sinA=3/5,cosB=-12/13,角A,B都为钝角,求sin(A-B)的值
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![sinA=3/5,cosB=-12/13,角A,B都为钝角,求sin(A-B)的值](/uploads/image/z/5197876-52-6.jpg?t=sinA%3D3%2F5%2CcosB%3D-12%2F13%2C%E8%A7%92A%2CB%E9%83%BD%E4%B8%BA%E9%92%9D%E8%A7%92%2C%E6%B1%82sin%28A-B%29%E7%9A%84%E5%80%BC)
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