In a ABC, the value of ^2 A+ ^2 (A+ 3 )+ ^2 (A- 3 ) is - Vidyadip

MCQ

In a $\triangle ABC$, the value of $\cos ^2 A+\cos ^2\left(A+\frac{\pi}{3}\right)+\cos ^2\left(A-\frac{\pi}{3}\right)$ is

A
$0$
B
$\frac{1}{2}$
✓
$\frac{3}{2}$
D
1

✓

Answer

Correct option: C.

$\frac{3}{2}$

(C)
$\cos ^2 A+\cos ^2\left(A+\frac{\pi}{3}\right)+\cos ^2\left(A-\frac{\pi}{3}\right)$
$=\frac{1}{2}(1+\cos 2 A)+\frac{1}{2}\left\{1+\cos \left(2 A+\frac{2 \pi}{3}\right)\right\}$$+\frac{1}{2}\left\{1+\cos \left(2 A-\frac{2 \pi}{3}\right)\right\}$
$=\frac{3}{2}+\frac{1}{2} \cos 2 A$$+\frac{1}{2}\left\{\cos \left(2 A+\frac{2 \pi}{3}\right)+\cos \left(2 A-\frac{2 \pi}{3}\right)\right\}$
$=\frac{3}{2}+\frac{1}{2} \cos 2 A+\cos 2 A \cos \frac{2 \pi}{3}$
$\ldots[\because \cos (A+B)+\cos (A-B)$$=2 \cos A \cos B ]$
$=\frac{3}{2}+\frac{1}{2} \cos 2 A-\frac{1}{2} \cos 2 A=\frac{3}{2}$

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