[Sinusoidal Function Period]: demonstrate period of $f(x)=\sin(k\,x)$

Question

I'm an Italian student (sorry for my english).

I have to demonstrate that the period $T$ of the general function $f(x)=\sin(k\, x)$ is equal to $2\,\pi/k$. I understand the idea behind the demonstration, but there is a particular unclear for me.

Dim. Knowing that $f(x)$ is a periodic function, must be exist $T\in\mathbb{R}^+$ such that:

$\sin(k\, x) = \sin[k\,(x + m\,T)]$

Knowing that sine function is periodic of period $T_{\sin}=n\,2\,\pi$, it can be written:

$\sin(k\, x) = \sin(k\,x + n\,2\,\pi)$

[stop]

The problem is at this point: in my study source (here) the coefficient that multiplies $2\,\pi$ is the same that multiplies $T$, that is $m$ and i don't understand why. Can you help me?

Thanks.

Answer 1

I think the definition you have is equivalent to the following: $f(x)$ is periodic with (principle/minimum) period $T$ if $T>0$ is the smallest positive value such that,

$$f(x)=f(x+T) \quad \text{for all } x$$

Notice that this would agree with setting $k=1$ in your definition and also you can use the above to show that for any integer $n\in\mathbb{Z}$ we have $f(x)=f(x+nT)$ for all $x$.

Now to figure out the period of $f(x)=\sin(kx)$ we can compute:

$$f(x+T) = \sin(k(x+T)) = \sin(kx+kT) \quad (1)$$

We know that the function $h(x)=\sin(x)$ has minimum period $2\pi$ and so we have $h(x+2\pi)=h(x)$ for all $x$ and further $2\pi$ is the smallest positive value with this property. Noting that $f(x)=h(kx)$ we can replace $x$ with $kx$ to obtain:

$$h(kx+2\pi)=h(kx) \quad \text{for all } x$$

which implies,

$$\sin(kx+2\pi) = \sin(kx) \quad \text{for all } x$$

Comparing with $(1)$ we must have $kT=2\pi$ from which we obtain $T=2\pi/k$.

Hope this helps :)