from Carathéodory Derivative definition to the derivative of $\sin(x)$

Because of definition of $\sin(x)$, it is very unlikely that we will find a nice representation of $\phi$ by elementary functions. This is why it isn't as easy as for $x^n$:
First, we recall that $\sin(x)$ is analytically defined as infinite series, $\sum\limits_{i=0}^\infty \frac{(-1)^n}{(2n+1)!}x^{2n+1} $. Then, $$ \sin(x) - \sin(a) = \sum\limits_{i=0}^\infty \frac{(-1)^n}{(2n+1)!}(x^{2n+1}-a^{2n+1}) = (x-a) \cdot \sum\limits_{i=0}^\infty \frac{(-1)^n}{(2n+1)!}(x^{2n} + ax^{2n-1} + ... + a^{2n}) = (x-a) \cdot \phi(x).$$ Also, we have $\phi(a) = \sum\limits_{i=0}^\infty \frac{(-1)^n}{(2n+1)!}((2n+1)a^{2n}) = \sum\limits_{i=0}^\infty \frac{(-1)^n}{(2n)!}a^{2n} = \cos(a). $ Notice that, unlike $x^n$-example, here (even if we overlook convergence problems) $\phi$ doesn't have a nice representation by elementary functions, so that's why it isn't as easy to do the same thing as for $x^n$ and get a Caratheodory derivative.

The Weierstrass-Caratheodory formulation asserts that $f\colon E\rightarrow\mathbb{R}$ is differentiable at a point $x_0\in E$ if and only if there exists a function $\phi$ continuous at $x_0$ such that $$f(x)=f(x_0)+\phi(x)(x-x_0)$$ and the derivative is given by $f^{\prime}(x_0)=\phi(x_0)$.

Take $f(x)=\sin(x)$, then suppose we have the following equation: $$\sin(x)=\sin(x_0)+\phi(x)(x-x_0)$$ Rearranging gives us: $$\phi(x)=\frac{\sin(x)-\sin(x_0)}{x-x_0}$$ As you can see, if you can show that this quotient has a limit as $x\rightarrow x_0$, then you are done. But this is exactly the same as finding the derivative of $\sin(x)$ from first principle.

I could go on and derive it, but it is an easy exercise to show that it has a limit by using double-angle formula :)