I don't get why sin((-\pi,\pi)=(-1,1). I thought because sin(\pi/2)=1 and since \pi/2\in (-\pi,\pi) and also because sin(-\pi/2)=-1 and since -\pi/2\in (-\pi,\pi) then sin((-\pi,\pi)=[-1,1].
If sin((0,\pi) then it equals (0,1] which is neither open nor closed. I'm not sure where my reasoning...