Question

Prove that:
$\cos 36°\cos 42°\cos 60°\cos 78°=\frac{1}{16}$

Answer 1

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