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][1] does not confirm that $\liminf 4^n u_n  =0$[![enter image description here][2]][2]


  [1]: https://www.wolframalpha.com/input/?i=%5Cprod_%7Bk%3D1%7D%5E%7Bn%7D%20%7C1%2Be%5E%7B2ik%20%7D%7C
  [2]: https://i.sstatic.net/5E7n3.jpg