Solving Cos(2πx)Sin(2πx) & Its Multiplied Variation

[kolycholy]

what would be the period of cos(2pi*x)*sin(2pi*x)??
calculator is telling me it's 2, but i somehow don't want to believe that



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

 

[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

 

[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.

 

[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).

 

[SGT]

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.

You're right

 

[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!

 

[HallsofIvy]

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.

 

[SGT]

Here is a plot.