Proving a function is peridoic

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Show that cos(x) + cos([itex]\alpha[/itex]x) is periodic if [itex]\alpha[/itex] is a rational number.


Ok So I don't think I have ever done a question proving periodicity. But by definition a function f is periodic if:

f(x + p) = f(x)

so then:

f(x+p) = cos(x +p) + cos([itex]\alpha[/itex](x+p))
= [cos(x)cos(p)-sin(x)sin(p)] + [cos([itex]\alpha[/itex]x)cos([itex]\alpha[/itex]p)-sin([itex]\alpha[/itex]x)sin([itex]\alpha[/itex]p)


and this is where I am stuck, so from what I've learned I have to solve for p is some fashion, I'm just not sure how,
 
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trap101 said:
Show that cos(x) + cos([itex]\alpha[/itex]x) is periodic if [itex]\alpha[/itex] is a rational number.

The first thing to do with a rational number is write it as [itex]\alpha = r/s[/itex] for co-prime integers r and s.

Ok So I don't think I have ever done a question proving periodicity. But by definition a function f is periodic if:

f(x + p) = f(x)

so then:

f(x+p) = cos(x +p) + cos([itex]\alpha[/itex](x+p))
= [cos(x)cos(p)-sin(x)sin(p)] + [cos([itex]\alpha[/itex]x)cos([itex]\alpha[/itex]p)-sin([itex]\alpha[/itex]x)sin([itex]\alpha[/itex]p)


and this is where I am stuck, so from what I've learned I have to solve for p is some fashion, I'm just not sure how,

You already know that [itex]\cos x[/itex] has period [itex]2\pi[/itex] and [itex]\cos(\alpha x)[/itex] has period [itex](2\pi/\alpha)[/itex]. So you need to find [itex]p[/itex] such that [itex]x + p = x + 2n\pi[/itex] and [itex]x + p = x + 2m\pi/\alpha[/itex] for integers n and m.
 
You have expressions for f(x) [given], and f(x+p) [derived above].
Now assume that alpha is rational, m/n, and then see if there is any period p which satisfies the condition f(x)=f(x+p).