sin(π/4+x)sin(π/4-x)
=sin[π/2-(π/4-x)]sin(π/4-x)
=cos(π/4-x)sin(π/4-x)
=1/2sin(π/2-2x)
=1/2cos2x=1/6
cos2x=1/3
x在(pai/2,pai),则2x在(pai,2pai)
所以:sin2x=-√[1-(cos2x)^2]=-2√2/3
sin4x=2sin2xcos2x=-2*1/3*2√2/3=-4√2/9