TITLE: Find better than $ 4^n\prod_{k=1}^{n-1}\cos^2(k)\sim e^{o(n)}$
QUESTION [2 upvotes]: Let $$u_n=\prod_{k=1}^{n-1}\cos^2(k)$$ then $$\frac1n \ln(u_n) = \frac1n\sum_{k=0}^{n-1}  \ln(\cos^2(k)) \underset{n\to\infty}\longrightarrow \frac1{2\pi} \int_0^{2\pi} \ln(\cos^2(x))\,{\rm d}x = -\ln(4)$$ We deduce that  $$\ln(u_n)\sim -n\ln(4)$$ . I think that the sequence $v_n=\ln(4^n u_n))$ have not limit , So there is no  constant c such that $u_n \sim c\,4^{-n}$.  Can we find better then $$ \prod_{k=1}^{n-1}\cos^2(k)\sim 4^{-n}e^{o(n)}$$ . What is $\limsup 4^n u_n$ and $\liminf 4^n u_n$ ?
additional comments
wolfram does not confirm that $\liminf 4^n u_n  =0$

REPLY [5 votes]: I presume that exchanging the cosine by the sine will not matter for the large-$n$ behavior of the product, so let me consider
$$a_n^2=4^n\prod_{k=1}^{n}\sin^2 k=\left(\prod_{k=1}^n\left|1-e^{k\alpha \pi i}\right|\right)^2\;\;\text{with}\;\;\alpha=2/\pi.$$
The convergence of $a_n$ was determined in this MO posting from five years ago, with lim inf $a_n$ equal to zero and lim sup $a_n$ equal to infinity for irrational $\alpha$.
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