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Prove that Cos²A + Cos²(A+120) + Cos² (A-120) = 3/2

$\cos^{2} A + \cos^{2} (A + 120 °) + \cos^{2} (A - 120 °) = \frac{1 + \cos 2 A}{2} + \frac{1 + \cos (2 A + 240 °)}{2} + \frac{1 + \cos (2 A - 240 °)}{2} = \frac{3}{2} + \frac{1}{2} [\cos 2 A + \cos (2 A + 240 °) + \cos (2 A - 240 °)] = \frac{3}{2} + \frac{1}{2} [\cos 2 A + 2 \cos \frac{4 A}{2} \cos \frac{480 °}{2}] = \frac{3}{2} + \frac{1}{2} [\cos 2 A + 2 \cos 2 A \cos (180 ° + 60 °)] = \frac{3}{2} + \frac{1}{2} [\cos 2 A + 2 \cos 2 A (- \cos 60 °)] = \frac{3}{2} + \frac{1}{2} [\cos 2 A - 2 \cos 2 A (\frac{1}{2})] = \frac{3}{2} + \frac{1}{2} (0) = \frac{3}{2}$

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