求证[sin(2A-B)/sinA]-2cos(A-B)=-sinBcscA

问题描述:

求证[sin(2A-B)/sinA]-2cos(A-B)=-sinBcscA

证明:左边等于[sinAcos(A-B)+cosAsin(A-B)-2sinAcos(A-B)]/sinA
=[cosAsin(A-B)-sinAcos(A-B)]/sinA
=sin(A-B-A)/sinA
=-sinB/sinA
=-sinBcscA=右边
问题得证