sinB+sinC=sinB+sin(pai/3-B)=sin(B+pai/3)这一步是怎么来的要详解由正弦定理得BC/sinA=2r,得sinA-根3/2∵BC是最长边,且三角形为非等边三角形∴A=2pai/3sinB+sinC=sinB+sin(pai/3-B)=sin(B+pai/3),又0

Question

sinB+sinC=sinB+sin(pai/3-B)=sin(B+pai/3) 这一步是怎么来的要详解
由正弦定理 得BC/sinA=2r,得sinA-根3/2∵BC是最长边,且三角形为非等边三角形∴ A=2pai/3sinB+sinC=sinB+sin(pai/3-B)=sin(B+pai/3),又 0

Answer 1

A+B+C=π A=2π/3 B+C=π/3 C=π/3-B
sinB+sinC=sinB+sin(π/3-B)= sinB+sinπ/3cosB-cosπ/3sinB
=sinπ/3cosB+cosπ/3sinB=sin(B+π/3)