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: 16

Answer

$cosh~2x = cosh^2~x+sinh^2~x$

Work Step by Step

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