在三角形ABC中,若角A,B,C所对的三边长分别为a,b,c,证明:(a2-b2)/c2=sin(A-B)/sinC

在三角形ABC中,若角A,B,C所对的三边长分别为a,b,c,证明:(a2-b2)/c2=sin(A-B)/sinC

由正弦定理:

a/sinA=c/sinC

a/c=sinA/sinC,两边同时乘以2cosB,左边分子分母同乘以c.得：

2ac*cosB/c²=2sinAcosB/sinC.

由余弦定理a²+c²-b²=2ac*cosB得：

(a²+c²-b²)/c²=2sinAcosB/sinC

两边同时减去1,可得:

(a²-b²)/c²=(2sinAcosB-sinC)/sinC

且有2sinAcosB-sinC=2sinAcosB-sin(A+B)
=2sinAcosB-(sinAcosB+cosAsinB)
=sinAcosB-cosAsinB
=sin(A-B)

则原式得证.

(a^2-b^2)/c^2=[(sinA)^2-(sinB)^2]/(sinC)^2 =sin(A-B)/sinC→ [(sinA)^2-(sinB)^2]/(sinC)=sin(A-B)→ (sinA)^2-(sinB)^2=sin(A-B)sinC→ (sinA)^2-(sinB)^2=sin(A-B)sin(A+B) sin(A-B)sin(A+B)=(sinA)^2(cosB)^2-(cosA)^2(sinB)^2 =(sinA)^2[1-(sinB)^2]-[1-(sinA)^2](sinB)^2= (sinA)^2-(sinA)^2(sinB)^2 -(sinB)^2+(sinA)^2(sinB)^2 =(sinA)^2-(sinB)^2

(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