This may be a silly question, but I cannot figure it out. I want to prove that

$ \Big\vert \frac{\sin(x)}{x} \Big\vert \leq 1 $ for $x\in[-1,0)\cup(0,1]$,

but I don't even know where to start.

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    $\begingroup$ In these answers, you will find several ways of seeing this bound and more. $\endgroup$ – Mark Viola Jul 14 '15 at 14:16

Hint By the MVT you have

$$\frac{\sin(x)-\sin(0)}{x-0} =\cos(c)$$ for some $c$.

  • $\begingroup$ Thank you! I knew it was easy, but I had no clue how to. Thanks! $\endgroup$ – user254489 Jul 14 '15 at 14:06
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    $\begingroup$ Well, this is potentially a circular argument inasmuch as in order to calculate the derivative of the sine function, one needs the limit of the left-hand side of the expression herein. $\endgroup$ – Mark Viola Jul 14 '15 at 14:18
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    $\begingroup$ This is potentially circular logic. One needs information about $\frac{\sin(x)}{x}$ to show that the derivative of $\sin$ is $\cos$. $\endgroup$ – Steven Gubkin Jul 14 '15 at 14:18
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    $\begingroup$ @Dr.MV it depends on how i define $sin/cos$ in the first place. If i do so via a power series i think we are safe here $\endgroup$ – tired Jul 14 '15 at 14:28
  • $\begingroup$ @tired Yes, and if the sine is defined in terms of the inverse function of $\int_0^x\frac{1}{\sqrt{1-t^2}}dt$ everything is fine. That is the purpose for my inserting the word "potentially." I believe that the answer ought to begin with the underlying assumption as to the definition used. It would enhance the answer and be more useful, don't you agree? $\endgroup$ – Mark Viola Jul 14 '15 at 14:32

If you draw a picture, it looks like we can use the triangle with points $A =(\cos(x),\sin(x))$, $B = (1,0)$, $C = (\cos(x),0)$. $AC$ has length $\sin(x)$, and $AB$ is the hypotenuse. So $AC$ is longer than $\sin(x)$. But then, since the shortest path connecting two points is a straight line, we must have $|AB| \leq x$. So we get $|\sin(x)| \leq |x|$ when $x$ is in the range $(-\pi/2, \pi/2)$.


Simple geometric proof:

Consider the unit circle, centred at $O$, with origin $A(1,0)$, and the point $M$ on the unit circle such that $\overset{\displaystyle\frown}{AM}=x$. Then $$\text{triangle } OAM \subset \text{circle sector }OAM $$ hence $$\text{area tr. }OAM=\frac12\lvert\sin x\rvert\le \text{area c. sect.}OAM=\frac12\lvert x\rvert.$$

  • $\begingroup$ How do you know the area formula for a sector? I think it also boils down to knowing this limit. $\endgroup$ – Steven Gubkin Jul 14 '15 at 15:22
  • $\begingroup$ That's high-school geometry: the area of a circle sector with angle $\theta$ and radius $r$ is $\frac12\theta\, r^2$. It's the same formula as the one for triangles: just think of a circle sector as a curvilinear triangle, with basis the arc and height a radius; .For $x=2\pi$ you get the area of a disk. Behind all this, there is the length of an arc as the limit of lrngths of inscribed polygon, that's all. It is the basis of a high schhol proof that the derivative of $\sin x$ is $\cos x$. $\endgroup$ – Bernard Jul 14 '15 at 15:36
  • $\begingroup$ Yes, I agree with your statement about the definition of arc length. This is why I think it is a bit weird to jump to areas, when the arc length gives it directly. $\endgroup$ – Steven Gubkin Jul 14 '15 at 15:41
  • $\begingroup$ Well, you have to consider also the triangulations derived from the polygonal lines. I didn't say it was te best proof if you want to go deep into the foundations. But if you accept a number of natural facts, that you know can be proved in detail, it is an intuitive proof. $\endgroup$ – Bernard Jul 14 '15 at 15:47

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