The problem is that the most important part of Carateodory theorem (and sometimes of many theorems) is not the exact statement but the main idea(s) in the proof. In this theorem's proof there is a key idea to remember, which is what sets to call measurable for the (complete) measure it defines. These are the sets that are partitioned additively (for the outer measure) by all other sets.
Another idea to remember could be to remember how to build an outer measure. But this is kind of natural, and therefore easy to remember. Learn the proof and you will know all the statements you have seen.
Here is a different approach that uses the integral definition of the arcsine function. We will deduce the limit of interest, $\lim_{h\to 0}\frac{\sin(h)}{h}=1$, without appeal to geometry or differential calculus. (Note that $\cos(h)-1=-2\sin^2(h/2)$)
Instead, we only rely on elementary analysis of continuous functions and their inverses along with simple properties of the Riemann integral. To that end, we now proceed.
We define the sine function, $\sin(x)$, as the inverse function of the function $f(x)$ given by
$$\bbox[5px,border:2px solid #C0A000]{f(x)=\int_0^x \frac{1}{\sqrt{1-t^2}}\,dt }\tag 1$$
for $|x|< 1$.
NOTE: It can be shown that the sine function defined as the inverse of $f(x)$ given in $(1)$ has all of the familiar properties that characterize the circular function $\sin(x)$.
It is straightforward to show that since $\frac{1}{\sqrt{1-t^2}}$ is positive and continuous for $t\in (-1,1)$, $f(x)$ is continuous and strictly increasing for $x\in (-1,1)$ with $\displaystyle\lim_{x\to 0}f(x)=f(0)=0$.
Therefore, since $f$ is continuous and strictly increasing, its inverse function, $\sin(x)$, exists and is also continuous and strictly increasing with $\displaystyle \lim_{x\to 0}\sin(x)=\sin(0)=0$.
From $(1)$, we have the bounds (SEE HERE)
$$\bbox[5px,border:2px solid #C0A000]{1 \le \frac{f(x)}x\le \frac{1}{\sqrt{1-x^2}}} \tag 2$$
for $x\in (-1,1)$, whence applying the squeeze theorem to $(2)$ yields
$$\lim_{x\to 0}\frac{f(x)}{x}=1 \tag 3$$
Finally, let $y=f(x)$ so that $x=\sin(y)$. As $x\to 0$, $y\to 0$ and we can write $(3)$ as
$$\lim_{y\to 0}\frac{y}{\sin(y)}=1$$
from which we have
$$\bbox[5px,border:2px solid #C0A000]{\lim_{y\to 0}\frac{\sin(y)}{y}=1}$$
as was to be shown!
NOTE:
We can deduce the following set of useful inequalities from $(2)$. We let $x=\sin(\theta)$ and restrict $x$ so that $x\in [0,1)$. In addition, we define new functions, $\cos(\theta)=\sqrt{1-\sin^2(\theta)}$ and $\tan(\theta)=\sin(\theta)/\cos(\theta)$.
Then, we have from $(2)$
$$\bbox[5px,border:2px solid #C0A000]{y\cos(y)\le \sin(y)\le y\le \tan(y)} $$
which are the familiar inequalities often introduced in an introductory geometry or trigonometry course.
Best Answer
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.