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Where did the i in the denominator go? Part B To prove sin (iz) = i sin h (z) LHS=sin (iz) € eiliz) –e-i(iz) e--e 2i 2! S= i h(z) = i · e?-e- 2 e-e-z e-e-z e--e 2 2! 2! Since, LHS=RHS. It can be proved that sin i sin h (-) II

Where did the i in the denominator go? Part B To prove sin (iz) = i sin h (z) LHS=sin (iz) € eiliz) –e-i(iz) e--e 2i 2! S= i h(z) = i · e?-e- 2 e-e-z e-e-z e--e 2 2! 2! Since, LHS=RHS. It can be proved that sin i sin h (-) II

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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Where did the i in the denominator go? Part B To prove sin (iz) = i sin h (z) eiliz) –e-i(iz) LHS=sin (iz) € 2i 2! S= e-e-z ez -e-z e--e? i h(z) = i · 2 2 2! 2! Since, LHS=RHS. It can be proved that sin (iz) = i sin h (z)
Transcribed Image Text:Where did the i in the denominator go? Part B To prove sin (iz) = i sin h (z) eiliz) –e-i(iz) LHS=sin (iz) € 2i 2! S= e-e-z ez -e-z e--e? i h(z) = i · 2 2 2! 2! Since, LHS=RHS. It can be proved that sin (iz) = i sin h (z)
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