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I do not understand this consideration: > By considering a triangle with hypotenuse 1 and a very small “opposite” side, it’s not hard to see geometrically that

I do not understand this consideration: > By considering a triangle with hypotenuse 1 and a very small “opposite” side, it’s not hard to see geometrically that sin(x)≈x and cos(h)=x when x is small I fail to see how you can "see" finer than sin(h) -> 0 & cos(h) -> 1 From the limit definitions you actually need : * (1-cos(h)) / h -> 0 * sin(h)/h -> 1 (which correspond to the derivatives at 0).

lupire 2y ago
Your limit definition is the same as the part you quoted, so it's not clear what your question is. I also don't see what you are quoting. Curvature is inverse of radius. Decreasing angle is equivalent to increasing radius, and this decreasing curvature. This, as angle decreases, the curve becomes close to a straight line, and that straight line approaches a vertical line. As usual, 3B1B created a quintessential visualization and explanation.https://m.youtube.com/watch?v=S0_qX4VJhMQ https://m.youtube.com/watch?v=S0_qX4VJhMQ
- enizor2 2y ago
  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.html 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.
philsnow 2y ago
That one line was the part that stood out to me the most as well, but: If you zoom in sufficiently at x = 0, f(x) = sin(x) looks indistinguishable from f(x) = x, whereas g(x) = cos(x) looks indistinguishable from g(x) = 1. (also, sin(x) is negative approaching 0 from the left and positive approaching 0 from the right)
- enizor2 2y ago
  > If you zoom in sufficiently at x = 0, f(x) = sin(x) looks indistinguishable from f(x) = x, whereas g(x) = cos(x) looks indistinguishable from g(x) = 1. You can't use the plot as you only know the triangle definition yet. (And "looks indistinguidable" is rather handwavy). > (also, sin(x) is negative approaching 0 from the left and positive approaching 0 from the right) That only tells you that sin(0)=0
dullcrisp 2y ago
Came here to say the same thing. But I suppose if you extend cosine into negative values you'll see it has a maximum at zero so its derivative must be zero. Don't know about sin'(0) offhand but you'd think it wouldn't be hard.

lupire 2y ago

Your limit definition is the same as the part you quoted, so it's not clear what your question is. I also don't see what you are quoting. Curvature is inverse of radius. Decreasing angle is equivalent to increasing radius, and this decreasing curvature. This, as angle decreases, the curve becomes close to a straight line, and that straight line approaches a vertical line. As usual, 3B1B created a quintessential visualization and explanation.https://m.youtube.com/watch?v=S0_qX4VJhMQ https://m.youtube.com/watch?v=S0_qX4VJhMQ

enizor2 2y ago
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.html 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.

philsnow 2y ago

That one line was the part that stood out to me the most as well, but: If you zoom in sufficiently at x = 0, f(x) = sin(x) looks indistinguishable from f(x) = x, whereas g(x) = cos(x) looks indistinguishable from g(x) = 1. (also, sin(x) is negative approaching 0 from the left and positive approaching 0 from the right)

enizor2 2y ago
> If you zoom in sufficiently at x = 0, f(x) = sin(x) looks indistinguishable from f(x) = x, whereas g(x) = cos(x) looks indistinguishable from g(x) = 1. You can't use the plot as you only know the triangle definition yet. (And "looks indistinguidable" is rather handwavy). > (also, sin(x) is negative approaching 0 from the left and positive approaching 0 from the right) That only tells you that sin(0)=0

dullcrisp 2y ago

Came here to say the same thing. But I suppose if you extend cosine into negative values you'll see it has a maximum at zero so its derivative must be zero. Don't know about sin'(0) offhand but you'd think it wouldn't be hard.