sinA=3/5,cosB=-12/13,角A,B都为钝角,求sin(A-B)的值

sinA=3/5,cosB=-12/13,角A,B都为钝角,求sin(A-B)的值
sinA=3/5,cosB=-12/13,角A,B都为钝角,求sin(A-B)的值

sinA=3/5,cosB=-12/13,角A,B都为钝角,求sin(A-B)的值
∵角A,B都为钝角
∴cosA=-√(1-sin²A)=-4/5
sinB=√(1-cos²B)=5/13
sin(A-B)=sinAcosB-cosAsinB
=3/5×(-12/13)-(-4/5)×5/13
=-36/65+20/65
=-16/65

sin²A+cos²A=1
钝角则cosA<0
所以cosA=-4/5
同理
sinB=5/13
所以原式=sinAcosB-cosAsinB=-16/65