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?

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.
