How to show $ \Big\vert \frac{\sin(x)}{x} \Big\vert $ is bounded by $1$?

Question

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.

Answer 1

Hint By the MVT you have

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

Answer 2

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 $(-\pi/2, \pi/2)$.

Answer 3

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 $$\text{area tr. }OAM=\frac12\lvert\sin x\rvert\le \text{area c. sect.}OAM=\frac12\lvert x\rvert.$$