(vi) (−3,−2π3)(-3,-\dfrac{2\pi}{3})(−3,−32π) Here r=−3<0r=-3<0r=−3<0 and θ=−2π3\theta=-\dfrac{2\pi}{3}θ=−32π. Since r<0r<0r<0, use angle: −2π3+π=π3-\frac{2\pi}{3}+\pi=\frac{\pi}{3}−32π+π=3π x=−3cos(−2π3)=32y=−3sin(−2π3)=332\begin{aligned} x&=-3\cos\left(-\frac{2\pi}{3}\right)=\frac{3}{2}\\ y&=-3\sin\left(-\frac{2\pi}{3}\right)=\frac{3\sqrt{3}}{2} \end{aligned}xy=−3cos(−32π)=23=−3sin(−32π)=233