已知：sin(π/4+α)xsin(π/4-α)=1/6, 且α∈(π/2,π)，求sin2α

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

α∈(π/2,π)
所以2α∈(π,2π)
所以sin2α<0
因为sin(2α)^2+(cos2α)^2=1
所以sin2α=-2√2/3

sin[2(π/4+α)]=sin(π/2+2α)
由诱导公式
sin(π/2+x)=cosx
所以sin(π/2+2α)=cos2α

诱导公式
sin(π/4+α)sin(π/4-α)=1/6
sin(π/4+α)cos(π/4+α)=1/6
2sin(π/4+α)cos(π/4+α)=1/3
sin[2(π/4+α)]=1/3
cos2α=1/3

α∈(π/2,π)
则2α∈(π,2π)
则sin2α<0

由(cos2α)^2+(sin2α)^2=1
sin2α=-2√2/9