(a) Find a rectangular equation of the plane curve whose parametric equations are $$x( t) =\cos ( 2t) \qquad y( t) =\sin t\quad -\dfrac{\pi }{2}\leq t\leq \dfrac{\pi }{2}$$

(b) Graph the rectangular equation.

(c) Determine the restrictions on $x$ and $y$ so the graph corresponding to the rectangular equation is identical to the plane curve described by $$x=x( t)\qquad y=y( t)\quad -\dfrac{\pi }{2} \leq t\leq\dfrac{\pi}{2}$$

(d) Graph the plane curve whose parametric equations are $$x=\cos ( 2t)\qquad y=\sin t\quad -\dfrac{\pi }{2}\leq t\leq \dfrac{\pi }{2}$$

Figure 6

Solution (a) To eliminate the parameter $t$, we use a trigonometric identity that involves $\sin t$ and $\cos ( 2t)$, namely, $\sin ^{2}t=\dfrac{1-\cos ( 2t) }{2}$. Then $$\begin{equation*} y^{2}\underset{\color{#0066A7}{\underset{\hbox{\(y( t) =\sin t\)}}{{\uparrow }}}}{=}\sin ^{2}t=\dfrac{1-\cos (2t) }{2}\underset{\color{#0066A7}{\underset{\hbox{\(x( t)=\cos ( 2t)\)}}{{\uparrow }}}}{=}\dfrac{1-x}{2}\\ \end{equation*}$$

(b) The curve represented by the rectangular equation $y^{2}=\dfrac{1-x}{2}$ is the parabola shown in Figure 6(a).

(c) The plane curve represented by the parametric equations does not include all the points on the parabola. Since $x( t) =\cos ( 2t)$ and $-1\leq \cos ( 2t) \leq 1$, then $-1\leq x\leq 1$. Also, since $y( t) =\sin t$, then $-1\leq y\leq 1$. Finally, the curve is traced out exactly once in the counterclockwise direction from the point $( -1,-1)$ (when $t=-\dfrac{\pi }{2}$) to the point $( -1,1)$ (when $t=\dfrac{\pi }{2}$).

(d) The plane curve represented by the given parametric equations is the part of the parabola shown in Figure 6(b).