Use de Moivre's theorem to show that
$$\sin 8 \theta = 8 \sin \theta \cos \theta \left( 1 - 10 \sin ^ { 2 } \theta + 24 \sin ^ { 4 } \theta - 16 \sin ^ { 6 } \theta \right) .$$
Use the equation \(\frac { \sin 8 \theta } { \sin 2 \theta } = 0\) to find the roots of
$$16 x ^ { 6 } - 24 x ^ { 4 } + 10 x ^ { 2 } - 1 = 0$$
in the form \(\sin k \pi\), where \(k\) is rational.