证明2sinB/(cosA+cosB)=tan[ (A+B）/ 2 ]—tan[ (A—B) / 2 ]
问题描述：
证明2sinB/(cosA+cosB)=tan[ (A+B）/ 2 ]—tan[ (A—B) / 2 ]
答
证明：由“半角公式”：tan(A/2)=(1-cosA)/sinA.可知：
右边={[1-cos(A+B)]/sin(A+B)}-{[1-cos(A-B)]/sin(A-B)}
=[sin(A-B)-cos(A+B)sin(A-B)-sin(A+B)+sin(A+B)cos(A-B)]/[sin(A+B)sin(A-B)](通分)
=[sin(A+B)cos(A-B)-cos(A+B)sin(A-B)+sin(A-B)-sin(A+B)]/[sin(A+B)sin(A-B)]
=2[sin[(A+B)-(A-B)]+2cosAsin(-B)]/[cos(2B)-cos(2A)](积化和差,和差化积)
=2[sin(2B)-2sinBcosA]/{[2cos²B-1]-[2cos²A-1]}
=[2sinBcosB-2sinBcosA]/[cos²B-cos²A]
=2sinB[cosB-cosA]/[(cosB+cosA)(cosB-cosA)]=2sinB/(cosB+cosA)=左边.