Solving the Expanded Cosine Series of y=sin(x) in (0,180)

[prasanaharani]

expand the function y=sinx in a series of cosines in the interval (0 to 180)
i want to know only the value of f(x) for solving this.what is the value of
f(x).

 

[prasanaharani]

expand the function y=sinx in a series of cosines in the interval (0 to 180)
i want to know only the value of f(x) for solving this.what is the value of
f(x).

 

[chroot]

I suggest you post the entire problem, exactly as it was given to you. Next, show us your attempts at a solution, and where you've gotten stuck.

- Warren

 

[prasanaharani]

to solve this problem first we need to know the value of An and A0
ie the series is A0/2+SUMATION n=1 to infinity (An cosnx)where An is given by An=2/pie integral of -pie to +pie f(x)cosnx dx
now what is the value of f(x) to substitute in that place to solve it.pls tell me.

 

[prasanaharani]

To solve this problem first we need to know the value of An and A0
ie the series is A0/2+SUMATION n=1 to infinity (An cosnx)where An is given by An=2/pie integral of -pie to +pie f(x)cosnx dx
now what is the value of f(x) to substitute in that place to solve it.pls tell me.

 

[CompuChip]

[tex]f(x) = \sin(x)[/tex]?
Why do you have two names (y and f(x) for the same thing?)
At least, that's what you said in your first (two) post(s).

 

[matematikawan]

cos(nx) is an even function. So you have to even extend the function[tex]f(x) = \sin(x)[/tex] first.
It will be a periodic function of period [tex]2L = \pi[/tex].
Then the coefficient
[tex]a_n = \frac{2}{\pi/2} \int{\sin(x)\cos(2nx) dx}[/tex]
integrate from 0 to [tex]\frac{\pi}{2} [/tex]
which simplify to
[tex]a_n = -\frac{2}{\pi (4n^2-1)} [/tex]

The Fourier series is then
[tex]\sin(x) = \frac{2}{\pi} -\frac{4}{\pi}(\frac{\cos(2x)}{1.3} + \frac{\cos(4x)}{3.5} + . . . ) [/tex]