# A very weird problem

1. Sep 24, 2006

### kolycholy

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)??

2. Sep 24, 2006

### SGT

$$cos 2\pi x sin 2\pi x = \frac{1}{2}sin 4\pi x$$,
so the period is $$\frac{4\pi}{2\pi} = 2$$

3. Sep 24, 2006

### Timbuqtu

That's almost correct, but the period is 1/2: if x increases from 0 to 1/2, then $$4 \pi x$$ increases from 0 to $$2 \pi$$.

4. Sep 24, 2006

### antonantal

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: Sep 24, 2006
5. Sep 24, 2006

### SGT

You're right

6. Sep 24, 2006

### kolycholy

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. Sep 25, 2006

### HallsofIvy

Staff Emeritus
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. Sep 25, 2006

### SGT

Here is a plot.

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