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: 8


$\cosh(-x)= \cosh x$

Work Step by Step

Since, we know $\cosh(x) = \dfrac{e^{x} + e^{-x}}{2}$ Thus, $\cosh(-x) =\dfrac{e^{-x} + e^{-(-x)}}{2}$ or, $ =\dfrac{e^{-x} + e^{x}}{2}$ or, $= \cosh x$ Hence, it has been proved that $\cosh(-x)= \cosh x$
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