Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 8 - Review Exercises - Page 411: 77


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

Work Step by Step

$y = sin~t$ $y^2 = sin^2~t$ 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 $[-1,1]$
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