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

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