Suppose $f(x)=sinx,\quad 0<x<π$.
Can you do a half-range sine expansion on f(x)? I tried, but I got $a_0=a_n=b_n=0$.
If you requrie me to show my steps (i.e. I should have not gotten $a_0=a_n=b_n=0$), please leave a comment. I am not very good with tex so I'd rather avoid it if I can.
If you do an odd expansion, then the extension is $\sin x$ itself. And, indeed, you have $$ b_1=\frac2\pi\int_0^\pi\sin^2x\,dx=\frac2\pi\,\frac\pi2=1. $$
If you do an even expansion, then your function is $|\sin x|$ on $(-\pi,\pi)$ and its coefficients are $$ a_0=\frac2\pi\int_0^\pi\sin x\,dx=\frac4\pi, $$ $$ a_n=\frac2\pi\int_0^\pi\sin x\,\cos nx\,dx=\frac2\pi\,\frac{1+(-1)^k}{1-k^2}. $$ so the Fourier series is $$ \frac4\pi-\frac4\pi\,\sum_{k=1}^\infty\frac{\cos 2kx}{4k^2-1} $$