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Prove that:

Question:

Prove that:

$\cos 78^{\circ} \cos 42^{\circ} \cos 36^{\circ}=\frac{1}{8}$

Solution:

$\mathrm{LHS}=\cos 78^{\circ} \cos 42^{\circ} \cos 36^{\circ}$

$=\frac{\left(2 \cos 78^{\circ} \cos 42^{\circ}\right)}{2} \cos 36^{\circ}$

$=\frac{\cos \left(78^{\circ}+42^{\circ}\right)+\cos \left(78^{\circ}-42^{\circ}\right)}{2} \times \cos 36^{\circ}$

$[2 \cos A \cos B=\cos (A+B)+\cos (A-B)]$

$=\frac{1}{2}\left(\cos 120^{\circ}+\cos 36^{\circ}\right) \cos 36^{\circ}$

$=\frac{1}{2}\left(-\cos \left(180^{\circ}-120^{\circ}\right)+\cos 36^{\circ}\right) \cos 36^{\circ}$

$=\frac{1}{2}\left(-\cos 60^{\circ}+\cos 36^{\circ}\right) \cos 36^{\circ}$

$=\frac{1}{2}\left(-\frac{1}{2}+\frac{\sqrt{5}+1}{4}\right) \frac{\sqrt{5}+1}{4}$

$=\frac{1}{2} \times \frac{\sqrt{5}-1}{4} \times \frac{\sqrt{5}+1}{4}$

$=\frac{1}{8}$

= RHS

Hence proved.

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