\section*{Problem 4 - 041214}

Without using a table of numbers or a slide rule, the product

\[
x=\cos 20^{\circ} \cdot \cos 40^{\circ} \cdot \cos 60^{\circ} \cdot \cos 80^{\circ}
\]

is to be calculated.
}

With the law \(2 \cos \alpha \cos \beta=\cos (\alpha-\beta)+\cos (\alpha+\beta)\) and the knowledge that \(\cos 60^{\circ}=\frac{1}{2}\) and \(\cos 90^{\circ}=0\):

\[
\begin{aligned}
x & =\cos 20^{\circ} \cdot \cos 40^{\circ} \cdot \cos 60^{\circ} \cdot \cos 80^{\circ}=\frac{1}{2} \cdot\left(\cos 20^{\circ}+\cos 60^{\circ}\right) \cdot \frac{1}{2} \cdot \cos 80^{\circ} \\
& =\frac{1}{4} \cdot \cos 20^{\circ} \cdot \cos 80^{\circ}+\frac{1}{8} \cdot \cos 80^{\circ}=\frac{1}{8} \cdot\left(\cos 60^{\circ}+\cos 100^{\circ}+\cos 80^{\circ}\right) \\
& =\frac{1}{8} \cdot\left(\frac{1}{2}+2 \cdot \cos 10^{\circ} \cdot \cos 90^{\circ}\right)=\frac{1}{16}
\end{aligned}
\]

It follows that \(x=\frac{1}{16}\).