Question

Prove that ?

`cos^2x +cos^2(x+pi/3)+cos^2(x-pi/3)=3/2`

Answer 1

Proof below...

Explanation:

We can use our knowledge of additional formulae...

`cos(A+B) = cosAcosB - sinAsinB`

`cos^2 (x+pi/3) = ( cosxcos(pi/3) - sinx sin(pi/3) )^2`

`= (1/2cosx - sqrt(3)/2 sinx )^2 = 1/4cos^2x -sqrt(3)/2 sinxcosx +3/4 sin^2 x`

`cos^2 (x-pi/3) = ( cosxcos(pi/3) + sinxsin(pi/3) )^2`

`= (1/2cosx + sqrt(3)/2 sinx )^2 = 1/4cos^2x + sqrt(3)/2 sinxcosx +3/4cos^2 x`

`=> cos^2x + cos^2(x-pi/3) + cos^2(x+pi/3)`

`= cos^2x + 1/2cos^2x + 3/2 sin^2 x = 3/2cos^2x + 3/2sin^2x`

`-= 3/2 ( cos^2 x + sin^2 x ) = color(blue)(3/2`

Using the identity `sin^2 theta + cos^2 theta -= 1`

Answer 2

Another approach.

Explanation:

We'll use 1) `2cos^2x=1+cos2x`
2) `cosA+cosB=2cos((A+B)/2)*cos((A-B)/2)`

`LHS=cos^2x+cos^2(x+60°)+cos^2(x-60°)`

`=1/2[2cos^2x+2cos^2(x+60°)+2cos^2(x-60°)]`

`=1/2[1+cos2x+1+cos2*(x+60°)+1+cos2*(x-60°)]`

`=1/2[3+cos2x+cos(2x+120°)+cos(2x-120°)]`

`=3/2+1/2*[cos2x+2*cos((2x+120°+2x-120)/2)*cos((2x+120°-(2x-120°))/2)]`

`=3/2+1/2*[cos2x+2cos(2x)*cos120°]`

`=3/2+1/2[cos2x+2cos2x*(-1/2)]`

`=3/2+1/2*0=3/2=RHS`