I am trying to prove that $\frac{d^n\theta}{dx^n}$ where $x = 1-\cos\frac{\theta}{2}$ exists $\forall n > 0$. I found that $\frac{d\theta}{dx} = \frac{2}{\sin\frac{\theta}{2}}$ and based on this answer I think I only need to prove that $\sin\frac{\theta}{2}$ is smooth (notice is a function of $\theta$ and its firsts derivatives looks like fun).
I also found that $\cos^{-1}(x) = \frac{\pi}{2} + \sum_{k=0}^\infty \frac{x^{1+2k}\left(\frac{1}{2}\right)_k}{k!+2kk!}$ (at wolfram alpha). But I am not sure that the existence of its Taylor Series implies its smoothness. I thought that by linearity of differentiation and because all terms in the series are polynomials (smooth funcions) there could be a way to make it work.
Is there a simple (elegant) way to prove its smoothness? Until now my only approach is to find a pattern of the n-th derivative, however I am actively trying to avoid that because of this.
edit:
I am trying to prove that $\theta = 2 \cos^{-1}(1-x)$ is smooth (in other words, has infinitely many derivatives $\exists \frac{d^n\theta}{dx^n}\, \forall n>0$)
I tried different approaches, but based on the Taylor Expansion of $\cos^{-1}$ I found that $\theta = 2 \cos^{-1}(1-x)$ can be expressed by polynomials (which are smooth functions) and therefore its derivatives.
I have never worked with Taylor Series, so I am not really confident about this last statement.
Side note: My original expression was $x=1-\cos\frac{\theta}{2},\, -1 \le x \le 1$ (which implies $\theta = 2 \cos^{-1}(1-x)$) and because of that I derived $\frac{d\theta}{dx} = \frac{2}{\sin\frac{\theta}{2}}$ applying $\frac{d}{dx}$ to both sides.
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