在三角形ABC中cosA=3/5tanB=2求sin2(A+B)的值

cosA=3/5 sinA=√(1-cos^2A)=4/5tanB=2 cosB=1/√(1+tan^2B)=√5/5 sinB=2√5/5sin(A+B)=sinAcosB+sinBcosA=(4/5)(√5/5)+(2√5/5)(3/5)=2√5/5cos(A+B)=cosAcosB-sinAsinB=(3/5)(√5/5)-(4/5)(2√5/5)=-√5/5∴sin2...