Prove that, cosα cos(120+α) cos(240+α)= $\frac{1}{4}$ cos3α
$cos\alpha cos(120 + \alpha )cos(240 + \alpha )$ = $\frac{1}{4}cos3\alpha$
Solution:
cosα cos(120°+α) cos(240°+α)
= $\frac{1}{2}$ cosα{2 cos(120°+α) cos(240°+α)}
= $\frac{1}{2}$ cosα {cos(120°+α+240°+α)+cos(120°+α-240°-α)}
= $\frac{1}{2}$ cosα {cos(360°+2α)+cos(-120°)}
= $\frac{1}{2}$ cosα (cos2α+cos120°)
= $\frac{1}{2}$ cosα (cos2α-$\frac{1}{2}$)
= $\frac{1}{2}$ cosα cos2α -$\frac{1}{2}$ cosα
=$\frac{1}{4}$ (2cosα cos2α) -$\frac{1}{2}$ cosα
= $\frac{1}{4}$ {cos(α+2α) +cos(α-2α)} -$\frac{1}{2}$ cosα
= $\frac{1}{4}$ (cos3α +cosα) – $\frac{1}{2}$ cosα
= $\frac{1}{4}$ cos3α + $\frac{1}{4}$ cosα – $\frac{1}{2}$ cosα
= $\frac{1}{4}$ cos3α [Proved]