Proof that $\sin {x}$ is infinitely continuously differentiable over $[m,n]$

Question

I am trying to prove that $\sin {x}$ is infinitely continuously differentiable over $[m,n]$ where $m$ and $n$ are real numbers. Here is my attempt at doing so. Is my proof complete? If not, what can I do to improve it? Thank you in advance.

Since,

$\frac{d}{dx}\sin{x} = \cos{x}$,

$\frac{d^2}{dx^2}\sin{x} = -\sin{x}$,

$\frac{d^3}{dx^3}\sin{x} = -\cos{x}$,

and

$\frac{d^4}{dx^4}\sin{x} = \sin{x}$,

the derivatives of $\sin{x}$, are periodic. Since the first four derivatives of $\sin{x}$ are continuous over $[m,n]$ where $m$ and $n$ are real numbers, $\sin{x}$ must be differentiable an infinite amount of times over $[m,n]$.

Answer 1

A more formal way to show this is by induction. We know that $f(x) = \sin(x)$ is continuous. Also, $f'(x) = \cos(x)$ is continuous. Now, assume that $f^{(2n-1)}(x) = (-1)^{n+1}\cos(x)$ for all $n = 1,2,...$. Then, $f^{(2(n+1)-1)}(x) = (-1)^{n+1}*(-\cos(x)) = (-1)^{(n+1)+1}\cos(x)$, which is continuous. This proves that all odd derivatives are continuous. For even derivatives, we just take any odd derivative and differentiate it once: $\frac{d}{dx}f^{(2n-1)}(x) = (-1)^{n+1}*(-\sin(x)) = (-1)^{n+2}\sin(x)$, which is also continuous. So, for all $n \geq 0$, $f^{(n)}(x)$ is continuous.