deduce that $\cos 6° \cos42° \cos66° \cos78°= \frac{1}{16}$ > Prove that $$4 \cos\theta \cos(\frac{\pi}{3}-\theta) \cos(\frac{\pi}{3}+\theta)= \cos 3\theta$$ and deduce that $$\cos 6° \cos42° \cos66° \cos78°= \frac{1}{16}$$ I have proved by using $2 \cos A \cos B= \cos (A+B)+ \cos (A-B) $ But I cant deduce $1/16$ * * * P.S. we have to use prove that to deduce $1/16$

$$\begin{align} \cos6\cos42\cos66\cos78&=\frac{1}{\cos54\cos18}\left(\cos6\cos54\cos66\right)\left(\cos18\cos42\cos78\right)\\\ &=\frac{1}{\cos54\cos18}\frac{\cos18}{4}\frac{\cos54}{4}\\\ &=\frac{1}{16} \end{align}$$