在三角形ABC中，a,b,c分别是角A,B,C的对边，设a+c＝2b,A-C＝π/3，求sinB的值。
热心网友
正弦定理:a/sinA=......=2R---a=2RsinA;b=2RsinB;C=2RsinC和差化积公式:sinA+sinB=2sin[(A+B)/2]cos[(A-B)/2]a+c=2b---2RsinA+2RsinC=4RsinB---sinA+sinC=2sinB---2sin[(A+C)/2]cos[(A-C)/2]=4sin(B/2)cos(B/2)---cos(B/2)cos[(A-C)/2]=2sin(B/2)cos(B/2)[因为0sin(B/2)=1/2*cos[(A-C)/2]=1/2*cos(π/6)=√3/4---cos(B/2)=√{1-[sin(B/2)]^2}=√(1-3/16)=√13/4---sinB=2sin(B/2)cos(B/2)=2*√3/4*√13/4=√39/8
热心网友
a+c＝2b == sinA + sinC = 2*sinB,(正弦定理)== 2*sin(A+C)/2 * cos(A-C)/2 = 2*(2*sinB/2cosB/2)A-C＝π/3，(A+C)/2 = pai/2 - B/2== sinB/2 = (genhao3)/4== cosB/2 = (genhao13)/4 sinB = 2*sinB/2cosB/2 = (genhao39)/8