sinA*cosA=(sinB)^2=(1-cos2B)/2
sinA+cosA=2sin2B,那么4(sin2B)^2=1+2sinA*cosA=2-cos2B
又(sin2B)^2=1-(cos2B)^2
解方程4-4(cos2B)^2=2-cos2B
cos2B=1±√33)/8
sinA*cosA=(sinB)^2=(1-cos2B)/2
sinA+cosA=2sin2B,那么4(sin2B)^2=1+2sinA*cosA=2-cos2B
又(sin2B)^2=1-(cos2B)^2
解方程4-4(cos2B)^2=2-cos2B
cos2B=1±√33)/8