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We know that Fourier series of odd function consists of sine terms only. What additional condition on symmetry will ensure that sine coefficients with even indices is zero i.e.; $ \ b_{2n}=0 $. Give an example.

Answer: I do not know the reason but I got an example. The example is below:

$$f(x)=\begin{cases}-1&-1\leq x<0\\ 0&x=0\\ 1&0<x\leq 1\end{cases}$$

This function is odd because $ f(-x)=-f(x) $.

$b_n=2 \int_{0}^{1} f(x) \sin (n \pi x)dx=\frac{2}{n \pi}[(-1)^n-1] $ .

This proves that $ b_{2n}=0 $.

But I need the reason why $ b_{2n} =0 $?

MAS
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Given we already assumed $f$ odd, the additional condition is $f(1-x)=f(x)$ on $(0,1)$, i.e., the half of $f$ is an even function if we consider $1/2$ as the center of symmetry.

The reason is that $\sin(\pi n (1-x)) = (-1)^{n-1} \sin (\pi n x)$ for all integer $n$. Hence every linear combination of odd-numbered sines satisfies $f(1-x)=f(x)$, and every linear combination of even-numbered sines satisfies $g(1-x) = -g(1-x)$. These two kinds of functions are orthogonal.

Formally: if $f$ is odd and $f(1-x) = f(x)$, then $$ \int_{-1}^1 f(x)\sin (\pi n x)\,dx = 2 \int_{0}^1 f(x)\sin (\pi n x)\,dx = 2 \int_{0}^1 f(1-x)\sin (\pi n (1-x))\,dx = 2 (-1)^{n-1}\int_{0}^1 f(x)\sin (\pi n x)\,dx = (-1)^{n-1}\int_{-1}^1 f(x)\sin (\pi n x)\,dx $$ hence the integral is zero when $n$ is even.