Question
Prove that: $\cos6^\circ\cos42^\circ\cos66^\circ\cos78^\circ=\frac{1}{16}$

Answer

$\text{LHS}=\cos6^\circ\cos42^\circ\cos66^\circ\cos78^\circ$ $=\frac{1}{4}(2\cos6^\circ\cos66^\circ)(2\cos42^\circ\cos78^\circ)$ $=\frac{1}{4}(\cos72^\circ+\cos60^\circ)(\cos120^\circ+\cos36^\circ)$ $[\because2\cos\text{A}\cos\text{B}=\cos(\text{A+B})+\cos(\text{A}-\text{B})]$ $=\frac{1}{4}\big\{\cos(90^\circ-72^\circ)+\frac{1}{2}\big\}\Big\{-\frac{1}{2}+\frac{\sqrt{5}+1}{4}\Big\}$ $=\frac{1}{4}\big(\sin18^\circ+\frac{1}{2}\big)\Big(-\frac{1}{2}+\frac{\sqrt{5}+1}{4}\Big)$ $=\frac{1}{4}\Big(\frac{\sqrt{5}-1}{4}+\frac{1}{2}\Big)\Big(\frac{\sqrt{5}+1}{4}+\frac{1}{2}\Big)$ $=\frac{1}{4}\Big(\frac{\sqrt{5}-1+2}{4}\Big)\Big(\frac{\sqrt{5}+1-2}{4}\Big)$ $=\frac{1}{64}(\sqrt{5}+1)(\sqrt{5}-1)$ $=\frac{1}{64}(5-1)$ $=\frac{1}{16}=\text{RHS}$ Hence proved

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