Sines! Stop bothering me!

For $k$ from $0$ to $6$ , the complex numbers $e^{i \frac{2k\pi}{7}}$ are roots of polynomial $z^{7} - 1$ .

Also $z^{7} - 1 = (z-1) (z^{6} + z^{5} + z^{4} + z^{3} + z^{2} + z + 1)$ so

$\frac{z^{7} - 1}{z-1} = z^{6} + z^{5} + z^{4} + z^{3} + z^{2} + z + 1$ for $z \neq 1$ and

$\frac{z^{7} - 1}{z-1} = \prod \limits_{k=1}^{6} (z - e^{i \frac{2k\pi}{7}})$

So for $z = 1$ we have $\prod \limits_{k=1}^{6} (1 - e^{i \frac{2k\pi}{7}}) = 1 + 1 + 1 + 1 + 1 + 1 + 1 = 7$

The $\sin$ trigonometric function is related with complex number by $\sin{\alpha} = \large \frac{e^{i \alpha} - e^{- i \alpha}}{2 i}$ So

$\large \prod \limits_{k=1}^{6} \sin{\frac{k \pi}{7}} = \prod \limits_{k=1}^{6} \frac{e^{i \frac{k \pi}{7}} - e^{- i \frac{k \pi}{7}}}{2 i} = \prod \limits_{k=1}^{6} \frac{e^{- i \frac{k \pi}{7}} (e^{i \frac{2 k \pi}{7}} - 1)}{2 i} =$

$\large = \frac{(-1)^6}{2^6 i^6} \prod \limits_{k=1}^{6} e^{- i \frac{k \pi}{7}} \prod \limits_{k=1}^{6} (1 - e^{i \frac{2 k \pi}{7}}) = \frac{7}{ 64 i^6} e^{- i \frac{\pi}{7} \sum \limits_{k=1}^{6} k} = \frac{7}{ 64 i^6} e^{- i \frac{\pi \cdot 6 \cdot 7}{2 \cdot 7}} =$

$\large = \frac{7}{ 64 i^6} (e^{- i \frac{\pi}{2}})^6 = \frac{7 i^6}{ 64 i^6} = \frac{7}{64}$

So the solution is

$\large 7 + 64 =$