Prove that \( \cos10^\circ\cos30^\circ\cos50^\circ\cos70^\circ=\frac{3}{16} \)
Solution
\[
\begin{aligned}
&\cos10^\circ\cos30^\circ\cos50^\circ\cos70^\circ\\[4pt]
&=\frac{\sqrt3}{2}\cos10^\circ\cos50^\circ\sin20^\circ
\end{aligned}
\]
Using identity: \(2\cos A\cos B=\cos(A+B)+\cos(A-B)\)
\[
\begin{aligned}
&=\frac{\sqrt3}{4}(\cos60^\circ+\cos40^\circ)\sin20^\circ\\[4pt]
&=\frac{\sqrt3}{4}\left(\frac12+\cos40^\circ\right)\sin20^\circ\\[4pt]
&=\frac{\sqrt3}{8}\sin20^\circ+\frac{\sqrt3}{4}\cos40^\circ\sin20^\circ
\end{aligned}
\]
Using identity: \(2\sin A\cos B=\sin(A+B)-\sin(B-A)\)
\[
\begin{aligned}
&=\frac{\sqrt3}{8}\sin20^\circ+\frac{\sqrt3}{8}(\sin60^\circ-\sin20^\circ)\\[4pt]
&=\frac{\sqrt3}{8}\times\frac{\sqrt3}{2}\\[4pt]
&=\frac{3}{16}
\end{aligned}
\]