sin(pi/4+2a)sin(pi/3-2a)=1/4,a属于(pi/4,pi/2),求2sin^2a+tana-cota-1

sin(π/4-2a)
=cos[π/2-(π/4-2a)]
=cos(π/4+2a)
所以2sin(π/4+2a)cos(π/4+2a)=sin2(π/4+2a)=2*1/4
sin(π/2+4a)=1/2
cos4a=1/2
所以2(cos2a)^2-1=1/2
(cos2a)^2=3/4
π/4<a<π/2
π/2<2a<π
所以cos2a<0
所以cos2a=-√3/2
sin2a>0
cos4a=1-2(sin2a)^2=1/2
所以sin2a=1/2

tana-cota=sina/cosa-cosa/sina=[(sina)^2-(cosa)^2]/sinacosa=-cos2a/(1/2*sin2a)=-2cos2a/sin2a
(sina)^2=(1-cos2a)/2
所以原式=(1-cos2a)-2cos2a/sin2a-1
=1+√3/2-2*(-√3/2)/(1/2)-1
=5√3/2