How to prove that $d \sin(x)/dx = \cos(x)$ without circular logic such as L'Hôpital's rule? How do I prove that the derivative of $\sin$ is $\cos$  without resorting to L'Hôpital's rule (circular logic)?
This part is easy:
$$
\begin{align*}
\sin'(x) &= \lim_{\Delta x \to 0} \frac{\sin(x + \Delta x) - \sin(x)}{\Delta x}  \\
\sin'(x) &= \lim_{\Delta x \to 0} \frac{\cos(x)\sin(\Delta x) + \sin(x) \cos(\Delta x) - \sin(x)}{\Delta x}  \\
\sin'(x) &= \lim_{\Delta x \to 0} \left(\cos(x)\frac{\sin(\Delta x)}{\Delta x} + \sin(x)\frac{\cos(\Delta x) - 1}{\Delta x}\right)  \\
\end{align*}
$$
but where do I go from here?
 A: Starting off where you finished and credits to this pdf:
$$\sin'(x) = \lim_{\Delta x \to 0} \left(\cos x \frac{\sin \Delta x}{\Delta x} + \sin x \frac{\cos \Delta x - 1}{\Delta x} \right)$$ $$= \cos x \lim_{\Delta x \to 0} \frac{\sin \Delta x}{\Delta x} - \sin x \lim_{\Delta x \to 0} \frac{1 - \cos \Delta x}{\Delta x}$$

Part I: prove that $\displaystyle \lim_{\Delta x \to 0} \frac{\sin \Delta x}{\Delta x} = 1$.

Assuming that $OA = 1$ (i.e. it's a unit circle), we have $\sin \theta = BC$. We then note that
[Area of triangle $AOD$] > [Area of sector $AOC$] > [Area of triangle $AOC$]
which means
$$\frac{1}{2} \frac{\sin \theta}{\cos \theta} > \frac{1}{2} \theta > \frac{1}{2} \sin \theta$$
From this we get $$\cos \theta < \frac{\sin \theta}{\theta} < 1$$
and since $\displaystyle \lim_{\theta \to 0} \cos \theta = 1$, we have that $\displaystyle \lim_{\theta \to 0} = 1$.

Part II: prove that $\displaystyle \lim_{\Delta x \to 0} \frac{1 - \cos \Delta x}{\Delta x} = 0$.
$$\lim_{\Delta x \to 0} \frac{1 - \cos \Delta x}{\Delta x} = \lim_{\Delta x \to 0} \frac{(1 - \cos \Delta x) (1 + \cos \Delta x)}{(1 + \cos \Delta x)\Delta x} = \lim_{\Delta x \to 0} \frac{1 - \cos^2 \Delta x}{(1 + \cos \Delta x)\Delta x}$$
$$ = \lim_{\Delta x \to 0} \frac{\sin^2 \Delta x}{(1 + \cos \Delta x)\Delta x} = \lim_{\Delta x \to 0} \frac{\sin \Delta x}{\Delta x} \frac{\sin \Delta x}{1 + \cos \Delta x}$$
We have proven in part I that $\displaystyle \lim_{\Delta x \to 0} \frac{\sin \Delta x}{\Delta x} = 1$, and we note that $\displaystyle \frac{\sin \Delta x}{1 + \cos \Delta x} = \frac{0}{1+1} = 0$.

We finally put all this together to get that $$\boxed{\displaystyle \sin'(x) = \cos x \lim_{\Delta x \to 0} \frac{\sin \Delta x}{\Delta x} - \sin x \lim_{\Delta x \to 0} \frac{1 - \cos \Delta x}{\Delta x} = \cos x \cdot 1 - \sin x \cdot 0 = \cos x}$$
A: Read this post, as it is very helpful: http://www.wyzant.com/resources/lessons/math/calculus/derivative_proofs/sinx
Since you are unsatisfied and do not understand the proof, I will further explain. As we learned in precalc, $\lim_{x\rightarrow 0} \frac{\sin x}{x} = 1$ and $\lim_{x\rightarrow 0} \frac{1- \cos x}{x} = 0 $. Please search a proof of these identities in another question if still confused, as that is a much simpler problem that can be solved using the squeeze theorum. Make these substitutions and you will have your answer. 
A: In my introduction to limits, we used a lemma
\begin{equation}
 \lim_{x\rightarrow 0} \frac{\sin x}{x} = 1.
\end{equation}
This was proved by drawing a picture and using the definition of the sine function and the definition of radians, along with the geometric "axiom" "the shortest distance between two points is a straight line segment".
That lemma can then be used to calculate $\lim_{x\rightarrow 0} \frac{1-\cos x}{x}$ and $\sin'(x)$.
