## Precalculus (6th Edition) Blitzer

Consider the provided function, $x=3\sin t$, $y=4\cos t+2$ Consider the first function, $x=3\sin t$ The interval is given from $t=0$ to $t=2\pi$. Therefore, put values of $t$ in the function as $0\le t\le 2\pi$ Consider second function, $y=4\cos t+2$ The interval is given from $t=0$ to $t=2\pi$. Therefore, put values of $t$ in the function as $0\le t\le 2\pi$ For, $t=\frac{\pi }{2}$ \begin{align} & x=3\sin t \\ & =3\cdot \sin \frac{\pi }{2} \\ & =3 \end{align} And, \begin{align} & y=4\cos \left( t+2 \right) \\ & =4\cos \left( \frac{\pi }{2}+2 \right) \\ & =2 \end{align} For, $t=\pi$ \begin{align} & x=3\sin t \\ & =3\cdot \sin \pi \\ & =0 \end{align} And, \begin{align} & y=4\cos \left( t+2 \right) \\ & =4\cos \left( \pi +2 \right) \\ & =-2 \end{align} For, $t=\frac{3\pi }{2}$ \begin{align} & x=3\sin t \\ & =3\cdot \sin \frac{3\pi }{2} \\ & =-3 \end{align} And, \begin{align} & y=4\cos \left( t+2 \right) \\ & =4\cos \left( \frac{3\pi }{2}+2 \right) \\ & =2 \end{align} For, $t=0\text{ or 2}\pi$ \begin{align} & x=3\sin t \\ & =3\cdot \sin 0 \\ & =0 \end{align} And, \begin{align} & y=4\cos \left( t+2 \right) \\ & =4\cos \left( 0+2 \right) \\ & =6 \end{align}