Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.11 - Hyperbolic Functions - 3.11 Exercises - Page 264: 8


cosh(-x) = cosh(x)

Work Step by Step

Use the definition of the hyperbolic cosine. Then, find $\cosh{(-x)}$ and re-arrange to arrive at $\cosh x$ $$cosh(x) = \frac{e^{x} + e^{-x}}{2}$$ cosh(-x) = $\frac{e^{-x} + e^{-(-x)}}{2}$ = $\frac{e^{-x} + e^{x}}{2}$ = cosh(x)
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