Question

在△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

证明：
三角形ABC中
a/sinA=b/sinB=c/sinC=2R
左边=(a^2-b^2)/c^2
=(sin^2A-sin^2B)/sin^2C
=(sinA+sinB)(sinA-sinB)/sin^2C
=2sin(A+B)/2cos(A-B)/2*2cos(A+B)/2sin（A-B）/2/sin^2C
=2sin(A+B)/2cos(A+B)/2*2sin[(A-B)/2]cos[(A-B)/2/sin^2C
=sin(A+B)sin（A-B）/sin^2C
=sin(π-C)sin（A-B）/sin^2C
= =sinCsin（A-B）/sin^2C
=sin（A-B）/sinC
右边=sin（A-B）/sinC
左边=右边
所以(a^2-b^2)/c^2=sin(A-B)/sinC.

(中间使用的是正弦函数的和差化积)

Answer 2

正弦定理知
a/c=sinA/sinC,b/c=sinB/sinC
∴(a²-b²)/c²=(sin²A-sin²B)/sin²C=(1-cos2A-1+cos2B)/2sin²C
=(cos2B-cos2A)/2sin²C=2sin(A+B)sin(A-B)/2sin²C=sin(A-B)/sinC
其中sin(A+B)=sin(π-A-B)=sinC
cos2B=cos[(A+B)-(A-B)]=cos(A+B)cos(A-B)+sin(A+B)sin(A-B)
cos2A=cos[(A+B)+(A-B)]=cos(A+B)cos(A-B)-sin(A+B)sin(A-B)
也就是和差化积

Answer 3

由正弦定理a/sinA=b/sinB=c/sinC=2R
sin(A-B)/sinC=(sinAcosB-sinBcosA)/sinC=(acosB-bcosA)/c
再由余弦定理cosB=(a^2+c^2-b^2)/2ac,cosA=(b^2+c^2-a^2)/2bc
化简后sin(A-B)/sinC=(acosB-bcosA)/c=
[(a^2+c^2-b^2)-(b^2+c^2-a^2)]/2c^2=(a^2-b^2)/c^2