## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 3 - Section 3.11 - Hyperbolic Functions - 3.11 Exercises: 21

#### Answer

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

#### Work Step by Step

Since x $\gt$ 0, we can say that sinh x = $\sqrt{ (cosh^{2}x - 1)}$ Substitute cosh x = $\frac{5}{3}$ to get sinh x = $\sqrt{ (\frac{5}{3}^{2}-1)}$ = $\sqrt {(\frac{25}{9}-1)}$ = $\sqrt{ (\frac{25-9}{9})}$ = $\sqrt {\frac{16}{9}} = \frac{4}{3}$

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