How to show a function is even/odd

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Homework Statement


Hi,
I'm new to this site, I've had a look around and there are a lot 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}<br /> -\sin x&\mbox{ for }-\pi \leq x< \frac{-\pi}{ 2}\\<br /> \sin x &\mbox{ for } \frac{-\pi}{2} \leq x \leq \frac{\pi}{2}\\<br /> -\sin x &\mbox{ for } \frac{\pi}{2}<x<\frac{\pi}{2}<br /> \end{array}\right.[/tex]

[tex]\mbox{ and }f(x + 2 \pi) = f(x) \mbox{<br /> for all other values of x, is an odd function.}[/tex]

Homework Equations



I know an odd function is definded as [tex]f(-x) = -f(x)[/tex]

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!
 
Last edited:
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If f(-x)=f(x), then the function is even. If f(-x)=-f(x), then the function is odd.
 
robphy said:
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?
 
mcfc said:
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.