Inverse identity Show that cosh (cosh x) = |x| by using theformula cosht = In(t + Vr - 1) and considering the casesx 2 0 and x < 0.%3D

Given : cosh-1(coshx)=x for all x and cosh-1t=ln (t+t2-1) Claim :…

Answered: Inverse identity Show that cosh (cosh…

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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Question

Inverse identity Show that cosh (cosh x) = |x| by using the
formula cosht = In(t + Vr - 1) and considering the cases
x 2 0 and x < 0.
%3D

Expert Solution

Step 1

$G i v e n : \cos h^{- 1} (\cos h x) = |x| f o r a l l x a n d \cos h^{- 1} t = \ln (t + \sqrt{t^{2} - 1}) C l a i m : T o P r o v e : \cos h^{- 1} (\cos h x) = |x| f o r a l l x C a s e : I W h e n x \geq 0 : \ln [\cos h x + \sqrt{\cos h^{2} x - 1}] = \ln [\cos h x + \sqrt{\sin h^{2} x}] = \ln [\cos h x + \sin h x]$

Step 2

$\ln [\cos h x + \sqrt{\cos h^{2} x - 1}] = \ln [\cos h x + \sin h x] = \ln [\frac{e^{x} + e^{- x}}{2} + \frac{e^{x} - e^{- x}}{2}] = \ln [[\frac{2 e^{x}}{2}] = \ln [e^{x}] = x$