正弦定理转换a+c=2R(SinA+SinC)=4/√3Sin((A+C)/2)Cos((A-C)/2)是怎么得到的?

Question

正弦定理转换
a+c=2R(SinA+SinC)
=4/√3Sin((A+C)/2)Cos((A-C)/2)
是怎么得到的?

Answer 1

2R=b/sinB
A=(A+C)/2+(A-C)/2
C=(A+C)/2-(A-C)/2
sinA+sinC
=sin[(A+C)/2+(A-C)/2]+sin[(A+C)/2-(A-C)/2]
=sin(A+C)/2cos(A-C)/2+cos(A+C)/2sin(A-C)/2+sin(A+C)/2cos(A-C)/2-cos(A+C)/2sin(A-C)/2
=2sin(A+C)/2cos(A-C)/2