Question

在三角形ABC中,角A,B,C的对边分别为a,b,c求证(a^2-b^2)/c^2=sin(A-B)/sinC

Answer 1

(a^2-b^2)/c^2=(a+b/c)(a-b/c)
根据正弦定理：
(a+b/c)(a-b/c)
=(sinA+sinB/sinC)(sinA-sinB/sinC)
分别处理，用和化为积公式：
sinA+sinB/sinC=2sin(A+B/2)cos(A-B/2)/sin(A+B)
=2sin(A+B/2)cos(A-B/2)/2sin(A+B/2)cos(A+B/2)
=cos(A-B/2)/cos(A+B/2)
同理：a-b/c=sin(A-B/2)/sin(A+B/2)
所以原式=sin(A-B/2)cos(A-B/2)/sin(A+B/2)cos(A+B/2)
=sin(A-B)/sin(A+B)=sin(A-B)/sinC