Question

sin²b-sin²c=sin a,求b-c

在△ABC中,已知(sin A+sin B+sin C)(sin A+sin B-sin C)=3sin A×sin B,a＜b,a cos A+b cos B=c cos C.
在△ABC中,已知(sin A+sin B+sin C)(sin A+sin B-sin C)=3sin A×sin B,a＜b,a cos A+b cos B=c cos C,求△ABC的各内角的大小.

Answer 1

由正弦定理
a/sinA=b/sinB=c/sinC=2R
因为(sin A+sin B+sin C)(sin A+sin B-sin C)=3sin A×sin B
(a+b+c)(a+b-c)=3ab
(a+b)²-c²=3ab
a²+b²+2ab-c²=3ab
a²+b²-c²=ab
cosC=(a²+b²-c²)/2ab=1/2
C=60°
同样由正弦定理
a cos A+b cos B=c cos C
sinAcosA+sinBcosB=sinCcosC
sin2A+sin2B=2sinCcosC
2sin(A+B)cos(A-B)=2sinCcosC
2sin120°cos(A-B)=2sin60°cos60°
cos(A-B)=cos60°
因为 a