已知sin(π/4-α)=5/13,α∈(0,π/4) 求cos2α/cos(π/4+α)

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已知sin(π/4-α)=5/13,α∈(0,π/4) 求cos2α/cos(π/4+α)

已知sin(π/4-α)=5/13,α∈(0,π/4) 求cos2α/cos(π/4+α)
已知sin(π/4-α)=5/13,α∈(0,π/4) 求cos2α/cos(π/4+α)

已知sin(π/4-α)=5/13,α∈(0,π/4) 求cos2α/cos(π/4+α)
sin(Π/4 -x)=5/13 √2/2(cosx-sinx)=5/13
x∈(0,Π/4),
cos(Π/4 -x)=12/13 √2/2(cosx+sinx)=12/13 相乘,得
1/2(cos^2x-sin^2x)=60/169
cos2x=120/169
cos(Π/4 +x)=sin(Π/4 -x)=5/13
(cos2x)/[cos(Π/4 +x)]=(120/169)/(5/13)=24/13

两端平方得 [sin(π/4-a)]^2=25/169 ,即 [1-cos(π/2-2a)]/2为锐角,因此 cos(2a)=√[1-(sin2a)^2]=120/169 。

∵α∈(0,π/4)
∴π/4-α∈(0,π/4)
∴cos(π/4-α)=√[1-sin²(π/4-α)]=√[1-﹙5/13﹚²]=12/13
cos2α=sin﹙π/2-2α﹚=sin[2﹙π/4-α﹚]=2sin(π/4-α)cos(π/4-α)
=2×﹙5/13﹚×﹙12/13﹚=120/169
cos(π/4+α)=sin[π...

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∵α∈(0,π/4)
∴π/4-α∈(0,π/4)
∴cos(π/4-α)=√[1-sin²(π/4-α)]=√[1-﹙5/13﹚²]=12/13
cos2α=sin﹙π/2-2α﹚=sin[2﹙π/4-α﹚]=2sin(π/4-α)cos(π/4-α)
=2×﹙5/13﹚×﹙12/13﹚=120/169
cos(π/4+α)=sin[π/2-﹙π/4 +α)]=sin(π/4-α)=5/13
∴cos2α/cos(π/4+α)=﹙120/169﹚/﹙5/13﹚
=24/13

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