A very weird problem

  • Thread starter kolycholy
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  • #1
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what would be the period of cos(2pi*x)*sin(2pi*x)??
calculator is telling me it's 2, but i somehow dont wanna believe that



also, what would be the period of cos(2pi*m*x)*sin(2pi*m*x)??
 

Answers and Replies

  • #2
SGT
[tex]cos 2\pi x sin 2\pi x = \frac{1}{2}sin 4\pi x[/tex],
so the period is [tex]\frac{4\pi}{2\pi} = 2[/tex]
 
  • #3
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That's almost correct, but the period is 1/2: if x increases from 0 to 1/2, then [tex] 4 \pi x[/tex] increases from 0 to [tex]2 \pi[/tex].
 
  • #4
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As a rule, the period of sin(n*x) or cos(n*x) is 2*pi/n. More generally, if f(x) is a periodic function with period T, than f(n*x) has period T/n. You can deduce this simply by thinking that when you multiply the argument of a periodic function by a number you actually multiply its speed of oscillation by that number, meaning that you divide its period by that number.
So the period of cos(2pi*x) is 1. The same for sin(2pi*x). To calculate the period of cos(2pi*x)*sin(2pi*x) you can write these two functions in the complex form and see that their product has the same period as cos(4pi*x) which is 0.5. Do the same for cos(2pi*m*x)*sin(2pi*m*x).
 
Last edited:
  • #5
SGT
Timbuqtu said:
That's almost correct, but the period is 1/2: if x increases from 0 to 1/2, then [tex] 4 \pi x[/tex] increases from 0 to [tex]2 \pi[/tex].
You're right
 
  • #6
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i agree with you people that the period should be 1/2.
but when i look at the graph, it seems like it's 2!
can anybody plot for me and see if they are getting the period 1/2 or 2, just based on the plot?
thanks!
 
  • #7
HallsofIvy
Science Advisor
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I graphed it on a TI 93 and it looks to me like the period is clearly 1/2. Of course, since 2 is a multiple of 1/2, 2 is a period.
 
  • #8
SGT
Here is a plot.
 

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