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