If cos α + 2Sin α = - 5, then Tan α=______ ．

If cos α + 2Sin α = - 5, then Tan α=______ ．

It is known that 5sin (α + φ) = - 5 (where Tan φ = 12), that is, sin (α + φ) = - 1, so α + φ = 2K π - π 2, α = 2K π - π 2 - φ (K ∈ z), so tan α = Tan (− π 2 − φ) = 1tan φ = 2

Given that tan2 α = & frac34; (π / 2 < α < π), find the square of [2cos α / 2 + sin α - 1] / [√ 2cos (α + π / 4)]

tan2α=3/4
2tanα/(1-tan^2α) = 3/4
8tanα = 3 - 3tan^2α
3tan^2α + 8tanα - 3 = 0
(tanα+3)(3tanα-1) = 0
∵π／2＜α＜π
∴tanα＜0
∴3tanα-1＜0
∴tanα+3 = 0
∴tanα = -3
[2cos^2(α/2)+sinα-1]／[√2cos(α+π/4)]
= [(cosα + 1)+sinα-1]／[√2(cosαcosπ/4-sinαsinπ/4)]
= (cosα + sinα)／(cosα - sinα)
= (1+tanα)/(1-tanα)
= (1-3)/(1+3)
= -1/2

If cos (π 2 − α) = 2cos (3 π 2 + β), 3sin (3 π 2 − α) = − 2Sin (π 2 + β), and 0 ＜ α ＜ π, 0 ＜ β ＜ π, the values of α and β can be obtained

∵ cos (π 2 - α) = sin α, cos (3 π 2 + β) = sin β, sin (3 π 2 - α) = - cos α, sin (π 2 + β) = cos β, the known two equations are transformed as: sin α = 2Sin β ①, - 3cos α = - 2cos β ②, ① 2 + ② 2: sin 2 α + 3cos 2 α = 2 (sin 2 β + cos 2 β) = 2, and sin 2 α + cos 2 α = 1, 0 ＜ α＜ π, 0 ＜ β＜ π, ∵ sin 2 α = cos 2 α = 12, that is sin α = 22, sin β = 12, ∵ α = π 4, β = π 6 or α=3π4，β=5π6．

If 3sin ^ 2 (α) + 2Sin ^ 2 (β) = 2Sin α, the value range of COS ^ 2 (α) + cos ^ 2 (β) is

3sin ^ 2 (α) + 2Sin ^ 2 (β) = 2Sin α sin ^ 2 (α) + (2-2cos ^ 2 (α)) + (2-2cos ^ 2 (β)) = 2Sin α cos ^ 2 (α) + cos ^ 2 (β) = [sin ^ 2 (α) - 2Sin α + 4] / 2 let sin α = t because 2Sin ^ 2 (β) = 2Sin α - 3sin ^ 2 (α) ≥ 02t-3t ^ 2 ≥ 02 / 3 ≥ t ≥ 0cos ^ 2 (α) + cos ^ 2 (β) = [