2T,Q.14 Prove thatcos 2 x + cos 2 (x+ä¸)+cos 2 (x-ä¸) = 3

1.	2T,Q.14 Prove thatcos 2 x + cos 2 (x+ä¸)+cos 2 (x-ä¸) = 3
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

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

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