Question

Prove that:
$\cos 6°\cos 42°\cos 66°\cos 78°=\frac{1}{16}$

Answer 1

$$\begin{gathered} \mathrm{LHS}=\cos 6°\cos 42°\cos 66°\cos 78° \\ =\frac{1}{4}\left(2\cos 6°\cos 66°\right)\left(2\cos 42°\cos 78°\right) \\ =\frac{1}{4}\left(\cos 72°+\cos 60°\right)\left(\cos 120°+\cos 36°\right)\left[\because 2\cos A\cos B=\cos \left(A+B\right)+\cos \left(A-B\right)\right] \\ =\frac{1}{4}\left\{\cos \left(90°-72°\right)+\frac{1}{2}\right\}\left\{-\frac{1}{2}+\frac{\sqrt{5}+1}{4}\right\} \end{gathered}$$
$$\begin{gathered} =\frac{1}{4}\left(\sin 18°+\frac{1}{2}\right)\left(-\frac{1}{2}+\frac{\sqrt{5}+1}{4}\right) \\ =\frac{1}{4}\left(\frac{\sqrt{5}-1}{4}+\frac{1}{2}\right)\left(\frac{\sqrt{5}+1}{4}-\frac{1}{2}\right) \\ =\frac{1}{4}\left(\frac{\sqrt{5}-1+2}{4}\right)\left(\frac{\sqrt{5}+1-2}{4}\right) \\ =\frac{1}{64}\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right) \\ =\frac{1}{64}\left(5-1\right) \\ =\frac{1}{16}=\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$