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.

I'm not sure if this is how you want the proof, correct me if I'm wrong:
$$\frac{\sin(x+h)-\sin(x)}{h}=\frac{\sin(x)\cos(h)+\sin(h)\cos(x)-\sin(x)}{h}$$
Geometrically speaking, if the angle $h$ is small, then $\cos(h)$ is very nearly equal to the radius $1$, while $\sin(h)$ can be approximated by the arc of length $\text{radius}\times\text{angle}$ (two parallel sides of a triangle) which is $1\times h$, so:
$$\frac{\sin(x+h)-\sin(x)}{h}=\frac{\sin(x)+h\cos(x)-\sin(x)}{h}=\cos(x)$$
