已知,sin(π/4 2α)*sin(π/4-2α),α∈(π/4,π/2)求2sin²α tanα-1/tanα-1

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已知,sin(π/4 2α)*sin(π/4-2α),α∈(π/4,π/2)求2sin²α tanα-1/tanα-1

已知,sin(π/4 2α)*sin(π/4-2α),α∈(π/4,π/2)求2sin²α tanα-1/tanα-1
已知,sin(π/4 2α)*sin(π/4-2α),α∈(π/4,π/2)求2sin²α tanα-1/tanα-1

已知,sin(π/4 2α)*sin(π/4-2α),α∈(π/4,π/2)求2sin²α tanα-1/tanα-1
题目应该是已知sin(π/4+2α)·sin(π/4-2α)=1/4,α∈(π/4,π/2),求2sin^2α+tanα-1/tanα-1的值
sin(π/4+2α)sin(π/4-2α)
=1/2*(cos(π/4+2α-π/4+2α)-cos(π/4+2α+π/4-2α))
=1/2*(cos(4α)-cos(π/2))
=1/2*cos(4α)
=1/4
所以cos(4α)=1/2
因为α属于(π/4,π/2)
所以π