α属于（0,π/2）且2sin²α-sinαcosα-3cos²α=0求sin(α+π/4)/sin2α+cos2α+1的值?

α属于（0,π/2）且2sin²α-sinαcosα-3cos²α=0求sin(α+π/4)/sin2α+cos2α+1的值?
α属于（0,π/2）且2sin²α-sinαcosα-3cos²α=0求sin(α+π/4)/sin2α+cos2α+1的值?

α属于（0,π/2）且2sin²α-sinαcosα-3cos²α=0求sin(α+π/4)/sin2α+cos2α+1的值?
2sin²α - sinαcosα - 3cos²α = 0
(2sinα - 3cosα)(sinα + cosα) = 0
因为α ∈ (0,π/2),sinα + cosα ≠ 0,所以
sinα = (3/2)cosα
1 - cos²α = (9/4)cos²α
cosα = 2√13/13
sin(α + π/4) / (sin2α + cos2α + 1)
= [sinαcos(π/4) + cosαsin(π/4)] / [2sinαcosα + 2cos²α - 1 + 1]
= (√2/4)[(3/2)cosα + cosα] / [(3/2)cos²α + cos²α]
= (√2/4)/cosα
= (√2/4)/(2√13/13)
= √26/8

由2sin²α-sinαcosα-3cos²α=0
(2sinα-3)(sinα+1)=0
得sinα=-1
cosα=1
sin(α+π/4)/sin2α+cos2α+1
=(√2sinα+√2cosα)/2(sin2α+2cos²α)
=(-√2+√2)/2(sin2α+2cos²α)
=...

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由2sin²α-sinαcosα-3cos²α=0
(2sinα-3)(sinα+1)=0
得sinα=-1
cosα=1
sin(α+π/4)/sin2α+cos2α+1
=(√2sinα+√2cosα)/2(sin2α+2cos²α)
=(-√2+√2)/2(sin2α+2cos²α)
=0

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