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: 75


$y = \frac{3}{5}~\sqrt{x^2+25}$, where $x$ in $(-\infty, \infty)$

Work Step by Step

$x = 5~tan~t$ $x^2 = 25~tan^2~t$ $x^2 = 25~(sec^2~t-1)$ $sec^2~t = \frac{x^2}{25}+1$ $sec~t = \sqrt{\frac{x^2+25}{25}}$ We can replace this expression in the equation with $y$: $y = 3~sec~t$ $y = 3~\sqrt{\frac{x^2+25}{25}}$ $y = \frac{3}{5}~\sqrt{x^2+25}$ Since $t$ in $(-\frac{\pi}{2}, \frac{\pi}{2})$, then $x$ in $(-\infty, \infty)$ $y = \frac{3}{5}~\sqrt{x^2+25}$, where $x$ in $(-\infty, \infty)$
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