1. The problem statement, all variables and given/known data Hi, I'm new to this site, I've had a look around and there are alot of useful sections, particularly the section with math and science learning materials. Anyway, I need to show that the following function is odd [tex]f(x)=\left\{\begin{array}{ccc} -\sin x&\mbox{ for }-\pi \leq x< \frac{-\pi}{ 2}\\ \sin x &\mbox{ for } \frac{-\pi}{2} \leq x \leq \frac{\pi}{2}\\ -\sin x &\mbox{ for } \frac{\pi}{2}<x<\frac{\pi}{2} \end{array}\right.[/tex] [tex]\mbox{ and }f(x + 2 \pi) = f(x) \mbox{ for all other values of x, is an odd function.}[/tex] 2. Relevant equations I know an odd function is definded as [tex] f(-x) = -f(x)[/tex] 3. The attempt at a solution In the interval [tex]-\pi\leq x < {-\pi \over 2} \mbox{ if I substiture } -\pi \mbox{ it becomes }-\sin(-x) = -\sin[-(-{\pi \over 2})] = -\sin({\pi \over 2})[/tex] Is that the correct way to solve it? But I'm not sure how to show it's odd in the other intervals!
This might be a useful variation of the definitions: If the function f(x) is even, then f(x)-f(-x)=0 for all x. If the function f(x) is odd, then f(x)+f(-x)=0 for all x.
To show it's odd: look at values in the intervals? [tex]-\sin(-\pi) - \sin({\pi }) = 0 [/tex] [tex] \sin({-\pi \over 2}) + \sin({\pi \over 2}) = 0[/tex] [tex]-\sin({3 \pi \over 4}) - \sin({-3 \pi \over 4}) = 0[/tex] do I need to show anything else?
You would have to show that it's true for every value in the interval, not just at a few random points. So you'd have to let [tex]a[/tex] be a random value in each interval, and then look at [tex]f(a)[/tex] and [tex]f(-a)[/tex]. Since the intervals are symmetric, once you've assigned an interval for [tex]a[/tex], it will be obvious what interval [tex]-a[/tex] is in and therefore which definition of the function you need to use.