若cosx=-4/5,-3π/2＜x＜-π,求sin(π/2-2x)-2sin^2(π/4-x)/1+tan(π/4-x)的值
问题描述：
若cosx=-4/5,-3π/2＜x＜-π,求sin(π/2-2x)-2sin^2(π/4-x)/1+tan(π/4-x)的值
答
x在第二象限,sinx>0
sin²x+cos²x=1
所以 sinx=3/5
tanx=sinx/cosx=-3/4
sin(π/2-2x)
=cos(2x)
=cos²x-sin²x
=7/25
2sin^2(π/4-x)
=1-cos[2(π/4-x)]
=1-cos(π/2-2x)
=1-sin2x
=1-2sinxcosx
=49/25
tan(π/4-x)
=(tanπ/4-tanx)/(1+tanπ/4tanx)
=(1+3/4)/(1-3/4)
=7
所以原式=-91/200