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
问题描述：
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
答
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