Question

三角形ABC中，a b c是其对边，已知：a+c=2b A-C=三分之派，求sinB

Answer 1

简单分析一下，详情如图所示

Answer 2

解：因a+c=2b,故由正弦定理有：sinA+sinC=2sinB
sinA+sinC=2[sin(A+C)/2]cos(A-C)/2
=2[sin(180-B)/2]*cos30°
=2sin(90-B/2)*√3/2
=(√3)cosB/2
故，2sinB=(√3)cosB/2
2*2sinB/2*cosB/2=(√3)cosB/2,因cosB/2<>0
故sinB/2=(√3)/4
又，sinB/2=±√[(1-cosB)/2]
cosB=1-2sin^2(B/2)=1-2*3/16=5/8
故，sinB=√(1-cos²
B)=√[1-(5/8)²
]=√39/8