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

问题描述:

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

∵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