Question

How do you prove cos^2a + cos^2(2π/3 + a) + cos^2(2π/3 - a) = 3/2? a = alpha Thanks!

Answer 1

We have:

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

We can use the sum and difference formulas for cosine to expand.

`(cosalpha)^2 + (cos((2pi)/3)cosalpha - sin((2pi)/3)sin alpha)^2 + (cos((2pi)/3)cosalpha + sin((2pi)/3)sinalpha)^2 = 3/2`

`(cosalpha)^2 + (-1/2cosalpha - (sqrt(3)/2sinalpha))^2 + (-1/2cosalpha + sqrt(3)/2sinalpha)^2 = 3/2`

`(cosalpha)^2 + 1/4cos^2alpha + 3/4sin^2alpha + sqrt(3)/4cosalphasinalpha + 1/4cos^2alpha + 3/4sin^2alpha - sqrt(3)/4cosalphasinalpha = 3/2`

`cos^2alpha + 1/2cos^2alpha + 6/4sin^2alpha = 3/2`

`3/2cos^2alpha + 3/2sin^2alpha = 3/2`

`3/2(cos^2alpha + sin^2alpha) = 3/2`

`3/2 = 3/2`

`LHS = RHS`

As required.

Hopefully this helps!