Cosine Series of $-\frac{1}{5}\sin x$ on $[0,\pi]$

Question

I am really confused as to why $\sin x$ has a Fourier cosine series. I thought that since $\sin x$ is an odd function, then $a_0$ and $a_n$ both equal $0$, and we find the coefficient for $b_n$. However, in the examples I have looked over, they are calculating $a_0$ and $a_n$ and subjecting $b_n$ to be equal to zero. Why is this? Maybe I am misunderstanding the how the interval ties into this. This problem is on the interval $[0,\pi]$. Please help me!

Also- I thought that Cosine series and Sine series are merely methods of shortcuts. If we know that $f(x)$ is an even function, then we only need to solve for $a_0$ and $a_n$. If we know that $f(x)$ is an odd function, then we only need to solve for $b_n$. Is this an incorrect way of thinking?

Thank you in advance!

Answer 1

On the figure below

• The arc of sine function in $0<x<\pi$ is drawn in red.

• The function drawn in black is the Fourier cosine series.

This makes you understand the meaning of the problem ?