If 0 ≤ α < 2 π, if sin α < 0 and Cos2 α < 0, what is the value range of α

If 0 ≤ α < 2 π, if sin α < 0 and Cos2 α < 0, what is the value range of α

If 0 ≤ α < 2 π, if sin α < 0, we get π < α < 2 π
And Cos2 α < 0 get 135 degrees < α < 315 degrees
Therefore, to sum up, π < α < 315 degrees

Let 0 ＜θ＜ 2 π, if sin θ＜ 0, Cos2 θ

sinx

Sin (α + π / 2) = 1 / 3, then Cos2 α=____ .

sin（α+π/2）
=cosa
=1/3,
So Sina = ± √ (1-1 / 9) = ± 2 √ 2 / 3;
Then Cos2 α = cos & # 178; a-SiN & # 178; a = 1 / 9-8 / 9 = - 7 / 9;
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If sin (π + θ) = - 1 / 3, then Cos2 θ

sin(π+θ)=-sinθ=-1/3
COS2θ=1-2sin²a=7/9

If α∈ (0, π) and cos α + sin α = - 1 / 3, then Cos2 α =?
Why is  2Sin α cos α = - 8 / 9
∴sin2α=-8/9
How did you get here

The solution is obtained by the formula sin 2 θ = 2 sin θ cos θ

Let a ∈ (0, π), sin α + cos α = 1 / 2, then Cos2 α =?

Sin α + cos α = 1 / 2, both sides Square: 1 + sin2a = 1 / 4
Sin2a = - 3 / 4, so 2A ∈ (π, 2 π), so a ∈ (π / 2, π),
So Cos2 α = ± √ (1-sin2a ^ 2) = ± √ 7 / 4

Given α∈ (0, π), sin α + cos α = 1 / 2, find Cos2 α

Sin α + cos α = 1 / 2, find Cos2 α:
(sinα+cosα)^2=(sinα)^2+(cosα)^2+2sinαcosα=1/4
So 1 + 2Sin α cos α = 1 / 4
Then sin2 α = 2Sin α cos α = - 3 / 4
Because (sin2 α) ^ 2 + (Cos2 α) ^ 2 = 1
So Cos2 α = + / - (root 7) / 4