Derivative of $\sin(x)$: how to avoid circular reasoning? Dad here, helping his kid out with math.
I get it that to determine the derivative of $\sin(x)$ you do:
$$\frac{\sin(x+h) - \sin(x)}h = \frac{\cos(x)\sin(h)}h \to \cos(x)$$
However I assume that $\sin(h) = h$ for $h\to 0$. This correct because the "angle" (not sure what the correct word is in English) of $\sin(x) $ is $1$ for $x=0$. But this is what I'm trying to prove in the first place. So a circular reference. How to solve this?
 A: The fact that $\sin(h) \approx h$ for small $h$ is usually proven non-circularly (although circles are involved :D) by resorting to the following inequalities
$$\begin{split}
\sin (h) &\leqslant h \leqslant \tan (h),\\
\cos (h) &\geqslant \frac 1 2,
\end{split} \tag 1$$
both valid when $0 < h < \pi/3$. You can see they hold by drawing a unit circle; here's a helpful picture for the first one:
$\hspace{1cm}\hspace{1cm}\hspace{1cm}\hspace{1cm}\hspace{1cm}$
Indeed, when $0 < h < \pi/3$, we obtain, by rearranging terms in inequality $(1\text a)$ that
$$0 \leqslant \sin (h) - h \leqslant \tan (h) - \sin (h ). \tag2$$
However, by the definition of $\tan$ and the trigonometric identity $1-\cos(h) = 2\sin^2(h/2)$,
$$\tan (h) - \sin (h )= \frac{\sin (h)}{\cos (h)}(1-\cos (h)) = \frac{\sin (h)}{\cos (h)}\cdot 2 \sin^2\left( \frac h 2 \right). $$
By employing inequality $(1\text a)$ again in the forms $\sin(h/2) \leqslant h/2$ and $\sin(h) \leqslant h$, and inequality $(1\text b)$ for the cosine at the bottom, we see that
$$\frac{\sin (h)}{\cos (h)}\cdot 2 \sin^2\left( \frac h 2 \right) \leqslant \frac h {1/2} \cdot 2 \left(\frac h 2\right)^2 = h^3. \tag 3$$
Merging inequalities $(2)$ and $(3)$ entails that $0\leqslant \sin(h) - h \leqslant h^3$, that is (adding $h$ to all sides) $h \leqslant \sin(h) \leqslant h^3 + h$, that is (dividing everything by $h$) $$1 \leqslant \frac{\sin(h)}{h} \leqslant 1+ h^2,$$ which ultimately implies that
$$\lim_{h\to 0^+} \frac{\sin(h)}{h} = 1 $$
by the squeeze theorem (both $1$ and $1+h^2$ clearly tend to $1$ as $h\to 0^+$).
The complete limit (as $h\to 0$ from both sides) follows from the fact that $\sin(-h) = -\sin (h)$.
A: It should rather be
$$\lim_{h \to 0}\frac{\sin(x+h) - \sin(x)}h =\lim_{h \to 0} \frac{2 \cos (x+h/2) \sin(h/2)}{h} =\cos x \lim_{h \to 0} \frac{\sin(h/2)}{h/2}=\lim_{h \to 0} \frac{h/2-h^3/48+O(h^5)}{h/2}\cos x=\cos x$$
Here we have used $\sin z=x-z^3/3!+h^5/5!+...$
