Question

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

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

Answer 1

The correct option is A

$\frac{3}{2}$

Explanation for correct option:

Simplify the series using Trigonometric identity:

$$\begin{gathered} {\cos }^2\left(\alpha \right)+{\cos }^2\left(\alpha +120°\right)+{\cos }^2\left(\alpha -120°\right) \\ ={\cos }^2\left(\alpha \right)+{\left[\cos \left(\alpha +120°\right)\right]}^2+{\left[\cos \left(\alpha -120°\right)\right]}^2 \\ ={\cos }^2\left(\alpha \right)+{[\cos \left(\alpha \right)\cos \left(120°\right)-\sin \left(\alpha \right)\sin \left(120°\right)]}^2+{[\cos \left(\alpha \right)\cos \left(120°\right)+\sin \left(\alpha \right)\sin \left(120°\right)]}^2[\because \cos \left(C+D\right)=\cos \left(C\right)\cos \left(D\right)-\sin \left(C\right)\sin \left(D\right)] \\ ={\cos }^2\left(\alpha \right)+{[\cos \left(\alpha \right)\times \frac{-1}{2}-\sin \left(\alpha \right)\frac{\sqrt{3}}{2}]}^2+{[\cos \left(\alpha \right)\frac{-1}{2}+\sin \left(\alpha \right)\frac{\sqrt{3}}{2}]}^2[\because \cos \left(C-D\right)=\cos \left(C\right)\cos \left(D\right)+\sin \left(C\right)\sin \left(D\right)] \\ ={\cos }^2\left(\alpha \right)+\frac{{\cos }^2\left(\alpha \right)}{4}+\frac{3{\sin }^2\left(\alpha \right)}{4}+\frac{\sqrt{3}\sin \left(\alpha \right)\cos \left(\alpha \right)}{2}+\frac{{\cos }^2\left(\alpha \right)}{4}+\frac{3{\sin }^2\left(\alpha \right)}{4}-\frac{\sqrt{3}\sin \left(\alpha \right)\cos \left(\alpha \right)}{2} \\ =\frac{3{\cos }^2\left(\alpha \right)}{2}+\frac{3{\sin }^2\left(\alpha \right)}{2} \\ =\frac{3}{2}[{\cos }^2\left(\alpha \right)+{\sin }^2\left(\alpha \right)] \\ =\frac{3}{2} \end{gathered}$$

Therefore, Option(A) is the correct answer.