If α ∈ (0, π) and cos α + sin α = - 1 / 3, then Cos2 α=

If α ∈ (0, π) and cos α + sin α = - 1 / 3, then Cos2 α=

∵cosα+sinα=-1/3
∴(cosα+sinα)²=1/9
∴sin²α+cos²α+2sinαcosα=1/9
∴2sinαcosα=-8/9
∴sin2α=-8/9
∵α∈(0,π)
∴2α∈(0,2π)
∴cos2α=±√(1-sin²2α)
=±√17/9
Revise it
∵cosα+sinα=-1/3,α∈(0,π)
∴cosα<0,sinα>0
And | cos α | sin α
∴α∈(3π/2,π)
∴2α∈(3π/2,2π)
∴cos2α>0
Ψ Cos2 α = - √ 17 / 9 (rounding off)
To sum up,
cos2α=√17/9