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In Jeffrey Bub's Bananaworld $^1$, there's a remark (p. 101) that

$$\sin^2 \frac{\pi}{4n} \;\leq\; \frac{1}{n^2} \;,$$

where $n$ is any positive integer.

It's been decades since I studied trig, although I have used it occasionally with the help of review and references. An appendix to an earlier chapter in Bananaworld reviews some useful trigonometric identities, but I can't see how to derive the above statement from them. I've scanned through the Wikipedia List of trigonometic identities to see whether I could see any obvious quick way of deriving Bub's claim, but I didn't succeed. I am sure this isn't a difficult problem, but I don't see how to get a solution, and I'm not sure where to look without doing extensive study. It's kind of an unusual trigonometric inequality, it seems to me, since one side is not trigonometic.

Since the sine expression only applies to positive values less than or equal to $\pi/4$, all I need to understand is why

$$\sin \frac{\pi}{4n} \;\leq\; \frac{1}{n} \;.$$

For $n=1$, it's obvious that $\sin \pi/4 \leq 1$. Similarly, for $n=2$, I can just see by looking at a figure that $\sin \pi/8 \leq 1/2$. As $n$ increases, both $\sin \pi/(4n)$ and $1/n$ decrease, and when I calculate values for $\sin \pi/(4n)$ and $1/n$ for many values of $n$, $\sin \pi/(4n)$ is always smaller. However, I don't see how to prove that it must be smaller.

That's as far as I've gotten. It must not be very hard to show that Bub's inequality is correct, since he didn't even provide a hint as to why it holds. The book requires a little bit of mathematical sophistication, but not a lot of specific mathematical knowledge beyond basic algebra. I'm willing to do the work to figure out why the inequality holds if I have some idea where to start, so a hint might be all I need. (A full proof is fine, too, if someone prefers to provide that.) Thanks.

$^1$ Bananaworld is a book on quantum entanglement. The inequality is a small step in a short argument.

Mars
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1 Answers1

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It helps to first establish that if $x\ge0$ then $\sin^2 x\le x^2$. We shall use the fact that if $x\ge0$ then $\sin x \le x$ (this follows from geometric considerations involving the unit circle, or from the mean value theorem). Note that $$ x^2-\sin^2x=(x-\sin x)(x+\sin x) \, . $$ For $x\ge0$, the term in the first set of brackets is nonnegative. The term in the second pair of brackets must also be nonnegative for $x\ge0$, but this is harder to see. On the interval $[0,\pi]$, both $x$ and $\sin x$ are nonnegative, so $x+\sin x$ is nonnegative. On the interval $[1,\infty)$, we have $x\ge 1$ and $\sin x\ge -1$, so $x+\sin x\ge0$. Thus on the interval $[0,\pi]\cup[1,\infty)=[0,\infty)$, we have $x+\sin x\ge0$. Since the product of two nonnegative numbers is nonnegative, it follows that $x^2-\sin^2x$ is nonnegative for $x\ge0$, meaning that $\sin^2 x\le x^2$. Setting $x=\pi/(4n)$ (where $n$ is a positive integer), we see that $$ \sin^2\left(\frac{\pi}{4n}\right)\le \frac{\pi^2}{16n^2} \, . $$ Now use the fact that $\pi<4$ to conclude.

Joe
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