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.

What  is required here is not a proof of $\sin'=\cos$ without using $$\lim_{\phi\to0}{\sin\phi\over\phi}=1\tag{1}\ ,$$ but a proof of the basic limit $(1)$ using the "geometric definition" of sine provided by the OP. To this end we shall prove "geometrically" that
$$\sin\phi<\phi\leq\tan\phi\qquad\left(0<\phi<{\pi\over2}\right)\ .\tag{2}$$
The inequalities $(2)$ imply
$$\cos\phi\leq{\sin\phi\over\phi}<1\qquad\left(0<\phi<{\pi\over2}\right)\ ,$$
so that $(1)$ follows from $\lim_{\phi\to0}\cos\phi=1$ and the squeeze theorem.
Comparing segment length to arc length immediately shows that $$\sin\phi=2\sin{\phi\over2}\cos{\phi\over2}\leq2\sin{\phi\over2}<\phi\ .$$ In order to prove that $\phi\leq\tan\phi$ we somehow have to use how the length of curves is defined. I'm referring to the following figure. If $0\leq\alpha<\beta<{\pi\over2}$ then
$$s'=\tan\beta-\tan\alpha={\sin(\beta-\alpha)\over\cos\beta\cos\alpha}=2\sin{\beta-\alpha\over2}\>{\cos{\beta-\alpha\over2}\over\cos\beta}\>{1\over\cos\alpha}>2\sin{\beta-\alpha\over2}=s\ .$$
It follows that the length $L_P$ of any polygonal approximation $P$ to the circular arc $AB$ is $\ <\tan\phi$, and this implies
$$\phi:=\sup_PL_P\leq\tan\phi\ .$$

