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How did the negative sign appear? To prove cos (iz) = cos h (z) = cos since et+e¬* eix +e-ix cos h (x) and cos (x) = 2 2 LHS=cos (iz) e(12) +e-i(iz) -e-+e? 2 2 RHS=cos h (z) : e+e=? Since, LHS=RHS.

How did the negative sign appear? To prove cos (iz) = cos h (z) = cos since et+e¬* eix +e-ix cos h (x) and cos (x) = 2 2 LHS=cos (iz) e(12) +e-i(iz) -e-+e? 2 2 RHS=cos h (z) : e+e=? Since, LHS=RHS.

Calculus: Early Transcendentals
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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How did the negative sign appear? To prove cos (iz) = cos h (z) since cos h (x) : e+e-* and cos (x) eix +e-ix LHS=cos (iz) : eiliz) +e¬i(iz) -e-?+e? 2 e+e-z RHS=cos h (z) 2 Since, LHS=RHS. It can be proved that cos (iz) = cos h (z)
Transcribed Image Text:How did the negative sign appear? To prove cos (iz) = cos h (z) since cos h (x) : e+e-* and cos (x) eix +e-ix LHS=cos (iz) : eiliz) +e¬i(iz) -e-?+e? 2 e+e-z RHS=cos h (z) 2 Since, LHS=RHS. It can be proved that cos (iz) = cos h (z)
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