Question

在△ABC中，a,b,c分别是A,B,C的对边，a+c=2b,A-C=π/3

①求证：A=（2π）/3-B/2，C=π/3-B/2;
②求sin（B/2）及sinB的值。

Answer 1

A+C=π-B （1）
A-C=π/3 （2）
（1）+（2）
2A=4π/3-B
A=2π/3-B/2
（1）-（2）
2C=2π/3-B
A=π/3-B/2
(2)a+c=2b
由正弦定理
sinA+sinC=2sinB
sin(2π/3-B/2)+sin(π/3-B/2)=4sin(B/2)cos(B/2)
展开，得
√3cos(B/2)=4sin(B/2)cos(B/2)
sin(B/2)=√3/4
B/2为锐角
cos(B/2)=√13/4
sinB=2*√3/4 *√13/4
=√39/8

Answer 2

(1)A+B+C=π 和A-C=π/3==〉A=（2π）/3-B/2，C=π/3-B/2;
(2)正玄定理a/sinA=b/sinB=c/sinc有
a+c/(sinA+sinc)=b/sinB=2b/2sinB
由a+c=2b有sinA+sinc=2sinB
将A=（2π）/3-B/2，C=π/3-B/2代入上式并展开
sin2π/3cos(1/2B)-cos2π/3sin(1/2B)+sinπ/3cos(1/2B)-cosπ/3sin(1/2B)=2sinB
√3cos(1/2B)=2sinB=2sin(1/2B)cos(1/2B)
√3cos(1/2B)=2sinB=4sin(1/2B)cos(1/2B)
sin(1/2B)=(√3)/4
因为C=π/3-B/2>0 所以B/2<π/3 cos(1/2B)=)=√13/4
sinB=2sin(1/2B)cos(1/2B)=2*(√3)/4*√13/4=√39/8