I"ve been trying to lớn see why it could be true but so far have been unable to come up with any success.

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Could someone show a way khổng lồ prove this?


PS. Intuitively, this seems lớn make sense since an arc is always bigger than its sine in magnitude. But how lớn show this rigorously?



The way lớn make this rigorous is via the Mean Value theorem. For any $x there is a $cin (x,y)$ s.t.$$sin(x)-sin(y) = cos(c)(x-y).$$But since $|cos(z)|leq 1$ for any $zin riz15.combbR$, you can take the absolute value và immediately get the result.

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$$ eginaligned |sin(x)-sin(y)|&=left|int_y^x cos(t)dt ight|\&leqslantint_min(x,y)^max(x,y)underbrace_leqslant 1dt\&leqslantmax(x,y)-min(x,y)\&=|x-y| endaligned$$



Using the identity$$sin x - sin y = 2 sin fracx-y2 cos fracx+y2$$and the fact that $sin$ is an odd function it will indeed follow from inequality$$sin t le t riz15.comrm~~for~~ t ge 0$$

There are many analytical ways of solving this, my favourite being the comparison of derivatives.

Geometrically you can argue that $sin t$ (blue segment) is shorter than red which is shorter than an arc of length $t$:


(This works for $t , what includes $t=1$ which is the maximal value of $sin t$).

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