Question

If $\sin ^{2}\left(10^{\circ}\right) \sin \left(20^{\circ}\right) \sin \left(40^{\circ}\right) \sin \left(50^{\circ}\right) \sin \left(70^{\circ}\right)=\alpha-$ $\frac{1}{16} \sin \left(10^{\circ}\right)$, then $16+\alpha^{-1}$ is equal to _________.

Answer 1

$\sin 10^{\circ}\left(\frac{1}{2} \cdot 2 \sin 20^{\circ} \sin 40^{\circ}\right) \cdot \sin 10^{\circ} \sin \left(60^{\circ}-10^{\circ}\right) \sin \left(60^{\circ}+10^{\circ}\right)$
$\sin 10^{\circ} \frac{1}{2}\left(\cos 20^{\circ}-\cos 60^{\circ}\right) \cdot \frac{1}{4} \sin 30^{\circ}$
$\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{2} \cdot \sin 10^{\circ}\left(\cos 20^{\circ}-\frac{1}{2}\right)$
$=\frac{1}{32}\left(2 \sin 10^{\circ} \cos 20^{\circ}-\sin 10^{\circ}\right)$
$=\frac{1}{32}\left(\sin 30^{\circ}-\sin 10^{\circ}-\sin 10^{\circ}\right)$
$=\frac{1}{32}\left(\frac{1}{2}-2 \sin 10^{\circ}\right)$
$=\frac{1}{64}\left(1-4 \sin 10^{\circ}\right)$
$=\frac{1}{64}-\frac{1}{16} \sin 10^{\circ}$
Hence $\alpha=\frac{1}{64}$
$16+\alpha^{-1}=80$