What is the exact value of the product $\sin 10^{\circ} \sin 30^{\circ} \sin 50^{\circ} \sin 70^{\circ}$?
It is known that

$$
\sin 30^{\circ}=\frac{1}{2}
$$

Using the relationship $\sin \alpha=\cos \left(90^{\circ}-\alpha\right)$,

$$
\sin 10^{\circ} \sin 50^{\circ} \sin 70^{\circ}=\cos 80^{\circ} \cos 40^{\circ} \cos 20^{\circ}
$$

Multiply the right side by $\sin 20^{\circ}$ and use the relationship $\cos \alpha \sin \alpha=\frac{1}{2} \sin 2 \alpha$ three times, then use $\sin \alpha=\sin \left(180^{\circ}-\alpha\right)$:

$$
\begin{gathered}
\cos 80^{\circ} \cos 40^{\circ} \cos 20^{\circ} \sin 20^{\circ}=\frac{1}{2} \cos 80^{\circ} \cos 40^{\circ} \sin 40^{\circ}= \\
=\frac{1}{4} \cos 80^{\circ} \sin 80^{\circ}=\frac{1}{8} \sin 160^{\circ}=\frac{1}{8} \sin 20^{\circ}
\end{gathered}
$$

Therefore,

$$
\cos 80^{\circ} \cos 40^{\circ} \cos 20^{\circ}=\frac{1}{8}
$$

since $\sin 20^{\circ} \neq 0$. From comparing (1), (2), and (3),

$$
\sin 10^{\circ} \sin 30^{\circ} \sin 50^{\circ} \sin 70^{\circ}=\frac{1}{16}
$$

Remark. Using the relationship $\sin 2 \varphi=2 \sin \varphi \cos \varphi=2 \sin \varphi \cdot \sin \left(90^{\circ}-\varphi\right)$, we have $\sin 10^{\circ}=2 \cdot \sin 5^{\circ} \cdot \sin 85^{\circ}$, $\sin 30^{\circ}=2 \cdot \sin 15^{\circ} \cdot \sin 75^{\circ}$, etc. Substituting these into (4) and multiplying by $\sin 45^{\circ}=\sqrt{2} / 2$, we get

$$
\sin 5^{\circ} \cdot \sin 15^{\circ} \cdot \sin 25^{\circ} \cdot \sin 35^{\circ} \cdot \sin 45^{\circ} \cdot \sin 55^{\circ} \cdot \sin 65^{\circ} \cdot \sin 75^{\circ} \cdot \sin 85^{\circ}=\frac{\sqrt{2}}{512}
$$

which is a surprising relationship.