Prove $[\sin x]' = \cos x$ without using $\lim\limits_{x\to 0}\frac{\sin x}{x} = 1$ I came across this question: How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$?
From the comments, Joren said:

L'Hopital Rule is easiest: $\displaystyle\lim_{x\to 0}\sin x = 0$ and $\displaystyle\lim_{x\to 0} = 0$, so $\displaystyle\lim_{x\to 0}\frac{\sin x}{x} = \lim_{x\to 0}\frac{\cos x}{1} = 1$.

Which Ilya readly answered:

I'm extremely curious how will you prove then that $[\sin x]' = \cos x$

My question:  is there a way of proving that $[\sin x]' = \cos x$ without using the limit $\displaystyle\lim_{x\to 0}\frac{\sin x}{x} = 1$. Also, without using anything else $E$ such that, the proof of $E$ uses the limit or $[\sin x]' = \cos x$.

All I want is to be able to use L'Hopital in $\displaystyle\lim_{x\to 0}\frac{\sin x}{x}$. And for this, $[\sin x]'$ has to be evaluated first.

Alright... the definition that some requested. 
Def of sine and cosine: Have a unit circumference in the center of cartesian coordinates. Take a dot that belongs to the circumference. Your dot is $(x, y)$. It relates to the angle this way: $(\cos\theta, \sin\theta)$, such that if $\theta = 0$ then your dot is $(1, 0)$.
Basically, its a geometrical one. Feel free to use trigonometric identities as you want. They are all provable from geometry.

There is a very nice Khan Academy video on it. I won't get into too much detail but basically you construct the following 
Then you take limits of the inequality $$1\geq|\frac{\sin\theta}{\theta}|\geq|\cos\theta|$$
to get
$$\lim_{\theta->0}1\geq|\frac{\sin\theta}{\theta}|\geq|\cos\theta|$$
$$1\geq|\frac{\sin\theta}{\theta}|\geq|1$$
and and therefore by the squeeze theorem we have $\lim_{\theta->0}\frac{\sin\theta}{\theta}=1$.
Having proved that for yourself you can then use the definition of a derivative that is $$f'(x)=\lim_{h->0}\frac{f(x+h)-f(x)}{h}$$ for $f(x)$ beeing $\cos(x)$. Hope that helps.
