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I need to prove the following:

Let $\delta > 0$. Then $$ \sin \pi x \geq \frac{\pi}{2}\delta\;. $$ holds for $x \in [\delta, 1-\delta]$.

I tried to deduce the inequality using the definition of the sine as a power series, however, to no avail. Is there any quick way or hint to deduce this inequality?

Thanks a lot!

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The inequality apparently holds for all $x \in [\delta, 1 - \delta]$, i.e. for all $\pi x \in [\delta \pi, \pi - \delta \pi]$, which is symmetric around $\pi/2$. So maybe you need a power series centered at $\pi/2$? –  TMM May 3 '12 at 19:13
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3 Answers

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Here is the start of a solution using power series. To fill in the arguments it may be helpful to assume that $0 < \delta \leq 1/2$, which is reasonable since the the interval $[\delta,1-\delta]$ is empty for $\delta > 1/2$.

As TMM noted, $\sin(\pi x)$ is symmetric about $x=1/2$ (where it has a local maximum), so for $x \in [\delta,1-\delta]$,

$$ \begin{align} \sin(\pi x) &\geq \sin(\pi \delta) \\ &= \pi \delta - \frac{\pi^3\delta^3}{3!} + \frac{\pi^5\delta^5}{5!} - \frac{\pi^7\delta^7}{7!} + \cdots \\ &\geq \pi \delta - \frac{\pi^3 \delta^3}{3!}. \end{align} $$

Now, just show that

$$ \pi \delta - \frac{\pi^3\delta^3}{3!} - \frac{\pi\delta}{2} > 0 $$

to get the result.

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Thank you, this helped me a lot! –  milos May 3 '12 at 20:07
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I will provide a sketch (no power series, however). You should supply the details yourself. By symmetry it suffices to show the inequality for $x\in[\delta,1/2]$.

We first consider the case $x\in[\delta,1/3]$. Consider the function given by $f(x)=\sin(\pi x)-\frac{\pi}2 x$. It is not hard to see that $f(0)=0$. Now differentiate $f$ and show that its derivative is strictly positive when $x\in (0,1/3)$. We conclude that $f(x)$ is positive when $x\in(0,1/3)$, and so $$ \sin(\pi x) \geq \frac{\pi}2 x \geq \frac{\pi}2 \delta .$$

Now consider the case $x\in[1/3,1/2]$. We have $$\sin(\pi x)\geq \frac{\sqrt3}2 \geq \frac{\pi}4 \geq \frac{\pi}2x \geq \frac{\pi}2\delta .$$

This proves the result.

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By looking at the graph of $x\mapsto\sin(\pi x)\ $ $(0\leq x\leq 1)$ one immediately sees that $$0<\delta\leq x\leq{1\over2}$$ implies $$\sin(\pi x)\geq 2x\geq {\pi\over2} x\geq{\pi\over2}\delta\ ,$$ and the rest is symmetry.

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