Why is the odd extension of $\sin(x)$ simply equal to $\sin(x)$? I want to expand function $f : (0, \pi) \to\mathbb{R}$ defined by $f(x) = \sin(x)$ as a Fourier sine series.
I know that for $f^{\text{odd}}$ implies that $f(x) = -f(-x)$ so $\sin(x) = -\sin(-x)$ for interval $[-\pi, \pi]$. For an odd function $a_0$ and $ a_n = 0$ So it remains to calculate $b_n$ which would be the integral $$\frac{2}{\pi}\int_{0}^{\pi} f(x) \sin(n\pi x/\pi)  dx  = \frac{2}{\pi}\int_{0}^{\pi} \sin(x) \sin(nx)  dx$$ 
Is that the same as $\sin(x)$? This isn't what is written in the solutions, it simply states that $f^{\text{odd}}\sim \sin(x)$ is the solution so i'm wondering if there's a point i'm missing that makes answering this question much easier.
Would appreciate the help.
 A: The idea is that if cannot write $\sin x$ on Fourier serie because... that's yet the Fourier serie of the sine! It's itself!
Of course there are lots of way to proof that, if $n \neq 1$, then 
$$
 \int_0^{\pi} \sin(x) \sin(nx) \operatorname d x = 0
$$
I'll show you that a more general fact:

If $m$ and $n$ are positive integers such that $m \neq n$, then 
  $$
I_{m,n}=  \int_0^{\pi} \sin(mx) \sin(nx) \operatorname d x = 0
$$

Let's integrate by parts, we get 
$$
\begin{split}
I_{m,n} &=  \left(-\frac{\cos{nx}}{n}\sin{mx}\right)\Bigg|_0^\pi- \frac{m}{n}\int_0^{\pi} \cos(mx) \cos(nx) \operatorname d x \\
&=\frac{m}{n}\underbrace{\int_0^{\pi} \cos(mx) \cos(nx) \operatorname d x}_{\mathcal F} \\
\end{split}
$$
If we integrate the other sine, we obtain
$$
I_{m,n}=\frac{n}{m}\int_0^{\pi} \cos(mx) \cos(nx) \operatorname d x = \frac{n}{m} \mathcal F
$$
Hence 
$$
\frac{n}{m} \mathcal F = \frac{m}{n} \mathcal F
$$
therefore
$$
(n^2-m^2) \mathcal F=0
$$
If $n^2=m^2$ then $m=n$ ($m$ and $n$ are positive integers) and that's absurd, since we suppose that $m \neq n$. So we obtain $\mathcal F =0$ and, finally
$$I_{m,n}=0$$ 
In your case $m=1$ and we discovered that $b_n=\frac{2}{\pi}I_{1,n}=0$ for every $n \neq 1$, but what happened when $n=1$?
Hint: You can use the rule
$$
\sin^2{(nx)} = \frac{1-\cos{nx}}{2}
$$
