已知7sinA=3sin(A+B),求证2tan(A+B/2)=5tanB/2.

Question

已知7sinA=3sin(A+B),求证2tan(A+B/2)=5tanB/2.

Answer 1

证明
因为7sinA=3sin(A+B)
所以7sin(A+0.5B-0.5B)=3sin(A+0.5B+0.5B)
利用和角公式
7sin(A+0.5B)cos(0.5B)+7cos(A+0.5B)sin(0.5B)=3sin(A+0.5B)cos(0.5B)-3cos(A+0.5B)sin(0.5B)
4sin(A+0.5B)cos(0.5B)=10cos(A+0.5B)sin(0.5B)
左右两边同时除以2cos(A+0.5B)cos(0.5B)
得证