In Exercises 24-34 sketch the parametric curve by eliminating the parameter.

$x=\sin t,y=\cos 2t,t\in [0,2\pi ]$

The parametric curve $x=\sin t,y=\cos 2t,t\in [0,2\pi ]$

$y=-2x^2+1.$Consider the parametric equations $x=\sin t,y=\cos 2t,t\in [0,2\pi ]$.

The objective is to sketch the parametric curve by eliminating the parameters.

Take the equation for example.$y=\cos 2t.$

$y=1-2{\sin }^{2}t\left[since\cos 2t=1-2{\sin }^{2}t\right]$

Substitute $x=\sin t$in the equation.

Take $y=1-2{\sin }^{2}t$

$y=1-2x^2$By substitution

$y=-2x^2+1$

In order to draw the graph of the equation, assume $x=-1,0,1,2$

Substitute $x=-1$in the equation $y=-2x^2+1.$

Then,

$$\begin{gathered} y=-2(-1{)}^2+1 \\ y=-2+1 \\ y=-1 \\ (x,y)=(-1,-1) \end{gathered}$$

Substitute $x=0$in the equation

Then,

$$\begin{gathered} y=-2(0{)}^2+1 \\ y=0+1 \\ (x,y)=(0,1) \end{gathered}$$

Substitute $x=1$in the equation $y=-2x^2+1$.

Then.

$$\begin{gathered} y=-2(1{)}^2+1 \\ y=-2+1 \\ y=-1 \\ (x,y)=(1,-1) \end{gathered}$$

Substitute $x=2$in the equation $y=-2x^2+1$.

Then,

$$\begin{gathered} y=-2(2{)}^2+1 \\ y=-2(4)+1 \\ y=-7 \\ (x,y)=(2,-7) \end{gathered}$$

The graphical representation by using the points $(-1,-1)(0,1)(1,-1)(2,-7)$is as follows,

Therefore, the required equation after eliminating the parameter is $y=-2x^2+1$