# Find the fourier sine series of cosine.

1. Nov 6, 2014

### pondzo

1. The problem statement, all variables and given/known data
Hi, so im doing some past exam papers and there was this question;

2. Relevant equations

3. The attempt at a solution
a0 and an both are equal to zero, this leaves only bn.

Since you can only use the sine series for an odd function, and cos(t) is even, does this mean i have to find the odd half expansion of cos(t) with L=π ? This would be f(t) = -cos(t) for [-π,0] and cos(t) for [0,π]. Or do I use L= π/2 and do the integral from 0 to π for bn?

Letting n=1 since it is the first partial sum we are talking about (can i do this?)then using the first method I get 16/(3π) and if i try the second method I get 8/(3π). But the answer is listed as 0.

Any help would be great thanks.

2. Nov 6, 2014

### ehild

You should write the relevant equations.
The problem text does not define f(t) in the interval -pi<t<0. If its Fourier series contains only sine terms, that suggests that the function is odd, f(t) = -cos(t) if -pi<t<0.
What is the formula for the Fourier coefficients of an odd function?

3. Nov 6, 2014

### pondzo

Hi ehild, sorry i wasn't quite sure how to typeset them.

I did write that if I perform the odd half expansion of cos(t) I get
f(t) = cos(t) for [0,π] and f(t) = -cos(t) for [-π,0]
Using this (and assuming I can let n=1 since it is this first partial sum, but im not sure if i can do this) then $b_1=\frac{1}{\pi}[\int_{-\pi}^0-\cos{t}\sin{2t}dt + \int_{0}^{\pi}\cos{t}\sin{2t}dt] = \frac{8}{3\pi}$

Where am I going wrong?

4. Nov 6, 2014

### pondzo

Oops, I found where i went wrong.
Instead of $b_1=\frac{1}{\pi}[\int_{-\pi}^0-\cos{t}\sin{2t}dt + \int_{0}^{\pi}\cos{t}\sin{2t}dt]$
It should be $b_1=\frac{1}{\pi}[\int_{-\pi}^0-\cos{t}\sin{t}dt + \int_{0}^{\pi}\cos{t}\sin{t}dt] = 0$

I would still like to know if it is correct to let n=1 in the bn integral to find b1. Or should you only sub in n=1 after you find the general form of bn?

5. Nov 7, 2014

### vela

Staff Emeritus
You can set n=1 before doing the integration. You might as well since all you were looking for was $b_1$.