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

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 $$[m,n]$$.

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.