已知sin(π/4+α)+sin(π/4-α)=根号2/3,（1）求sinα（2）求sin（α-π/4）/1-sin2α-cos2α

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已知sin(π/4+α)+sin(π/4-α)=根号2/3,（1）求sinα（2）求sin（α-π/4）/1-sin2α-cos2α

sin(π/4+α)+sin(π/4-α)
=sinπ/4cosα+cosπ/4sinα+sinπ/4cosα-cosπ/4sinα
=2sinπ/4cosα
=2*√2/2cosα
=√2cosα
√2cosα=√2/3
cosα=1/3
sinα=±2√2/3
sin（α-π/4）/(1-sin2α-cos2α )
=sin（α-π/4）/[1-sin2α-2(cosα)^2+1]
=sin（α-π/4）/[2-2(cosα)^2-sin2α]
=sin（α-π/4）/[2(sinα)^2-2sinαcosα]
=sin（α-π/4）/[2sinα(sinα-cosα)]
=sin（α-π/4）/[2√2sinα(√2/2sinα-√2/2cosα)]
=sin（α-π/4）/[2√2sinαsin(α-π/4)]
=1/(2√2sinα)
=1/[2√2*(±2√2/3)]
=1/(±8/3)
=±3/8

sin(π/4+α)+sin(π/4-α)=2(sin(π/4+a+π/4-a)/2)cos((π/4+a-π/4+a)/2)=2sinπ/4cosα=√2cosα=√2/3
cosα=1/3
sinα=±2√2/3
1-sin2α-cos2α =1-2sinacosa+2(sinα)^2-1=2(sinα)^2-2sinαcosα=2sinα(sinα-cosα)

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sin(π/4+α)+sin(π/4-α)=2(sin(π/4+a+π/4-a)/2)cos((π/4+a-π/4+a)/2)=2sinπ/4cosα=√2cosα=√2/3
cosα=1/3
sinα=±2√2/3
1-sin2α-cos2α =1-2sinacosa+2(sinα)^2-1=2(sinα)^2-2sinαcosα=2sinα(sinα-cosα)
=2√2sinαsin(α-π/4)
原式=sin(α-π/4)/(2√2sinαsin(α-π/4))
=1/[2√2*(±2√2/3)]=1/(±8/3)=±3/8

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