Disguised Trigo

Using the following identity ( eqn. 24 ),

$\begin{aligned} \prod_{k=1}^{n-1} \sin \left(\frac {k\pi}n \right) & = 2^{1-n}n & \small \color{#3D99F6} \text{Putting }n=7 \\ \prod_{k=1}^6 \sin \left(\frac {k\pi}7 \right) & = \frac 7{64} \\ \sin \left(\frac \pi 7\right) \sin \left(\frac {2\pi} 7\right) \sin \left(\frac {3\pi} 7\right)\color{#3D99F6}\sin \left(\frac {4\pi} 7\right) \sin \left(\frac {5\pi} 7\right)\sin \left(\frac {6\pi} 7\right) & = \frac 7{64} & \small \color{#3D99F6} \text{Note that }\sin (\pi - \theta) = \sin \theta \\ \sin \left(\frac \pi 7\right) \sin \left(\frac {2\pi} 7\right) \sin \left(\frac {3\pi} 7\right)\color{#3D99F6}\sin \left(\frac {3\pi} 7\right) \sin \left(\frac {2\pi} 7\right)\sin \left(\frac {1\pi} 7\right) & = \frac 7{64} \\ \left(\sin \left(\frac \pi 7\right) \sin \left(\frac {2\pi} 7\right) \sin \left(\frac {3\pi} 7\right)\right)^2 & = \frac 7{64} \\ \implies \sin \left(\frac \pi 7\right) \sin \left(\frac {2\pi} 7\right) \sin \left(\frac {3\pi} 7\right) & = \frac {\sqrt 7}8 \\ & \approx \boxed{0.331} \end{aligned}$

Although there are terms sin, and all sorts of trigonometrical stuffs but this can be done easily by complex numbers.

First of all use the n roots of unity to generate the phrase in the question using the modulus amplitude form. Then using de moiver's theorem find the terms in the question and finally multiply them to get the result √7/8 which on simplification gives the answer.