$f$ is an odd, $2\pi$ periodic function. Write explicitly Let $f$ be an odd $2\pi$ periodic function defined on $(0,\pi)$ by $f(x) = \frac12 (\pi - x)$. 
I am asked to "write $f$ explicitly". what does this mean? how to do that? thanks for help.
 A: I assume the question asks for an explicit formula for all the real numbers $(-\infty,\infty)$.
It is an odd function, meaning $f(x) = -f(-x)$ for all $x$, therefore $f(0) = -f(0) = 0$. Therefore we have its values for $(-\pi,\pi)$. Because of the periodicity, this automatically gives us the values for the function for all $x\neq \pi n$ where $n\in N$.
From periodicity and oddness we get $f(-\pi) = f(\pi) = -f(-\pi) = 0$. So we have the values set for a complete $2\pi$ period $(-\pi,\pi]$.
To write a truly explicit formula for $f(x)$ we have to give an explicit formula representing the annoying periodicity, meaning we'd want to deal only with $x -2\pi\lfloor \frac{x} {2\pi} \rfloor$. Note $x + 2\pi n -2\pi\lfloor \frac{x + 2\pi n} {2\pi} \rfloor = x + 2\pi n -2\pi(\lfloor \frac{x +n} {2\pi} \rfloor + n) = x -2\pi\lfloor \frac{x} {2\pi} \rfloor$.
This leaves us with the following function: 
$$f(x)=
\begin{cases}
     \hfill \frac{1}{2}\left(\pi-\left(x -2\pi\lfloor \frac{x} {2\pi} \rfloor\right)\right)    \hfill & \text{ if } x\neq2n\pi \\
      \hfill 0 \hfill & \text{ if } x=2n\pi \\
\end {cases}
 $$
Where $n$ is an integer. Note that $f(nπ)$ still equals $0$.
This function would also work: 
$$f(x)=
\begin{cases}
     \hfill -\arctan \left(\tan \left(\frac{x+\pi }{2}\right)\right)    \hfill & \text{ if } x\neq2n\pi \\
      \hfill 0 \hfill & \text{ if } x=2n\pi \\
\end {cases}
 $$
A: An explicit expression for the function is:
$$
f(x)=\left\{\begin{array}{l}
            f(x+2(\kappa+1)\pi), \ \ x\in\big(-2(\kappa+1)\pi,-2\kappa\pi\big), \\
            \frac{1}{2}(\pi-x), \ \ x\in(0,2\pi),   \\
            f(x-2\kappa\pi) \ \ x\in\big(2\kappa\pi,2(\kappa+1)\pi\big)
            \end{array}
      \right. \Rightarrow $$ 
$$ \Rightarrow 
f(x)=\left\{\begin{array}{l}
            -\frac{1}{2}\big(x+(2\kappa+1)\pi\big), \ \ x\in\big(-2(\kappa+1)\pi,-2\kappa\pi\big), \\
            \frac{1}{2}(\pi-x), \ \ x\in(0,2\pi),   \\
            \frac{1}{2}\big((2\kappa+1)\pi-x\big), \ \ x\in\big(2\kappa\pi,2(\kappa+1)\pi\big)
            \end{array}
      \right.
$$
for any real $x\neq\kappa\pi$ ($\kappa$ any non-negative integer). 
The function finally looks like: 
     
(with the integer multiples of $\pi$ excluded). 
Note that the geometric properties should now be obvious from the graph: the function is periodic, with period $2\pi$ and also odd, which is equivalent to the fact that its graph is symmetric with respect to the origin. 
