For equality to hold, note that $\theta$ cannot be an integer multiple of $\pi$ (or else $\sin =0$ and $\cos = \pm 1$ ). Let $z=e^{i \theta / 2} \neq \pm 1$. Then in terms of complex numbers, we want $\prod_{k=0}^{10}\left(1+\frac{2}{z^{2^{k+1}}+z^{-2^{k+1}}}\right)=\prod_{k=0}^{10} \frac{\left(z^{2^{k}}+z^{-2^{k}}\right)^{2}}{z^{2^{k+1}}+z^{-2^{k+1}}}$ which partially telescopes to $\frac{z+z^{-1}}{z^{2^{11}}+z^{-2^{11}}} \prod_{k=0}^{10}\left(z^{2^{k}}+z^{-2^{k}}\right)$. Using a classical telescoping argument (or looking at binary representation; if you wish we may note that $z-z^{-1} \neq 0$, so the ultimate telescoping identity holds), this simplifies to $\frac{\tan \left(2^{10} \theta\right)}{\tan (\theta / 2)}$. Since $\tan x$ is injective modulo $\pi$ (i.e. $\pi$-periodic and injective on any given period), $\theta$ works if and only if $\frac{\theta}{2}+\ell \pi=1024 \theta$ for some integer $\ell$, so $\theta=\frac{2 \ell \pi}{2047}$. The largest value for $\ell$ such that $\theta<\pi$ is at $\ell=1023$, which gives $\theta=\frac{2046 \pi}{2047}$