Question

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

Answer 1

$Using this, cos^2x = {1 + cos(2x)}/2$

$cos^2(x + π/3) = {1 + cos(2x + 2π/3)}/2 and cos^2(x - π/3) = {1 + cos(2x - 2π/3)}/2$

ii) Hence, left side of the given one is:

$= {1 + cos(2x)}/2 + {1 + cos(2x + 2π/3)}/2 + {1 + cos(2x - 2π/3)}/2$

$= (3/2) + (1/2)[cos(2x) + cos(2x + 2π/3) + cos(2x - 2π/3)]$

$= (3/2) + (1/2)[cos(2x) + 2cos(2x)*cos(2π/3)]$

[Since cos(A+B) + cos(A-B) = 2cosA*cosB]

$= (3/2) + (1/2)[cos(2x) - cos(2x)] [Since cos(2π/3) = -1/2]$

= 3/2 = Right side

If anable toh understand the answer see the attachment ❤️

Answer 2

Solution :

${\Rightarrow\sf cos^2+cos^2\Big(x+\dfrac{\pi}{3}\Big)+cos^2\Big(x-\dfrac{\pi}{3}\Big)}$

${\Rightarrow\sf \dfrac{1+cos\:2x}{2} + \Bigg[\dfrac{1+cos\:2\Big(x+\dfrac{\pi}{3}\Big)}{2}\Bigg]+ \Bigg[\dfrac{1+cos\:2\Big(x-\dfrac{\pi}{3}\Big)}{2}\Bigg]}$

${\Rightarrow\sf \dfrac{1}{2}\Bigg[1+cos\:2x+1+cos\:2\Big(x+\dfrac{\pi}{3}\Big)+1+cos\:2\Big(x-\dfrac{\pi}{3}\Big)\Bigg]}$

${\Rightarrow\sf \dfrac{1}{2}\Bigg[3+cos\:2x+cos\:2\Big(x+\dfrac{\pi}{3}\Big)+cos\:2\Big(x-\dfrac{\pi}{3}\Big)\Bigg]}$

${\Rightarrow\sf \dfrac{1}{2}\Bigg[3+cos\:2x+2cos\Bigg(\dfrac{2x+\dfrac{2\pi}{3}+2x-\dfrac{2\pi}{3}}{2}\Bigg)+cos\Bigg(\dfrac{2x+\dfrac{2\pi}{3}+2x-\dfrac{2\pi}{3}}{2}\Bigg)\Bigg]}$

${\Rightarrow\sf \dfrac{1}{2}\Bigg[3+cos \:2x+2cos \Bigg(\dfrac{4x}{2}\Bigg)cos \Bigg(\dfrac{\dfrac{4\pi}{3}}{2}\Bigg)\Bigg]}$

${\Rightarrow\sf \dfrac{1}{2} \Bigg[3+ cos\:2x + 2 cos \:2x \times cos \dfrac{2 \pi}{3}\Bigg]}$

${\Rightarrow\sf \dfrac{1}{2} \Bigg[3+cos\:2x + 2cos\: 2x \Big(cos\pi - \dfrac{\pi}{3}\Big) \Bigg]}$

${\Rightarrow \sf \dfrac{1}{2} \Bigg[3+cos\:2x + 2cos\:2x \Big(\dfrac{-1}{2}\Big)\Bigg]}$

${\Rightarrow \sf \dfrac{1}{2} \Bigg[3+cos\:2x - cos\:2x\Bigg] \Rightarrow \dfrac{3}{2}}$

Required Answer :

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