Question

cos'2x+cos'2(x+pi/3)+cos'2(x-pi/3)=3/2

Answer 1

Using this, cos²x = {1 + cos(2x)}/2 
cos²(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                            HENCE PROVED