Question

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

Answer 1

Given to prove :-

cos²x + cos²(x+π/3)+ cos²(x-π/3) = 3/2

Solution:-

Take L.H.S

As we know that π is 180° Substitute there

cos²x + cos²(x +180°/3) + cos²(x-180°/3)

cos²x + cos²(x+60°) + cos²(x-60°)

From Trigonometric  identities

sin²A + cos²A = 1

cos²A = 1-sin²A

So,

cos²(x-60°) = 1- sin²(x-60°)

cos²x + cos²(x+60°) + 1-sin²(x-60°)

1 + cos²x + cos²(x+60°) -sin²(x-60°)

cos²(x+60°) -sin²(x-60°)  in form of cos²A -sin²B

We all know that ,

cos(A+B) cos(A-B) = cos²A -sin²B  

Applying that formula

1 + cos²x + cos(x+60°+x-60°) . cos(x+60°-x+60°)

1 + cos²x + [cos 2x.cos(120°)]

We know that cos 2x in terms of cos

cos2x = 2cos²x - 1

and cos120° = -1/2

1+ cos²x +[  2cos²x-1×-1/2 ]

1 + cos²x+[ -2cos²x +1 /2]

Taking L.C.M

2 + 2cos²x -2 cos²x + 1 /2

2+1/2

3/2

So,

cos²x + cos²(x+π/3)+ cos²(x-π/3) = 3/2

Hence proved

Used formulae:-

cos(A +B) cos (A-B) = cos²A - sin²B

cos²A = 1-sin²A

cos2A = 2cos²A - 1

cos120° = -1/2

Answer 2

Answer :

cos²x + cos²(x + π/3) + cos²(x - π/2) = 3/2

Step-by-step explanation :

cos²x + cos²(x + π/3) + cos²(x - π/2) = 3/2

Considering L.H.S :

cos²x + cos²(x + π/3) + cos²(x - π/2)

(1 + cos2x)/2 + [1+cos2(x+π/3)]/2 + [1+cos2(x-π/3)]/2

1/2[1 + cos2x + 1 + cos2(x+π/3)] + 1 + cos2(x-π/3)

Let x = (2x+2π)/3 and y = (2x-2π)/3 :

Then we get :

1/2[3 + cos2x + 2cos(2x-2π/3+2x-2π/3)]cos[2x + 2π/3 - (2x-2π/3)]/2

1/2[3 + cos2x + cos(4x/2)cos(4π/3/2)]

1/2[3 + cos2x + 2cos2x cos(π - π/3)

1/2[3 + cos2x + 2cos2x (-cosπ3)

1/2[3 + cos2x + 2cos2x(-1/2)

1/2[3 + cos2x - cos2x]

3/2

∴ L.H.S = R.H.S

Formulae used :

cos2x = 2cos² - 1 = 1 + cos2x/2

cosx + cosy = 2cos(x+y/2)cos(x-y/2)

cos(π - θ) = -cosθ