Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 3 - Section 3.11 - Hyperbolic Functions - 3.11 Exercises - Page 266: 16

Answer

$cosh(x+y) = cosh~x~cosh~y+ sinh~x~sinh~y$

Work Step by Step

$cosh(x+y) = \frac{e^{x+y}+e^{-(x+y)}}{2}$ $cosh(x+y) = \frac{2e^{x+y}+2e^{-(x+y)}}{4}$ $cosh(x+y) = \frac{e^{x+y}+e^{-(x+y)}+e^{x+y}+e^{-(x+y)}}{4}$ $cosh(x+y) = \frac{e^{x+y}+e^{y-x}+e^{x-y}+e^{-(x+y)}+e^{x+y}-e^{y-x}-e^{x-y}+e^{-(x+y)}}{4}$ $cosh(x+y) = \frac{e^{x+y}+e^{y-x}+e^{x-y}+e^{-(x+y)}}{4}+ \frac{e^{x+y}-e^{y-x}-e^{x-y}+e^{-(x+y)}}{4}$ $cosh(x+y) = \frac{(e^x+e^{-x})~(e^{y}+e^{-y})}{4}+ \frac{(e^x-e^{-x})~(e^{y}-e^{-y})}{4}$ $cosh(x+y) = \frac{e^x+e^{-x}}{2}\cdot~\frac{e^{y}+e^{-y}}{2}+ \frac{e^x-e^{-x}}{2}\cdot \frac{e^{y}-e^{-y}}{2}$ $cosh(x+y) = cosh~x~cosh~y+ sinh~x~sinh~y$
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