Problem:
Solution:
The key trigonometry identity to use is the double angle formula:
$ \begin{eqnarray*} \cos(2t) &=& 2\cos^2(t) - 1 \\ y &=& 2x^2 - 1 \end{eqnarray*} $
This is a parabola with vertex $ (0, -1) $, it cuts the x-axis as $ (-\frac{1}{\sqrt{2}}, 0) $ and $ (\frac{1}{\sqrt{2}}, 0) $.
As $ t \in [-\pi, \pi ] $, $ x \in [-1, 1] $, $ y \in [-1, 1] $.
Solution:
The key trigonometry identity to use is the double angle formula:
$ \begin{eqnarray*} \cos(2t) &=& 2\cos^2(t) - 1 \\ y &=& 2x^2 - 1 \end{eqnarray*} $
This is a parabola with vertex $ (0, -1) $, it cuts the x-axis as $ (-\frac{1}{\sqrt{2}}, 0) $ and $ (\frac{1}{\sqrt{2}}, 0) $.
As $ t \in [-\pi, \pi ] $, $ x \in [-1, 1] $, $ y \in [-1, 1] $.
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