已知sin(π/4+α)·sin(π/4-α)=-1/4,α∈(π/4,π/2),求：（1）2sin^2α+(1/(secα·cscα))-1的值；（2）sin^4α+cos^4α的值

Question

已知sin(π/4+α)·sin(π/4-α)=-1/4,α∈(π/4,π/2),求：
（1）2sin^2α+(1/(secα·cscα)) -1的值；（2）sin^4α+cos^4α的值

Answer 1

1.2sin^2α=1-2cos2α
(1/(secα·cscα)) =sinα ·cosα =1/2sin2α
原式得1/2sin2α-2cos2α
2.sin^4α=(sin^2α)^2=1/4(1-cos2α)^2
cos^4α==(cos^2α)^2=1/4(1+cos2α)^2
原式得1/4(1-cos2α)^2+1/4(1+cos2α)^2 一化简就行了
条件sin(π/4+α)·sin(π/4-α)=-1/4可得1/2cos2α=-1/4,2α=2/3π,所以α=1/3π
带入1,得1/4根号3+1
带入2,得5/8
楼上
sin(π/4+α)*sin(π/4-α)
=sin(π/4+α)*cos(π/4-α)
错了