2π)
x = -2
y = -2sen (2π)
y = 0
Finally the equivalent point in Cartesian coordinates is the point:
(-2 0)
Step-by-step explanation:
