# Math Help - Extension of sin function

1. ## Extension of sin function

Folks,

Given $f(x)=sin x, \pi \le x \le 2\pi$ and 0 otherwise and
$F(x)=\begin{cases}f(x), x \ge 0 \\ -f(-x), x \le 0 \\ \end{cases}$

How do I complete this set up for, indicated by question mark

$F(x)=\begin{cases}f(x)=sin x, \pi \le x \le 2\pi \\ f(x)=0, -2\pi \le x \le -\pi \\ -f(-x)=?, \pi \le x \le 2\pi \\ -f(-x)=0, -2\pi \le x \le -\pi \end{cases}$

Ie, I want to use the same method that was applied to the following
$F(x)=\begin{cases}f(x)=xe^{-x}, x\ge 0 \\-f(-x)=-(-x)e^{-(-x)}, x \ge 0 \\ \end{cases}=\begin{cases}xe^{-x}, x\ge 0 \\ xe^x, x \le 0 \\ \end{cases}$

thanks

2. What you are writing simply doesn't make sense. F(x)= -f(-x) only for x<0 so you cannot have " $F(x)= -f(-x)$" for $\pi\le x\le 2\pi$. If $0\le x\le\pi$ then F(x)= 0. If $\pi\le x\le 2\pi$ then F(x)= sin(x). If $2\pi\le x$, F(x)= 0. For negative values of x, reverse those: if $-\pi\le x\le 0$ then $0\le -x\le \pi$ so F(x)= -f(-x)= 0. If $-2\pi\le x\le -\pi$ then $\pi\le -x\le 2\pi$ so F(x)= -f(-x)= -sin(-x)= sin(x). If $x\le -2\pi$ then $x\ge 2\pi$ so F(x)= -f(-x)= 0.

3. Originally Posted by HallsofIvy
What you are writing simply doesn't make sense. F(x)= -f(-x) only for x<0 so you cannot have " $F(x)= -f(-x)$" for $\pi\le x\le 2\pi$.
Ok, you say F(x)=-f(-x) only for x<0. This apply for all situations or only in the context of what was given in this example?