Answer
See the proof below.
Work Step by Step
Using the addition formulas, we have
$$\sinh 2x=\sinh (x+x)=\sinh x\cosh x+\sinh x\cosh x=2\sinh x\cosh x.$$
$$\cosh 2x=\cosh(x+x)=\cosh x\cosh x+\sinh x\sinh x=\cosh^2x+\sinh^2x.$$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
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
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