Finding the coeffiecients of a sine series for -3*cos(8*pi*x/L)

  • Thread starter Thread starter dnp33
  • Start date Start date
  • Tags Tags
    Series Sine
Click For Summary

Homework Help Overview

The discussion revolves around finding the coefficients of a sine series for the function -3*cos(8*pi*x/L). The original poster seeks assistance with the integral involved in this process.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the integral that needs to be solved and reference trigonometric identities that may simplify the problem. There is also mention of orthogonality relations for sine and cosine functions, with some participants expressing uncertainty about their derivation and applicability in the given context.

Discussion Status

The conversation includes attempts to clarify the use of trigonometric identities and the relevance of orthogonality relations. Some participants acknowledge their familiarity with these concepts but express a lack of understanding regarding their derivation, indicating a productive exploration of foundational ideas.

Contextual Notes

There is a noted constraint regarding the original poster's textbook coverage of the derivation of orthogonality relations, which may limit their understanding of the problem. Additionally, there is a discussion about the orthogonality of sine and cosine functions over different intervals.

dnp33
Messages
39
Reaction score
1

Homework Statement


all i really need to do is solve the integral:
[tex]\int-3*sin(\frac{n*pi*x}{L})*cos(\frac{8*pi*x}{L})[/tex]
from x=0 to x=L

Homework Equations


The Attempt at a Solution


I'm not very sure where to start with this. Any help would be appreciated.
 
Last edited:
Physics news on Phys.org
Seeing as you are finding a Fourier sine series, you really should know the orthogonality relations for sine and cosine by now, and how they are derived. Heres a first step: [itex]\sin x \cos y = \frac{1}{2} \left( \sin(x-y) + \sin (x+y) \right)[/itex].
 
hi dnp33! :smile:

use one of the standard trigonometric identities …

2sinAcosB = sin(A+B) + sin(A-B) :wink:
 
Gib Z said:
Seeing as you are finding a Fourier sine series, you really should know the orthogonality relations for sine and cosine by now, and how they are derived. Heres a first step: [itex]\sin x \cos y = \frac{1}{2} \left( \sin(x-y) + \sin (x+y) \right)[/itex].

I am aware of the orthogonality relations, however I am not familiar with how they are derived, that is not covered in my textbook unfortunately. I do know that sin and cos are orthogonal on the interval -L to L, but they are not orthogonal on the interval 0 to L (according to my text), so this doesn't help me unfortunately.
 
Worked perfectly! Thanks a lot.
tiny-tim said:
hi dnp33! :smile:

use one of the standard trigonometric identities …

2sinAcosB = sin(A+B) + sin(A-B) :wink:
 
dnp33 said:
I am aware of the orthogonality relations, however I am not familiar with how they are derived, that is not covered in my textbook unfortunately. I do know that sin and cos are orthogonal on the interval -L to L, but they are not orthogonal on the interval 0 to L (according to my text), so this doesn't help me unfortunately.

Actually, what I said would have helped you if you weren't so busy trying to be offended. It wasn't just knowing the orthogonality relations that helps here, but knowing how they are derived, as the same first ideas apply to this similar problem. The last sentence of my post was the same identity tiny-tim posted, but perhaps you didn't see it.
 
I'm sorry if I came off as offended and defensive. I wasn't trying to.

I did notice after that you had the same identity written down as tiny tim, but I had already posted twice and didn't want to go back and post again.

I'll respond more carefully in the future so this doesn't happen again.
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
Replies
12
Views
7K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 14 ·
Replies
14
Views
2K
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K