Question

一道正余弦定理的数学题

已知△ABC中，a+c=2b,A-C=60°，求sinB

Answer 1

A+B+C=180°
A-C=60°
所以C+60°+B+C=180°
C=60°-B/2
所以C小于60°
A=120°-B/2
因为a/sinA=b/sinB＝c/sinC (三角形特性)
设a/sinA=b/sinB＝c/sinC＝t
所以b＝t*sinB
c＝t*sinC=t*sin(60°-B/2)
a＝t*sinA＝t*sin(120°-B/2)
因为a+c=2b
所以t*sin(120°-B/2)+t*sin(60°-B/2)＝2t*sinB
sin(120°-B/2)+sin(60°-B/2)＝2sinB （和差化积公式，2倍角公式）
2sin(90°-B/2)cos30°＝4sin(B/2)cos(B/2)
√3cos（B/2）=4sin(B/2)cos(B/2)
sin（B/2）=√3/4
cos（B/2）=√{1-[sin（B/2）]^2}=√13/4 (0<B/2<90°,cos(B/2)>0)
sinB=2*sin（B/2）cos（B/2）＝2*（√3/4）*（√13/4）＝√39/8

Answer 2

根号下39 /8
由a+c=2b得sinA+sinC=2sinB
2sin(A+C)/2cos(A-C)/2=4sinB/2cosB/2
sinB/2=根号3/4
cosB/2=根号13/4
sinB=2sinB/2*cosB/2=根号下39 /8