Trajectory of $X_1(t) = (2\cos^2 x, \cos t \sin t)\;\; (0\le t \le \pi)$

Question

I need to draw trajectory of $X_1(t) = (2\cos^2 t, \cos t \sin t)\;\; (0\le t \le \pi)$

If one let $2\cos^2 x = x, \cos t \sin t =y$ then

$dx/dt = -4\cos t\sin t =-4y$ and one could derive x = -4yt+C.

However, with this approach doesn't make the t disappear so that one could get the pure relation of x and y.

any hint/adice ?

Answer 1

We don't have to differentiate.

$x=2\cos^2t$ and $y=\cos t\sin t$.

\begin{align} y^2&=\cos^2t\sin^2t\\ &=\cos^2t(1-\cos^2t)\\ &=\frac{x}{2}\left(1-\frac{x}{2}\right) \end{align}