convert the parametric equation $x = \cos (t) ; y = \sin (t)-t^2$ to rectangular form.

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I'm trying to convert the parametric equation $x = \cos (t) ; y = \sin (t)-t^2$ to rectangular form but have gotten stuck trying to eliminate the last $t$ from parametric form. I have gotten as far as: $y = \sqrt(1-x^2) - \arccos(x)^2$.

I know that $t = \arccos(x)$ but I have no idea how to square that quantity.

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It becomes more complicated:

$x=\cos t ⇒ t= Arc\ cos x $

$\sin t = \sqrt{1 - x^2}$

$y^2/25=sin^2 t - (32/5)t^2 \sin t+(256/25) t^4$

$x^2+y^2/25= cos^2t+ sin^2t - (32/5)t^2 \sin t+(256/25) t^4$

$x^2+y^2/25=1-(32/5)(Arc \cos x)^2 \sqrt{1-x^2}+(256/25)(Arc\cos x)^4$