Question

Prove that:
$\cos 78°\cos 42°\cos 36°=\frac{1}{8}$

Answer 1

$$\begin{gathered} \mathrm{LHS}=\cos 78°\cos 42°\cos 36° \\ =\frac{\left(2\cos 78°\cos 42°\right)}{2}\cos 36° \\ =\frac{\cos \left(78°+42°\right)+\cos \left(78°-42°\right)}{2}\times \cos 36° \\ \left[2\cos A\cos B=\cos \left(A+B\right)+\cos \left(A-B\right)\right] \\ =\frac{1}{2}\left(\cos 120°+\cos 36°\right)\cos 36° \end{gathered}$$

$$\begin{gathered} =\frac{1}{2}\left(-\cos \left(180°-120°\right)+\cos 36°\right)\cos 36° \\ =\frac{1}{2}\left(-\cos 60°+\cos 36°\right)\cos 36° \\ =\frac{1}{2}\left(-\frac{1}{2}+\frac{\sqrt{5}+1}{4}\right)\frac{\sqrt{5}+1}{4} \\ =\frac{1}{2}\times \frac{\sqrt{5}-1}{4}\times \frac{\sqrt{5}+1}{4} \\ =\frac{1}{8} \\ =\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$