Sin value

Let $I$ denote the value of the expression, since all the angles given are in the first two quadrants, sine of these angles are strictly positive, so $I > 0$ . With $\sin(\pi - A) = \sin(A)$ and $A = \frac\pi{14}$ . We have $I^2 = \sin(A) \sin(3A) \sin(5A) \sin(7A) \sin(9A) \sin(11A) \sin(13A)$ .

Consider the Chebyshev polynomial, $\sin(7x) = 1$ . then it has roots $A,3A,5A,7A,9A,11A,13A$ . With the properties of Chebyshev polynomial, we can expand $\sin(7x)$ to be

$\sin(7x) = -2^{7-1}\sin^7(x) + \sin(x) (\ldots )$

Thus by Vieta's formula, we have $I^2 = \dfrac {-1}{-2^{7-1}} \Rightarrow I = \boxed{\dfrac18}$ .