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

Answer: 80

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