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?

Thanks.

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


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


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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 $t$:

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


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