Prove that:
$\cos 78^{\circ} \cos 42^{\circ} \cos 36^{\circ}=\frac{1}{8}$
$\mathrm{LHS}=\cos 78^{\circ} \cos 42^{\circ} \cos 36^{\circ}$
$=\frac{\left(2 \cos 78^{\circ} \cos 42^{\circ}\right)}{2} \cos 36^{\circ}$
$=\frac{\cos \left(78^{\circ}+42^{\circ}\right)+\cos \left(78^{\circ}-42^{\circ}\right)}{2} \times \cos 36^{\circ}$
$[2 \cos A \cos B=\cos (A+B)+\cos (A-B)]$
$=\frac{1}{2}\left(\cos 120^{\circ}+\cos 36^{\circ}\right) \cos 36^{\circ}$
$=\frac{1}{2}\left(-\cos \left(180^{\circ}-120^{\circ}\right)+\cos 36^{\circ}\right) \cos 36^{\circ}$
$=\frac{1}{2}\left(-\cos 60^{\circ}+\cos 36^{\circ}\right) \cos 36^{\circ}$
$=\frac{1}{2}\left(-\frac{1}{2}+\frac{\sqrt{5}+1}{4}\right) \frac{\sqrt{5}+1}{4}$
$=\frac{1}{2} \times \frac{\sqrt{5}-1}{4} \times \frac{\sqrt{5}+1}{4}$
$=\frac{1}{8}$
= RHS
Hence proved.
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.