Answer
We use trigonometry identities:
$\cos 2t = {\cos ^2}t - {\sin ^2}t$
${\cos ^2}t = 1 - {\sin ^2}t$
to show that the projection of the curve onto the $xy$-plane is
$z = x - 2{x^3}$ for $ - 1 \le x \le 1$.
Work Step by Step
We have the parametrization of the curve given by ${\bf{r}}\left( t \right) = \left( {\sin t,\cos t,\sin t\cos 2t} \right)$. Thus, the coordinate functions are
$x = \sin t$, ${\ \ }$ $y = \cos t$, ${\ \ }$ $z = \sin t\cos 2t$
From trigonometry we have the identity: $\cos 2t = {\cos ^2}t - {\sin ^2}t$. Thus,
$z = \sin t\cos 2t = \sin t\left( {{{\cos }^2}t - {{\sin }^2}t} \right)$
$z = \sin t{\cos ^2}t - {\sin ^3}t$
Since ${\cos ^2}t = 1 - {\sin ^2}t$, it follows that
$z = \sin t\left( {1 - {{\sin }^2}t} \right) - {\sin ^3}t$
$z = \sin t - 2{\sin ^3}t$
Since $x = \sin t$ and $ - 1 \le \sin t \le 1$, therefore $z = x - 2{x^3}$ for $ - 1 \le x \le 1$.
Since the equation does not involve $y$, it is a projection of the curve onto the $xz$-plane.
