Show that $$\sin^2x\cdot\sin^22x\cdot\sin^24x\cdot\sin^28x\cdots\sin^22^nx\leq\frac{3^n}{4^n}$$

I understand the result of an arithmetic sequence $(\sin1^\circ)(\sin3^\circ)(\sin5^\circ)…(\sin89^\circ)$, how about the geometric sequence case?

  • 1
    $\begingroup$ See here $\endgroup$ Jul 8, 2020 at 10:04
  • $\begingroup$ @Nguyenhuyen_AG In the link, actually a hint is given. I considered the form of $(\sin x)^a \sin 2x \le \cdot$ and found that $a=2$ is the best constant. $\endgroup$
    – River Li
    Jul 15, 2020 at 6:33

1 Answer 1


We first prove that

$$ (\sin x)^4(\sin 2x)^2 \leq \left(\frac{3}{4}\right)^3. $$

Indeed, applying the double angle formula $\sin 2x = 2\sin x\cos x$ and substituting $t = \sin^2 x$, we have

$$ (\sin x)^4(\sin 2x)^2 = 4t^3(1-t) $$

and the right-hand side is maximized at $t = \frac{3}{4}$ with the value $(3/4)^3$ as desired. Now, returning to the original problem, the above inequality yields

\begin{align*} &(\sin x)^2 (\sin 2x)^2 \dots (\sin 2^n x)^2 \\ &= \Biggl[ (\sin x)^2 (\sin 2^n x)^4 \prod_{k=0}^{n-1} (\sin 2^k x)^4 (\sin 2^{k+1}x)^2 \Biggr]^{1/3} \\ &\leq \Biggl[ \prod_{k=0}^{n-1} \left(\frac{3}{4}\right)^3 \Biggr]^{1/3} \\ &= \left(\frac{3}{4}\right)^n \end{align*}

as required.

  • $\begingroup$ Thanks a lot. I have tried split it into $\sin^2x\sin2x$ first, but it doesn't work. The cubic operation is critical. $\endgroup$
    – Ziyi Guo
    Jul 8, 2020 at 9:47
  • $\begingroup$ @ZiyiGuo, I also went through various experiments until I found this version. It was quite tricky but enjoyable indeed. $\endgroup$ Jul 8, 2020 at 9:49
  • $\begingroup$ @ZiyiGuo Not sure what you meant. If you found the max of $ \sin ^2 x \sin 2x$ (eg by differentiating), that works too. However if you found the max of $\sin x \sin 2x$, then this approach doesn't work. $\endgroup$
    – Calvin Lin
    Jul 8, 2020 at 13:56
  • $\begingroup$ @CalvinLin I mean the split, not the max. First I split $\sin^2x\sin^2 2x\sin^24x\cdots=\sin^2x\sin2x\cdot \sin2x\sin4x\cdot\sin4x\cdots$, it does not work. I should split it as $\sin^6x\sin^6 2x\sin^64x\cdot=\sin^4x\sin^22x\cdot \sin^42x\sin^24x\cdots$ $\endgroup$
    – Ziyi Guo
    Jul 10, 2020 at 11:16

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