Question

三角形为锐角三角形,(sinA-sinB)(sinA+sinB)=sin(π/3-B)sin(π/3+B)，求A。

Answer 1

(sinA-sinB)(sinA+sinB)=sin(π/3-B)sin(π/3+B)

sin²A-sin²B=(sinπ/3cosB-cosπ/3sinB)(sinπ/3cosB+cosπ/3sinB)

sin²A-sin²B=(√3/2cosB-1/2sinB)(√3/2cosB+1/2sinB)

sin²A-sin²B=3/4cos²B-1/4sin²B

sin²A=sin²B+3/4cos²B-1/4sin²B=sin²B+3/4-3/4sin²B-1/4sin²B=3/4

sinA=√3/2

锐角三角形

∴A=π/3