MCQ
${\cos ^2}\alpha + {\cos ^2}(\alpha + 120^\circ ) + {\cos ^2}(\alpha - 120^\circ )$ is equal to
- ✓$3/2$
- B$1$
- C$1/2$
- D$0$
$ = {\cos ^2}\alpha + {\left\{ {\cos \,(\alpha + {{120}^o}) + \cos \,(\alpha - {{120}^o})} \right\}^2}$$ - 2\,\cos \,(\alpha + {120^o})\,\cos \,(\alpha - {120^o})$
$ = {\cos ^2}\alpha + {\left\{ {\,2\,\cos \,\,\alpha \,\cos \,{{120}^o}} \right\}^2} - 2\,\left\{ {{{\cos }^2}\alpha - {{\sin }^2}\,{{120}^o}} \right\}$
$ = {\cos ^2}\alpha + {\cos ^2}\alpha - 2\,{\cos ^2}\alpha + 2\,{\sin ^2}\,{120^o}$
$ = 2{\sin ^2}{120^o} $
$= 2 \times \frac{3}{4} $
$= \frac{3}{2}$.
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