I quote the second paragraph of the Derivatives section. (which was edited to a better, but not yet enough, sin(h)≈h and cos(h)≈1 when h is close to zero).

I perfectly understand that around 0, sin(x) ~ x and cos(x) = 1 + o(x) but it isn't obvious geometrically, unlike what the article implies.

From my point of view, increasing radius / decreasing curvature only gets you sin(x) -> 0 ; cos(x) -> 1, but that isn't enough to obtain the derivatives.

I found a geometric proof in [1] but that part is the longest and hardest of the page. I was wondering whether the author found a clearer way to express is.

[1] https://www.mathsisfun.com/calculus/derivatives-trig-proof.h...

EDIT: after looking at 3B1B's video, the "small" triangle d(sinΘ) by dΘ figure would be a better way to explain the derivative, rather than an "not hard to see geometrically" approximation that isn't enough to conclude.