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

In summary, the homework statement is asking for someone to find a Fourier sine series, and they need to know the orthogonality relations for sin and cos.
  • #1
dnp33
40
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:
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  • #2
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].
 
  • #3
hi dnp33! :smile:

use one of the standard trigonometric identities …

2sinAcosB = sin(A+B) + sin(A-B) :wink:
 
  • #4
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.
 
  • #5
Worked perfectly! Thanks alot.
tiny-tim said:
hi dnp33! :smile:

use one of the standard trigonometric identities …

2sinAcosB = sin(A+B) + sin(A-B) :wink:
 
  • #6
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.
 
  • #7
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.
 

1. What is a sine series?

A sine series is a mathematical representation of a periodic function using sine and cosine terms. It is used to approximate a given function and can be used to solve differential equations and other mathematical problems.

2. How do you find the coefficients of a sine series?

The coefficients of a sine series can be found by using the Fourier series formula, which involves integrating the given function over one period and then solving for the coefficients using the orthogonality properties of sine and cosine functions.

3. What is the purpose of finding the coefficients of a sine series?

The purpose of finding the coefficients of a sine series is to approximate a given function and represent it in terms of simpler sine and cosine functions. This can be useful in solving mathematical problems and analyzing periodic functions.

4. Why is the function -3*cos(8*pi*x/L) used in this example?

The function -3*cos(8*pi*x/L) is commonly used in examples of finding the coefficients of a sine series because it is a periodic function that can be easily represented using sine and cosine terms. It also allows for simplification of the calculations involved in finding the coefficients.

5. Can the coefficients of a sine series be found for any function?

Yes, the coefficients of a sine series can be found for any function as long as it is periodic and satisfies certain conditions. However, some functions may have more complicated coefficients and may require more advanced techniques to find them.

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