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sin 2πα cos αx-sin πα cos α + sin αx cos πα-cos 2πα sin αx da α sin α (2π- x)-sin α (π-x) da | α II

sin 2πα cos αx-sin πα cos α + sin αx cos πα-cos 2πα sin αx da α sin α (2π- x)-sin α (π-x) da | α II

College Algebra
College Algebra
7th Edition
ISBN:9781305115545
Author:James Stewart, Lothar Redlin, Saleem Watson
Publisher:James Stewart, Lothar Redlin, Saleem Watson
Chapter4: Exponential And Logarithmic Functions
Section: Chapter Questions
Problem 14CC
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I don't understand where sina(2pi-x) and -sina(pi-x) came from. Can you please explain it to me? Thank you 

12:12 AM Thu May 13 18% I AA bartleby.com M G Goog G Google 1-6 find the... A Classwork fo... G akshata mud... mer Mem... l inje... = bartleby Q Search for textbooks, step-by-step explanatio... Ask an Expert Math / Bundle: Differential Equations with Boundary-Value Problems, Loo... | In Problems 1-6 find the Fourier integral repr... i In Problems 1-6 find the Fourier integral representation of the given function. 2. f ( x ... = 4 cos axdx 27 = 4[- sin ax It sin 2πα-sin πα A (a) : = 4 And the value of B (a) is, 00 В (а) — f (x) sin axdx J. 2n 00 0 · sin axdx + (4) · sin axdx + 0 . sin axdx 2л 2n 4 sin axdx :4[ - cos axr || cos πα-cos 2πα B (a) = 4 Thus, the value of f (x) is, 00 sin 2πα- sin πα coS TA – cos 2na f(x) cos ax + 4 - sin ax| da 00 sin 2ra cos ax S1n πα COS αx+ S1n αx cOS πα-cos 2πα sin αχ da 4 00 sin a (2n – x) – sin a (n – x) da Hence, the Fourier integral representation of given function is sin a (2л — х) - sin a (л — х) 00 4 f(x) da. E Chapter 14.3, Problem 1E Chapter 14.3, Problem 3E > Was this solution helpful?
Transcribed Image Text:12:12 AM Thu May 13 18% I AA bartleby.com M G Goog G Google 1-6 find the... A Classwork fo... G akshata mud... mer Mem... l inje... = bartleby Q Search for textbooks, step-by-step explanatio... Ask an Expert Math / Bundle: Differential Equations with Boundary-Value Problems, Loo... | In Problems 1-6 find the Fourier integral repr... i In Problems 1-6 find the Fourier integral representation of the given function. 2. f ( x ... = 4 cos axdx 27 = 4[- sin ax It sin 2πα-sin πα A (a) : = 4 And the value of B (a) is, 00 В (а) — f (x) sin axdx J. 2n 00 0 · sin axdx + (4) · sin axdx + 0 . sin axdx 2л 2n 4 sin axdx :4[ - cos axr || cos πα-cos 2πα B (a) = 4 Thus, the value of f (x) is, 00 sin 2πα- sin πα coS TA – cos 2na f(x) cos ax + 4 - sin ax| da 00 sin 2ra cos ax S1n πα COS αx+ S1n αx cOS πα-cos 2πα sin αχ da 4 00 sin a (2n – x) – sin a (n – x) da Hence, the Fourier integral representation of given function is sin a (2л — х) - sin a (л — х) 00 4 f(x) da. E Chapter 14.3, Problem 1E Chapter 14.3, Problem 3E > Was this solution helpful?
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