Question

Prove :-
cos²x + cos² (x + π/3) + cos² (x - π/3) = 3/2 ​

Answer 1

Answer

$\sf \dashrightarrow LHS = {cos}^{2} x + {cos}^{2} \bigg(x + \dfrac{\pi}{3} \bigg) + {cos}^{2} \bigg( x - \dfrac{\pi}{3} \bigg)$

We know that cos²θ = 1 - sin²θ

$\sf \dashrightarrow {cos}^{2} x + {cos}^{2} \bigg(x + \dfrac{\pi}{3} \bigg) + 1 - {sin}^{2} \bigg( x - \dfrac{\pi}{3} \bigg)$

$\sf \dashrightarrow {cos}^{2} x + 1 + \bigg[{cos}^{2} \bigg( x + \dfrac{\pi}{3} \bigg) - {sin}^{2} \bigg( x - \dfrac{\pi}{3} \bigg) \bigg]$

$\sf \dashrightarrow {cos}^{2} x + 1 + \bigg[cos \bigg( x + \dfrac{\pi}{3} + x - \dfrac{\pi}{3} \bigg) - cos \bigg( x + \dfrac{\pi}{3} - x + \dfrac{\pi}{3} \bigg) \bigg]$

$\sf \dashrightarrow {cos}^{2} x + 1 + \bigg[cos2x.cos \dfrac{2 \pi}{3} \bigg]$

$\sf \dashrightarrow \dfrac{1 + cos2x}{2} + 1 - \dfrac{1}{2} cos2x \: \: \: \bigg( cos \dfrac{2 \pi}{3} = - \dfrac{1}{2} \bigg)$

$\sf \dashrightarrow \dfrac{1}{2} + \dfrac{1}{2} cos2x + 1 - \dfrac{1}{2} cos2x$

$\sf \dashrightarrow \dfrac{3}{2} = RHS$

$\sf LHS = RHS$

$\sf \dashrightarrow {cos}^{2} x + {cos}^{2} \bigg(x + \dfrac{\pi}{3} \bigg) + {cos}^{2} \bigg( x - \dfrac{\pi}{3} \bigg) = \dfrac{3}{2}$