已知sin(π/4+2a)*sin(π/4-2x)=1/4,a∈（π/4,π/2）,求2sina^2+tana-1/(tana)-1=
问题描述：
已知sin(π/4+2a)*sin(π/4-2x)=1/4,a∈（π/4,π/2）,求2sina^2+tana-1/(tana)-1=
答
sin(π/4+2a)=sinπ/4·cos√2a+cosπ/4·sin2a=√2/2(cos2a+sin2a)
sin(π/4-2a)=sinπ/4·cos√2a-cosπ/4·sin2a=√2/2(cos2a-sin2a)
∴sin(π/4+2a)·sin(π/4-2x)
=(1/2)(cos²2a-sin²2a)
=1/2cos4a
=1/4
cos4a=1/2
∵a∈（π/4,π/2）
∴4a∈（π,2π）
则4a=5π/3
2a=5π/6
sin2a=sin5π/6=1/2
cos2a=cos5π/6=-√3/2
cot2a=cos2a/sin2a=-√3
那么2sin²a+tana-1/(tana)-1
=-(1-2sin²a)+sina/cosa-cosa/sina
=-cos2a+2(sin²a-cos²a)/(2sina·cosa)
=-cos2a-2cos2a/sin2a
=-cos2a-2cot2a
=√3/2+2√3
=5√3/2