#### calcgirl

Joined Jun 8, 2008
4
I am having a difficult time finding the fourier series for (sin x)^2.
I am aware that I have to use the trig identity:
(sin x)^2=1/2-1/2(cos 2x) but I don't know where to go from here. ANy help would be greatly appreciated thanks!!

#### silvrstring

Joined Mar 27, 2008
159
Are there any limits (e.g., 0<=x<=1, or 0<=x<=2pi)?

I'm gonna guess 0<= x <= 2pi.

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#### calcgirl

Joined Jun 8, 2008
4
No limits are given in the problem but when looking at the graph it cycles every pi.

#### silvrstring

Joined Mar 27, 2008
159
Then you're using 0<=x<=2pi.

I'll see if I can come come up with something on paper to scan.

#### silvrstring

Joined Mar 27, 2008
159
Calcgirl, I can't come up with an example without having to do your problem.

Are you having a problem understanding how to solve a Fourier series problem, or are you having a problem integrating int((sin(x))^2 * cos(kx) dx) or int((sin(x))^2 *sin(kx) dx)?

#### calcgirl

Joined Jun 8, 2008
4
I understand how to do a typical fourier series problem such as if the function was a piecewise or the function x^2. But I was told that this problem can be used without any integration and that the double angle formula/trig identities were to be used. I guess I just don't know what else to do after I have my new equation:
(sin x)^2=1/2-1/2(cos 2x)

#### Ratch

Joined Mar 20, 2007
1,070

#### silvrstring

Joined Mar 27, 2008
159
Wow. Okay.

Are you sure you have to solve this for a Fourier series; I don't think it is possible, because the constant k in cos(kx) and sin(kx), will ruin the process when integrating. This happens whether you use double angle ids, or not.

The average is visually 1/2, and (1/2pi)int((sinx)^2) dx) equals 1/2. If your instructor told you integration need not be used, I'm wondering if he meant a Fourier series solution was needed at all. Even substituting a Taylor polynomial to the nth degree for sin^2(x) will not work.

Mark44 and hgmjr might be able to help. If not, I'll get back to you tomorrow.

Post if you figure it out.

Take care, and Welcome.

#### Mark44

Joined Nov 26, 2007
628
I am having a difficult time finding the fourier series for (sin x)^2.
I am aware that I have to use the trig identity:
(sin x)^2=1/2-1/2(cos 2x) but I don't know where to go from here. ANy help would be greatly appreciated thanks!!
A Fourier series is an infinite series involving terms in sin(nx) and cos(nx). Since one part of the function you're working with is 1/2 cos(2x), you're halfway home to your answer, meaning that the Fourier series for 1/2 cos(2x) is just 1/2 cos(2x). All you need to do for your problem is to find the Fourier series for 1/2.

You also need to know the interval over which to integrate, which you haven't mentioned, so I'm going to assume that it is [-pi/2, pi/2]. If it's a different interval, let me know, since the work that follows is using this assumption.

Now, because the function you're working with is even, there aren't going to be any sine terms. The series will look like this:
1/2 = a_0 + summation[a_n cos(2nt)], with the summation running from n = 1, 2, 3, ....

Here are the coefficients:
a_0 = 2/pi * integral[1/2 dt], integrated between 0 and pi/2

a_n = 4/pi * integral[1/2 cos (2nt) dt], integrated between 0 and pi/2, and for n = 1, 2, 3, ... You should be able to evaluate the integral one time for any value of n, and get an expression for a_n.

Calcgirl, see if this works for you. If you have problems, reply back.
Mark

#### silvrstring

Joined Mar 27, 2008
159
I think this might help. I attached it. Maybe your instructor wanted you to see that the trig identity could be proven with Fourier series.

You wouldn't know that you didn't need to integrate until after you did it, though; You didn't need to do anything for that matter. I guess he was hinting that the identity is in Fourier series form already.

All a(k) and b(k) after the work shown were 0.

Take care.

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#### Mark44

Joined Nov 26, 2007
628
Silvrstring,
Looks good to me. When I posted my reply earlier today, it completely escaped me that 1/2 - 1/2 cos(2x) already was a Fourier series. Doh!

#### silvrstring

Joined Mar 27, 2008
159
Yeah, me too! I hate when they play those games!

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