设△ABC的三个内角为A,B,C,三边长分别为a,b,c.求证:(a^2-b^2)/c^2=[sin(A-B)/sinC]
问题描述:
设△ABC的三个内角为A,B,C,三边长分别为a,b,c.求证:(a^2-b^2)/c^2=[sin(A-B)/sinC]
答
证:由正弦定理,得a=2RsinA,b=2RsinB,c=2RsinC三角形内,sin(A+B)=sinC所以:(a^2-b^2)/c^2=(sinAsinA-sinBsinB)/(sinCsinC)=(sinA+sinB)(sinA-sinB)/(sinCsinC)=2sin【(A+B)/2】cos【(A-B)/2】*2c...