limits - Prove $[sin x]' = cos x$ without using $limlimits_{xto 0}frac{sin x}{x} = 1$

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 $[\sin x]' = \cos x$.

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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.

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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.