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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?

agha
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    Since the function is even, the statement about $[-\pi,0)$ follows from the statement about $(0,\pi]$.

    You need to show that if $0<x<y\leq \pi$ then $y\sin x-x\sin y>0$.

     – Thomas Andrews Aug 23 '15 at 12:26

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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.

Jack D'Aurizio
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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).

agha
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  • How is this last inequality proven without the trigonometric circle? –  Aug 23 '15 at 14:20
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    @Jason, here: http://math.stackexchange.com/questions/98998/why-x-tanx-while-0x-frac-pi2 you can find some approaches. – agha Aug 23 '15 at 14:49
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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.

zhw.
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