Question

${\cos }^{2}A\theta +{\cos }^2(A+{120}^{\circ })+{\cos }^3(A-{120}^{\circ })=\frac{3}{2}$

Answer 1

L.H.S. = $\frac{1}{2}\left[\right(1+\cos 2A)+1+\cos (2A+{240}^{\circ }+1+\cos (2A-{240}^{\circ }$)]
$=\frac{1}{2}[3+\cos 2A+2\cos 2A\cos {240}^{\circ }$]
$=\frac{1}{2}[3+\cos 2A-\cos 2A]=\frac{3}{2}.$
[$\because \cos {240}^{\circ }=\cos ({270}^{\circ }-{30}^{\circ })=-sin{30}^{\circ }=-1/2]$