[sin(A+B)-2sinAcosB]/[2sinAsinB+cos(A+B)]如何化简

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[sin(A+B)-2sinAcosB]/[2sinAsinB+cos(A+B)]如何化简

[sin(A+B)-2sinAcosB]/[2sinAsinB+cos(A+B)]如何化简
[sin(A+B)-2sinAcosB]/[2sinAsinB+cos(A+B)]如何化简

[sin(A+B)-2sinAcosB]/[2sinAsinB+cos(A+B)]如何化简
[sin(A+B)-2sinAcosB]/[2sinAsinB+cos(A+B)]
=[sinAcosB+cosAsinB-2sinAcosB]/[2sinAsinB+cosAcosB-sinAsinB]
=(cosAsinB-sinAcosB)/(sinAsinB+cosAcosB)
=-sin(A-B)/cos(A-B)
=-tan(A-B)