If $f(x)$ has period $2\pi$, then $F(t) = f(\omega t) = f(\frac{2\pi}{T}t)$ has period T

Question

If $f(x)$ has period $2\pi$, then $F(t) = f(\omega t) = f(\frac{2\pi}{T}t)$ has period T (substitute $\frac{2\pi}{T}t = x,$ and $\omega = \frac{2\pi}{T}).$

This should be so easy to see, and yet I don't get it.

$f(x) = f(\frac{2\pi}{T}t).$ I should try to see that $f(\frac{2\pi}{T}t) = f(\frac{2\pi}{T}t + T).$

I know that $f(\frac{2\pi}{T}t) = f(\frac{2\pi}{T}t + 2\pi),$ because of the hypothesis. Then

$f(\frac{2\pi}{T}t + 2\pi) = f(2\pi(\frac{t + T}{T})) = f(\omega (t + T)).$ But that doesn't say anything to me...

Answer 1

$$F(x+T)=f(\frac{2\pi}{T}(x+T))=f(\frac{2\pi}{T}x + 2\pi)=f(\frac{2\pi}{T}x)=F(x)$$