已知sin(π/4+α)*sin(π/4-α)=1/4,α∈(π/4,π/2),求2sin^α+tanα-cotα-1的值

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已知sin(π/4+α)*sin(π/4-α)=1/4,α∈(π/4,π/2),求2sin^α+tanα-cotα-1的值
已知sin(π/4+α)*sin(π/4-α)=1/4,α∈(π/4,π/2),求2sin^α+tanα-cotα-1的值

已知sin(π/4+α)*sin(π/4-α)=1/4,α∈(π/4,π/2),求2sin^α+tanα-cotα-1的值
sin(π/4+α)*sin(π/4-α)
=sin(π/4+α)*cos(π/4-α)
=1/2sin(π/2+2α)
=1/2cos2α
1/2cos2α=1/4
cos2α=1/2 α∈(π/4,π/2),
所以2α=π/3
2sin^α+tanα-cotα-1
=2sin^α-1+sinα/cosα-cosα/sinα
=-cos2α+(sin^α-cos^α)/sinαcosα
=-cos2α-cos2α/(sin2α/2)
=-(cos2α+2cot2α)
=-(cosπ/3+2cotπ/3)
=-(1/2+2√3/3)
=-2√3/3-1/2

（根号2/2）^2*(cosa-sina)(cosa+sina)=1/2
所以C^2-S^2=1/2,又C^2+S^2=1 α∈(π/4,π/2)
所以C=根号3/2，S=1/2，即α=π/6
2sin^α+tanα-cotα-1
=1/2+根号3/3-根号3-1
=……

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