Trigonometry problem?

Question

If cos^2 θ = (m^2 - 1)/3 and tan^3 (θ/2) = tan α,
then prove that sin^(2/3) α + cos^(2/3) α = (2/m)^(2/3)

French,
Thank you for taking pains of solving this lengthy problem. But could you please explain the following steps?

sin^(2/3)(α) + cos^(2/3)(α)
= (2/(1+cos(θ)))*(1/(((1-cos(θ))/(1+cos(θ)...
= 2/((1-cos(θ))^3+(1+cos(θ))^3)^(1/3)

Answer 1

First, you have:
sin^(2/3)(α)
= cos^(2/3)(α)*tan^(2/3)(α)
= cos^(2/3)(α)*tan^2(θ/2)

Then, you take the expression from the left side:
sin^(2/3)(α) + cos^(2/3)(α)
= cos^(2/3)(α) * (tan^2(θ/2) + 1)
= cos^(2/3)(α) / cos^2(θ/2)
= 2*cos^(2/3)(α) / (1+cos(θ))
because cos(θ) = cos^2(θ/2) - sin^2(θ/2) = 2*cos^2(θ/2) - 1.

Besides, you have:
cos^2(α) = 1/(tan^2(α) + 1) = 1/(tan^6(θ/2)+1)
because tan(α) = tan^3(θ/2).

tan^2(θ/2) = (1/cos^2(θ/2) )- 1 = (2/(cos(θ)+1)) - 1
(same reason as before).

The last two expressions lead to:
cos^2(α) = 1/((2/(cos(θ)+1) - 1)^3+1).

Finally, you have:
sin^(2/3)(α) + cos^(2/3)(α)
= (2/(1+cos(θ)))*(1/(((1-cos(θ))/(1+cos(θ)))^3+1)^(1/3)
= 2/((1-cos(θ))^3+(1+cos(θ))^3)^(1/3)
= 2/(2*(1+3*cos^2(θ)))^(1/3) (if you calculate).

As cos^2(θ) = (m^2 - 1)/3, the result is:
sin^(2/3)(α) + cos^(2/3)(α)
= (2/m)^(2/3)

Hope this is clear!