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

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?

• See here Jul 8, 2020 at 10:04
• @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. Jul 15, 2020 at 6:33

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.

• Thanks a lot. I have tried split it into $\sin^2x\sin2x$ first, but it doesn't work. The cubic operation is critical. Jul 8, 2020 at 9:47
• @ZiyiGuo, I also went through various experiments until I found this version. It was quite tricky but enjoyable indeed. Jul 8, 2020 at 9:49
• @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. Jul 8, 2020 at 13:56
• @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 $\sin^6x\sin^6 2x\sin^64x\cdot=\sin^4x\sin^22x\cdot \sin^42x\sin^24x\cdots$ Jul 10, 2020 at 11:16