Question

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

Answer 1

由正弦定理，得：a/sinA=b/sinB=c/sinC
而a+c=2b，所以sinA+sinC=2sinB
而sinA=sin[(A+C)/2+(A-C)/2]，sinC=sin[(A+C)/2-(A-C)/2]
展开后相加，得：sinA+sinC=2sin[(A+C)/2]*cos[(A-C)/2]
而A+C=π-B，A-C=π/3，所以(A+C)/2=π/2-B/2，(A-C)/2=π/6
于是2sin(π/2-B/2)×cos(π/6)=2sinB
即2cos(B/2)×(√3/2)=2sinB=2×2sin(B/2)×cos(B/2)
即cos(B/2)×√3=4sin(B/2)×cos(B/2)
而0<B<π，所以0<B/2<π/2，所以cos(B/2)>0
那么√3=4sin(B/2)，所以sin(B/2)=√3/4
那么cos(B/2)=√[1-(√3/4)²]=√13/4
所以sinB=2sin(B/2)×cos(B/2)=2×(√3/4)×(√13/4)=√39/8

Answer 2

 柔弱依旧，你好：