Calculate without using tables:
$128 \sin^2 20^\circ \sin^2 40^\circ \sin^2 60^\circ \sin^2 80^\circ$
The given expression is: $\;\;$ $128 \sin^2 20^\circ \sin^2 40^\circ \sin^2 60^\circ \sin^2 80^\circ$ $\;\;\; \cdots \; (1)$
Consider the expression: $\;\;$ $\sin 20^\circ \sin 40^\circ \sin 60^\circ \sin 80^\circ$
$= \sin 20^\circ \times \left(\sin 80^\circ \sin 40^\circ\right) \times \dfrac{\sqrt{3}}{2}$
$= \dfrac{\sqrt{3}}{2} \times \sin 20^\circ \times \dfrac{1}{2} \left[\cos \left(80^\circ - 40^\circ\right) - \cos \left(80^\circ + 40^\circ\right)\right]$
$= \dfrac{\sqrt{3}}{4} \times \sin 20^\circ \left[\cos 40^\circ - \cos 120^\circ\right]$
$= \dfrac{\sqrt{3}}{4} \times \sin 20^\circ \left[\cos 40^\circ - \left(\dfrac{-1}{2}\right)\right]$
$= \dfrac{\sqrt{3}}{8} \left[2 \sin 20^\circ \cos 40^\circ + \sin 20^\circ\right]$
$= \dfrac{\sqrt{3}}{8} \left[\sin \left(20^\circ + 40^\circ\right) + \sin \left(20^\circ - 40^\circ\right) + \sin 20^\circ\right]$
$= \dfrac{\sqrt{3}}{8} \left[\sin 60^\circ + \sin \left(-20^\circ\right) + \sin 20^\circ\right]$
$= \dfrac{\sqrt{3}}{8} \left[\dfrac{\sqrt{3}}{2} - \sin 20^\circ + \sin 20^\circ\right]$
$\therefore$ $\sin 20^\circ \sin 40^\circ \sin 60^\circ \sin 80^\circ = \dfrac{3}{16}$ $\;\;\; \cdots \; (2)$
In view of expression $(2)$, expression $(1)$ becomes
$128 \times \left(\dfrac{3}{16}\right)^2 = \dfrac{9}{2}$