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?

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. 
