
#1
Aug2209, 03:49 PM

P: 18

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! 



#2
Aug2209, 05:24 PM

P: 200

If f(x)=f(x), then the function is even. If f(x)=f(x), then the function is odd.




#3
Aug2209, 05:38 PM

Sci Advisor
HW Helper
PF Gold
P: 4,108

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. 



#4
Aug2209, 06:16 PM

P: 18

How to show a function is even/oddlook 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? 



#5
Aug2209, 06:46 PM

P: 32




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