Question

Show that ${\cos }^2(A-{120}^o)+{\cos }^{2}A+{\cos }^2(A+{120}^o)=\frac{3}{2}$.

Answer 1

${\cos }^{2}A+{\cos }^2(A+{120}^o)+{\cos }^2(A-{120}^o)$

$=\frac{1+\cos 2A}{2}+\frac{1+\cos (2A+{240}^o)}{2}+\frac{1+\cos (2A-{240}^o)}{2}$

$=\frac{3}{2}+\frac{1}{2}[\cos 2A+\cos (2A+{240}^o)+\cos (2A-{240}^o\left)\right]$

$=\frac{3}{2}+\frac{1}{2}[\cos 2A+2\cos \frac{4A}{2}\cos \frac{{480}^o}{2}]$

$=\frac{3}{2}+\frac{1}{2}[\cos 2A+2\cos 2A\cos ({180}^o+{60}^o\left)\right]$

$=\frac{3}{2}+\frac{1}{2}[\cos 2A+\cos 2A(-\cos {60}^o\left)\right]$

$=\frac{3}{2}+\frac{1}{2}[\cos 2A-2\cos 2A\left(\frac{1}{2}\right)]$

$=\frac{3}{2}+\frac{1}{2}\left(0\right)=\frac{3}{2}$