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