Question

在△ABC中，a.b.c分别是角A。B。C所对的边，且 2Sin²（A+B）/2+cos2C=1,若a²=b²+1/2c²

求sin(A-B)

Answer 1

2Sin²（A+B）/2+cos2C=2sin²[(180-C)/2]+cos2C
=2cos²(C/2)+cos2C
=cosC+1+cos2C
=cosC+2cos²C
=1
(2cosC-1)(cosC+1)=0解得C=60°
由余弦定理：a²=b²+c² -2bccosA
b²=a²+c² -2accosB
将a²=b²+1/2c²分别代入上面两式，得
cosA=c/(4b)，cosB=3c/(4a)
由正弦定理：a/sinA=b/sinB=c/sinC
将C=60°代入上式，得
sinA=√3a/(2c)，sinB=√3b/(2c)
则sin(A-B)=sinAcosB-cosAsinB
=[√3a/(2c)]*[3c/(4a)]-[c/(4b)][√3b/(2c)]
=3√3/8-√3/8
=√3/4