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: 8


See the proof below.

Work Step by Step

Using the addition formulas, we have $$\sinh 2x=\sinh (x+x)=\sinh x\cosh x+\sinh x\cosh x=2\sinh x\cosh x.$$ $$\cosh 2x=\cosh(x+x)=\cosh x\cosh x+\sinh x\sinh x=\cosh^2x+\sinh^2x.$$
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