Question

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

Answer 1

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