Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - 6.7 Hyperbolic Functions - 6.7 Exercises - Page 489: 21

Answer

$\sinh x= \dfrac{4}{3}$

Work Step by Step

As we are given that $ x \gt 0$, so $\sinh x = \sqrt{ (cosh^{2}x - 1)}$ Plug $\cosh x = \dfrac{5}{3}$ Thus, $\sinh x =\sqrt{ (\frac{5}{3}^{2}-1)}$ or, $=\sqrt {\dfrac{25}{9}-1}$ or, $=\sqrt{ \dfrac{25-9}{9}}$ or, $=\sqrt {\dfrac{16}{9}} $ or, $= \dfrac{4}{3}$
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