Period of a composition of sinusoidal functions

physi_marc

I have two sinusoidal functions of the form f(x) = sin(sin(2pi*x)) and g(x) = cos(sin(2pi*x)).

How do I find the period of f and g analytically? Of course, I can use any graphing software to find that the period of f is 1 and the period of g is 1/2, but I want to do this analytically. How do I set an equation that I can solve to find the period?

I know that f(x) = f(x+T), so sin(sin(2pi*x)) = sin(sin(2pi*(x+T))) but I keep getting T=0 or something which includes arcsin(2pi). Maybe I'm on the right track, but I'm not sure how to handle complex numbers to get a real period.

Phyphor

It's been a while, but the usual way is to use trig identities to work with the sin(x+T) term IIRC, but I'm not sure you can for sin(sin(x)). One way to approach it is as sin(f(x)) - the period of the outer function will be related to the period of the inner function. I'm not sure if this is the best way or even what you should be doing, but it's what I'd try

In this case, the period of sin(2pi*x) is simply defined to be 1. Trig identities will give you this.

The period of sin(f(x)) can then be:

sin(f(x)) = sin(f(x+T))

We know f(x) = f(x+Tf), so T can't be more than Tf=1, but it could be less.


Also useful is f'(x) = 2pi*cos(2pi*x)
And f''(x) = -4pi²*sin(2pi*x)

Now, you can find minima, maxima, intercepts and concavity using these, but the function starts at 0, is increasing from [0,0.25] to 1, decreasing from [0.25,0.75] to -1, crossing zero at 0.5 and then finally increasing to zero from [0.75,1]. In each of these segments it's a 1:1 mapping, so each unique x in the period outputs a unique value in the range, within that segment.


sin(f(x))

The easy way is to just look at sin(f(x)) in those outputs.

sin(x) where x = [0,1] is increasing. [1,-1] is decreasing across 0. [-1,0] is increasing back to zero. No sub-periodicity.


cos(f(x))

cos(x) where x = [0,1] is decreasing. [1,0] is increasing to 1. [0, -1] is decreasing. [-1, 0] is increasing to 1. This is because cos(x) = cos(-x). Thus there's a period of 0.5 in this, so the full period is T = Tf/2 or 0.5

physi_marc

Thanks for the help!

I understand why cos(sin(x)) has a period of 1/2 now. I still would like an analytical way of finding this, but it'll do for now.