Prove that:\cos ^{2} \theta+\cos ^{2}\left(\frac{2 \pi}{3}-\theta\righ

1.	Prove that:\cos ^{2} \theta+\cos ^{2}\left(\frac{2 \pi}{3}-\theta\right)+\cos ^{2}\left(\frac{2 \pi}{3}+\theta\right)=\frac{3}{2}
Answer» i) By double angle identity, cos²θ = {1 + cos(2θ)}/2 Using this, cos²x = {1 + cos(2x)}/2cos²(x + π/3) = {1 + cos(2x + 2π/3)}/2 and cos²(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) = 2cosAcosB] = (3/2) + (1/2)[cos(2x) - cos(2x)] [Since cos(2π/3) = -1/2] = 3/2 = Right side [Proved]

Prove that:\cos ^{2} \theta+\cos ^{2}\left(\frac{2 \pi}{3}-\theta\right)+\cos ^{2}\left(\frac{2 \pi}{3}+\theta\right)=\frac{3}{2}

Answer»

i) By double angle identity, cos²θ = {1 + cos(2θ)}/2

Using this, cos²x = {1 + cos(2x)}/2cos²(x + π/3) = {1 + cos(2x + 2π/3)}/2 and cos²(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 [Proved]