cos^2A - cos^2B + sin^2C=2cosA *sinB *sinCABC为三角形,利用(cosA)平方-(cosB)平方=sin(A+B)*sin(B-A)证明cosA平方 - cosB平方 + sinC平方=2cosA *sinB *sinC
问题描述：
cos^2A - cos^2B + sin^2C=2cosA *sinB *sinC
ABC为三角形,利用(cosA)平方-(cosB)平方=sin(A+B)*sin(B-A)证明
cosA平方 - cosB平方 + sinC平方=2cosA *sinB *sinC
答
注意由A+B+C=180°可知sin(A+B)=sinC
左边=sin(A+B)sin(B-A)+sin^2C
=sinC*sin(B-A)+sin^2C
=sinC*[sin(B-A)+sinC]
=sinC*[sin(B-A)+sin(B+A)]
=sinC*(sinBcosA-cosBsinA+sinBcosA+cosBsinA)
=sinC*(2sinBcosA)
=2cosAsinBsinC
答
左边=sin(A+B)sin(B-A)+sin²C=sin(180-C)sin(B-A)+sin²C=sinCsin(B-A)+sin²C=sinC[sin(B-A)+sinC]=sinC[sin(B-A)+sin(B+A)]=sinC(sinBcosA-cosBsinA+sinBcosA+cosBsinA)=sinC*2sinBcosA=右边命题得证...