Prove the monotony of $\frac{\sin x}{x}$ I was wondering whether it is possible to prove that $f(x)=\frac{\sin x}{x}$ is strictly decreasing in $(0, \pi]$ and strictly increasing in $[-\pi , 0)$ without using any derivatives at all. If it is possible, how?
 A: By the Weierstrass product:
$$\frac{\sin x}{x}=\prod_{n\geq 1}\left(1-\frac{x^2}{n^2\pi^2}\right)$$
but for any $n\geq 1$ the function $f_n(x)=\left(1-\frac{x^2}{n^2\pi^2}\right)$ is non-negative and decreasing over $\left[0,\pi\right]$.
It follows that $f(x)=\frac{\sin x}{x}$ is non-negative and decreasing  over $\left[0,\pi\right]$, and we just need to notice that $f(x)$ is an even function to prove the claim.
A: Let $y=x+h$ where $h>0$. We want to show:
$$\frac{\sin(x+h)}{x+h}-\frac{\sin(x)}{x}<0$$
for $x,y \in (0,\pi]$. So we compute:
$$\frac{\sin(x+h)}{x+h}-\frac{\sin(x)}{x}=\frac{\sin(x+h)x-\sin(x)(x+h)}{x(x+h)}$$
Because $x(x+h)>0$ is is sufficient to show that:
$$\sin(x+h)x-\sin(x)(x+h)<0$$
Next we use formula for $\sin$ of sum:
$$\sin(x)\cos(h)x+\sin(h)\cos(x)x-\sin(x)(x+h)<0$$
But:
$$\cos(h) \leq 1$$
So it suffices to show:
$$\sin(x)x+\sin(h)\cos(x)x-\sin(x)(x+h)<0$$
Next:
$$\sin(x)x+\sin(h)\cos(x)x-\sin(x)(x+h)=\sin(h)\cos(x)x-h\sin(x)$$
So it's equivalent:
$$\frac{\cos(x)x}{\sin(x)}<\frac{h}{\sin(h)}$$
But $h>\sin(h)$, so it is sifficent to show:
$$\cos(x)x-\sin(x)<0$$
(Note that using derivatives gives us the same inequality to prove!) We can prove that by any method (for example, compare $x$ and $\tan(x)$ on $(0,\frac{\pi}{2})$, on $(\frac{\pi}{2},\pi)$ it is obvious).
A: Observe that
$$\frac{\sin x}{x} = \frac{1}{x}\int_0^x \cos t\, dt.$$
This is the average of $\cos t$ over $[0,x].$ Now $\cos t$ is strictly decreasing on $[0,\pi].$ So if $0\le x <y \le \pi,$ the average over $[0,y]$ is smaller than the average over $[0,x],$ simply because we are adding in strictly smaller values to the mix.
