Explanation Given Information: The provided function is y = 5 sin ( 1 2 x + 90 ° ) . Formula Used: For a given function y = A s c t ( B x + Q ) , where s c t = sin / cos / tan Here, A = Amplitude , Period = 360 B and Phase = Q B Calculation: Consider the provided function: y = 5 sin ( 1 2 x + 90 ° ) Now, compare the provided function y = 5 sin ( 1 2 x + 90 ° ) with the function y = A s c t ( B x + Q ) . So, it can be concluded that, Amplitude = 5 And, period = 360 1 2 = 360 ⋅ 2 = 720 And, phase = 90 1 2 = 90 ⋅ 2 = 180 Thus, The amplitude is 5 , period is 720 ° and phase is 180 ° left in the given function y = 5 sin ( 1 2 x + 90 ) . Graph: Consider the provided function: y = 5 sin ( 1 2 x + 90 ) Now, obtain the values of y by assuming corresponding values of x as: Substitute x = 0 in the function y = 5 sin ( 1 2 x + 90 ) . y = 5 sin ( 1 2 ⋅ 0 + 90 ) = 5 ⋅ sin ( 90 ) = 5 ⋅ 1 = 5 Substitute x = 30 in the function y = 5 sin ( 1 2 x + 90 ) . y = 5 sin ( 1 2 ⋅ 30 + 90 ) = 5 ⋅ sin ( 105 ) = 5 ⋅ 0.9659 = 4.8296 Substitute x = 60 in the function y = 5 sin ( 1 2 x + 90 ) . y = 5 sin ( 1 2 ⋅ 60 + 90 ) = 5 ⋅ sin ( 120 ) = 5 ⋅ 0

11th Edition

ISBN: 9781285199191

Author: Dale Ewen, C. Robert Nelson

Publisher: Cengage Learning

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Chapter 14.2, Problem 21E

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To calculate: The period and amplitude and phase of the given expression y=5sin(12x+90°) and graph the provided function.

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Chapter 14.2, Problem 20E

Chapter 14.2, Problem 22E

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Chapter 14 Solutions

Elementary Technical Mathematics

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The period and amplitude and phase of the given expression y = 5 sin ( 1 2 x + 90 ° ) and graph the provided function.

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To calculate: The period and amplitude and phase of the given expression y=5sin(12x+90°) and graph the provided function.

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Chapter 14 Solutions