求证[sin(2A-B)/sinA]-2cos(A-B)=-sinBcscA_数学解答

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

人气：419 ℃　时间：2020-06-22 20:11:47

解答

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