Long Enough

$\begin{aligned} P & = \left(1-\tan^2 \left(\frac x2 \right) \right)\left(1-\tan^2 \left(\frac x{2^2}\right) \right) \left(1-\tan^2 \left(\frac x{2^3}\right) \right) \cdots \left(1-\tan^2 \left(\frac x{2^{2011}}\right) \right) & \small \color{#3D99F6} \text{Since }\tan 2\theta = \frac {2\tan \theta}{1-\tan^2 \theta} \\ & = \frac {2\tan \left(\frac x2\right)}{\tan (x)} \times \frac {2\tan \left(\frac x{2^2}\right)}{\tan \left(\frac x2 \right)} \times \frac {2\tan \left(\frac x{2^3}\right)}{\tan \left(\frac x{2^2} \right)} \times \cdots \frac {2\tan \left(\frac x{2^{2011}}\right)}{\tan \left(\frac x{2^{2010}} \right)} \\ & = \frac {2 \color{#3D99F6}\cancel{\tan \left(\frac x2\right)}}{\tan (x)} \times \frac {2\color{#D61F06}\cancel{\tan \left(\frac x{2^2}\right)}}{\color{#3D99F6}\cancel{\tan \left(\frac x2\right)}} \times \frac {2 \color{#3D99F6}\cancel{\tan \left(\frac x{2^3}\right)}}{\color{#D61F06}\cancel{\tan \left(\frac x{2^2}\right)}} \times \cdots \frac {2\tan \left(\frac x{2^{2011}}\right)}{\color{#D61F06}\cancel{\tan \left(\frac x{2^{2010}} \right)}} \\ & = \frac {2^{2011}\tan \left(\frac x{2^{2011}} \right)}{\tan (x)} \end{aligned}$

Since $P = 2^{2011}\sqrt 3 \tan \left(\dfrac x{2^{2011}} \right) = \dfrac {2^{2011}\tan \left(\frac x{2^{2011}} \right)}{\tan (x)}$ , $\implies \tan (x) = \dfrac 1{\sqrt 3}$ $\implies x = \dfrac \pi 6$ and $\sin 2x = \sin \dfrac \pi 3 = \dfrac {\sqrt 3}2 \approx \boxed{0.866}$ .