Question

${\cos }^2\alpha +{\cos }^2\left(\alpha +{120}^{\circ }\right)+{\cos }^2\left(\alpha -{120}^{\circ }\right)$ is equal to

• A. $\frac{3}{2}$
• B. $1$
• C. $\frac{1}{2}$
• D. $0$

Answer 1

Answer: (A)

$\frac{3}{2}$

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

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

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

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

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

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

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

$=\frac{3}{2}+\frac{1}{2}\left[0\right]$

$=\frac{3}{2}+0$

$=\frac{3}{2}$

Hence, the correct option is A. $\frac{3}{2}$.