Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 7 - Exponential Functions - 7.9 Hyperbolic Functions - Exercises - Page 384: 7


See the proof below.

Work Step by Step

Using the definition of $\cosh x$, we have the following \begin{aligned} \cosh (x+y) &=\frac{e^{x+y}+e^{-(x+y)}}{2}=\frac{2 e^{x+y}+2 e^{-(x+y)}}{4} \\ &=\frac{e^{x+y}+e^{-x+y}+e^{x-y}+e^{-(x+y)}}{4}+\frac{e^{x+y}-e^{-x+y}-e^{x-y}+e^{-(x+y)}}{4} \\ &=\left(\frac{e^{x}+e^{-x}}{2}\right)\left(\frac{e^{y}+e^{-y}}{2}\right)+\left(\frac{e^{x}-e^{-x}}{2}\right)\left(\frac{e^{y}-e^{-y}}{2}\right) \\ &=\cosh x \cosh y+\sinh x \sinh y \end{aligned} which completes the proof.
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