高手帮帮忙 高中数学题

sin(π/4+2α)*sin(π/4-2α)=1/4
√2/2(sin2α+cos2α)*√2/2(cos2α-sin2α)=1/4
cos2α^2-sin2α^2=1又cos2α^2+sin2α^2=1
∴cos2α^2=1,sin2α^2=0
∵α∈(π/4,π/2),α=＝π/4
∴2(sinα)^2+tanα-1/（tanα-1）=2*1/2-1-1/(-1-1)=1/2

补充题
1.
sin(x/2)-2cos(x/2)=0
sin(x/2)=2cos(x/2)
tan(x/2)=2
(1)求tanx的值:tanx=2tan(x/2)/(1-tan(x/2)^2)=2*2/(1-4)=-4/3
(2)cos2x/[√2cos(π/4 +x)sinx]
=cos2x/[(sinx-cosx)sinx]
=[(cosx)^2-(sinx)^2]/[(sinx-cosx)sinx]
=-(sinx+cosx)/sinx
=-1-cotx
=-1-(-3/4)=-1/4
2.化简
[2(cosx)^4-2(cosx)^2+1/2]/{2tan(π/4 -x)*[sin(4/π+x)]^2}
=2[(cosx)^2-1/2]^2/{2（1-tanx)/(1+tanx)*[√2/2(cosx+sinx)]^2}
=2[(cosx)^2-1/2]^2/{2（cosx-sinx)/(cosx+sinx)*[√2/2(cosx+sinx)]^2}
=2[(cosx)^2-1/2]^2/[（cosx)^2-(sinx)^2]
=1/2[2(cosx)^2-1]^2/[（cosx)^2-(sinx)^2]
=1/2(cos2x)^2/(cos2x)
=cos2x/2

脑袋大了！！