Question

已知:cos(2α-β)=-11/14,sin(α-2β)=4根号3/7,π/4<α<π/2,0<β<π/4 求α+β

Answer 1

∵π/4<α<π/2,0<β<π/4
∴π/2<2α<π ,-π/4<-β<0
∴π/4<2α-β<π
∵cos(2α-β)=-11/14,
∴sin(2α-β)=√[1-sin²(2α-β)]=5√3/14

∵π/4<α<π/2, -π/4<-β<0
∴-π/2<-2β<0
∴-π/4<α-2β<π/2
∵sin(α-2β)=4√3/7,
∴cos(α-2β)=√[1-sin²(α-2β)]=1/7

∴cos(α+β)=cos[(2α-β)-(α-2β)]
=cos(2α-β)cos(α-2β)+sin(2α-β)sin(α-2β)
=-11/14*1/7+5√3/14*4√3/7
=49/98
=1/2
又π/4<α+β<3π/4
∴α+β=π/3