已知sina(π/4+a)*sin(π/4-a)=1/6,a属于（π/2,π),则sin4a/(1+cos^2a)的值
问题描述：
已知sina(π/4+a)*sin(π/4-a)=1/6,a属于（π/2,π),则sin4a/(1+cos^2a)的值
答
sin(π/4+a)=sin(π/4)*cosa+sina*cos(π/4)
sin(π/4-a)=sin(π/4)*cosa-sina*cos(π/4)
原式=[sin(π/4)*cosa+sina*cos(π/4)]*[sin(π/4)*cosa+sina*cos(π/4)]
=1/2*(cosa)^2-1/2*(sina)^2=1/6
即(cosa)^2-(sina)^2=1/3
又有(cosa)^2+(sina)^2=1
所以(cosa)^2=2/3
因为a属于（π/2,π),所以cosa=-√6/3,sina=√3/3
cos2a=(cosa)^2-(sina)^2=1/3
sin4a/(1+cos^2a)=2sin2a*cos2a/(1+2/3)=4sina*cosa*cos2a/(5/3)
=4*(√3/3)*(-√6/3)*(1/3)/(5/3)=-4√2/15