Question

do the question of attachment ​

Answer 1

We Have :-

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

Formula Used :-

$cos2x = 2cos^{2}x-1\\\\cos^{2}x=\dfrac{1+cos2x}{2} \\\\cos A + cos B= 2 cos \dfrac{A+B}{2}cos\dfrac{A-B}{2}$

Solution :-

$cos^{2}x + cos^{2}(x+\dfrac{\pi}{3} )+cos^{2}(x-\dfrac{\pi}{3})=\dfrac{3}{2} \\\\Taking\ LHS\\\\cos^{2}x + cos^{2}(x+\dfrac{\pi}{3} )+cos^{2}(x-\dfrac{\pi}{3})\\\\Using\ Identity\\\\\dfrac{1+cos2x}{2} +\dfrac{1+cos(2x+\dfrac{2\pi}{3} )}{2} +\dfrac{1+cos(2x-\dfrac{2\pi}{3} )}{2}\\\\\dfrac{1}{2} [ 3 + cos2x +cos(2x+\dfrac{2\pi}{3})+cos (2x-\dfrac{2\pi}{3})]$

$Using\ Identity$

$\dfrac{1}{2} [3+cos2x+2cos(\dfrac{2x+\dfrac{2\pi}{3}+2x-\dfrac{2\pi}{3} }{2})*cos( \dfrac{2x+\dfrac{2\pi}{3}-2x+\dfrac{2\pi}{3} }{2})]\\\\\dfrac{1}{2} [3+cos2x+2cos2x*cos\dfrac{2\pi}{3}]\\ \\\dfrac{1}{2}[3+cos2x+( 1+2cos\dfrac{2\pi}{3} )]\\\\\dfrac{1}{2} [3+cos2x(1+2cos(\dfrac{-\pi}{3}))]\\ \\\dfrac{1}{2}[3+cos2x(1-2*\dfrac{1}{2} )]\\ \\\dfrac{1}{2} [3+cos2x(1-1)]\\\\\dfrac{1}{2}[3+cos2x(0)]\\ \\\dfrac{1}{2}[3+0]\\ \\\dfrac{3}{2} \\\\LHS=RHS\\\\Hence\ Proved$