# Cosine series of sine

1. Oct 16, 2005

### tiagotorres

I tried to find the cosine series of the function $$f(x) = \sin x$$, using the equation below:

$$S(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos(nx)$$
where: $$a_n = \frac{2}{\pi} \int_{0}^{\pi} f(x) \cos(nx) dx$$

I found:

$$a_0 = \frac{4}{\pi}$$
$$a_n = \frac{2 }{\pi (1 - n^2)} (\cos(n \pi) + 1)$$

Therefore:

$$S(x) = \frac{2}{\pi} (1 - 2 \sum_{n=1}^{\infty} \frac{\cos(2nx)}{4n^2 - 1})$$

Making the graph of the first terms of the function above on my calculator, I noticed that this is actually $$| \sin x |$$, rather than just $$\sin x$$. Why does this happen? Is there a way of figuring out the sum not using a calculator?

Thanks

Last edited: Oct 16, 2005
2. Oct 16, 2005

### mathman

sinx is nonegative between 0 and pi, so sinx=|sinx| there.