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.
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:
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 $$ \sin^2{(nx)} = \frac{1-\cos{nx}}{2} $$