## Trigonometry (11th Edition) Clone

$f(-x)=-f(x)$
We examine the points where $x\in(0,\frac{\pi}{2})$. Here, we find that $f(x)\gt0$ Then we examine the points where $x\in(0,-\frac{\pi}{2})$ and find that $f(x)\lt0$ In another case, when $x\in(\frac{\pi}{2},\pi)$, $f(x)\lt0$. While $x\in(-\frac{\pi}{2},-\pi)$, $f(x)\gt0$ The graph has a symmetry through the point $O$, which means for a point where $x=a$ and $y=f(a)=b$, the point where $x=-a$ would have $y=f(-a)=-b$. In other words, $f(-x)=-f(x)$