Chapter 8 - Review Exercises - Page 406: 77

$x = 1- 2y^2$, where $y$ in $[-1,1]$

$y = sin~t$ $y^2 = sin^2~t$ We can replace this value in the other equation: $x = cos~2t$ $x = 1- 2~sin^2~t$ $x = 1- 2y^2$ Since $t$ in $(-\pi, \pi)$, then $y$ in $[-1,1]$ $x = 1- 2y^2$, where $y$ in $[-1,1]$