已知cos2a+sin2a(2sina-1)=2/5,a∈(π/2,π),则tan(a+π/4)的值为

题目

已知cos2a+sin2a(2sina-1)=2/5,a∈(π/2,π),则tan(a+π/4)的值为

答案

∵a*b=cos2a+sina(2sina-1)=cos2a+2(sina)^2-sina
=1-2(sina)^2+2(sina)^2-sina=1-sina=2/5
∴ sina=3/5
∵π/2＜a＜π,∴cosa＜0,
则cosa=-4/5
∴tana=-3/4
∴tan(a+π/4)=(tana+tanπ/4)/(1-tana*tanπ/4)
=(tana+1)/(1-tana)
=(-3/4+1)/(1+3/4)
=1/7