在三角形ABC中，求证：(a-c·cosB)/(b-c·cosA)=sinB/sinA
展开全部
a/sinA=b/sinB=c/sinC
a=csinA/sinC
b=csinB/sinC
(a-c·cosB)/(b-c·cosA)=(sinA/sinC-cosB)/(sinB/sinC-cosA)
=(sinA-sinCcosB)/(sinB-sinCcosA)
=(sinA-sin(A+B)cosB)/(sinB-sin(A+B)cosA)
=(sinA-sinAcos²B-cosAsinBcosB)/(sinB-sinAcosAcosB-cos²AsinB)
=(sinAsin²B-cosAsinBcosB)/(sinBsin²A-sinAcosAcosB)
=sinB(sinAsinB-cosAcosB)/[sinA(sinAsinB-cosAcosB)]
=sinB/sinA
a=csinA/sinC
b=csinB/sinC
(a-c·cosB)/(b-c·cosA)=(sinA/sinC-cosB)/(sinB/sinC-cosA)
=(sinA-sinCcosB)/(sinB-sinCcosA)
=(sinA-sin(A+B)cosB)/(sinB-sin(A+B)cosA)
=(sinA-sinAcos²B-cosAsinBcosB)/(sinB-sinAcosAcosB-cos²AsinB)
=(sinAsin²B-cosAsinBcosB)/(sinBsin²A-sinAcosAcosB)
=sinB(sinAsinB-cosAcosB)/[sinA(sinAsinB-cosAcosB)]
=sinB/sinA
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
推荐律师服务：
若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询
