在三角形ABC中，求证a²-b²/c²=sin(A-B)/sinC
1个回答
展开全部
a/sinA=b/sinB=c/sinC
sin(A+B)=sinC
(sin^2A-sin^2B)/sin^2C=(sinA+sinB)(sinA-sinB)/sin^2C=2sin(A+B)/2*cos(A-B)/2*2cos(A+B)/2sin(A-B)/2 /sin^2C=sin(A+B)sin(A-B)/sin^2C=sin(A-B)/sinC
sin(A+B)=sinC
(sin^2A-sin^2B)/sin^2C=(sinA+sinB)(sinA-sinB)/sin^2C=2sin(A+B)/2*cos(A-B)/2*2cos(A+B)/2sin(A-B)/2 /sin^2C=sin(A+B)sin(A-B)/sin^2C=sin(A-B)/sinC
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
