已知sin(π/4-α)sin(π/4+α)=(√2)/6,(0<α<π/2),求sin2α的值

sin(π/4+α)=cos[π/2-(π/4+α)]=cos(π/4-α)
所以sin(π/4-α)cos(π/4-α)=√2/6
(1/2)sin[2(π/4-α)]=√2/6
sin((π/2-2α)=√2/3
cos2α=√2/3

(sin2α)^2+(cos2α)^2=1
0<α<π/2
所以0<2α<π
所以sin2α>0
所以sin2α=√7/3

sin(π/4-α)sin(π/4+α)=√2/6
[sinπ/4cosα-cosπ/4sinα][sinπ/4cosα+cosπ/4sinα]=√2/6
√2/2(cosα-sinα)√2/2(cosα+sinα)=√2/6
1/2[(cosα)^2-(sinα)^2]=√2/6
cos2α=√2/3
0<α<π/2
0<2α<π
所以sin2α>0
sin2α=√[1-(cos2α)^2]=√7/3