Olá,
Heisenberg1.
Uma ideia é lembrar que [tex3]\sen(2\alpha)=2\sen(\alpha)\cos(\alpha).[/tex3]
Seja [tex3]P = \cos40^{\circ} \cdot \cos80^{\circ} \cdot \cos160^{\circ},[/tex3]
então
[tex3]\begin{align}
2\sen40^{\circ} \cdot P & = 2\sen40^{\circ} \cdot\cos40^{\circ} \cdot \cos80^{\circ} \cdot \cos160^{\circ} \\
& = \,\,\, \sen80^{\circ} \cdot \cos80^{\circ} \cdot \cos160^{\circ} \\
& = \,\,\, \frac{2}{2} \cdot \sen80^{\circ} \cdot \cos80^{\circ} \cdot \cos160^{\circ} \\
& = \,\,\, \frac{1}{2} \cdot \sen160^{\circ} \cdot \cos160^{\circ} \\
& = \,\,\, \frac{1}{4} \cdot \sen320^{\circ} \\
\end{align}[/tex3]
Mas [tex3]\sen40^{\circ} = - \sen320^{\circ},[/tex3]
de sorte que
[tex3]2\sen40^{\circ} \cdot P = - \frac{1}{4} \sen40^{\circ} [/tex3]
[tex3]\boxed{ P = -\frac{1}{8} }[/tex3]