Prove that \( \cos40^\circ\cos80^\circ\cos160^\circ=-\frac18 \)
Solution
\[
\begin{aligned}
&\cos40^\circ\cos80^\circ\cos160^\circ\\[4pt]
&=-\cos40^\circ\cos80^\circ\cos20^\circ
\end{aligned}
\]
Using identity: \(2\cos A\cos B=\cos(A+B)+\cos(A-B)\)
\[
\begin{aligned}
&=-\frac12(\cos100^\circ+\cos60^\circ)\cos20^\circ\\[4pt]
&=-\frac12\left(-\sin10^\circ+\frac12\right)\cos20^\circ
\end{aligned}
\]
Using identity: \(2\sin A\cos B=\sin(A+B)+\sin(A-B)\)
\[
\begin{aligned}
&=-\frac14\cos20^\circ+\frac14(\sin30^\circ-\sin10^\circ)\\[4pt]
&=-\frac14\cos20^\circ+\frac14\left(\frac12-\sin10^\circ\right)\\[4pt]
&=-\frac18
\end{aligned}
\]