sinB = sin(A + B - A) = sin(A+B)cosA - cos(A+B)sinA
2sin(2A+B) = 2sin(A + B + A) = 2sin(A+B)cosA + 2cos(A+B)sinA ,
∴sinB+2sin(2A+B)=0 = 3sin(A+B)cosA + cos(A+B)sinA ,
∵A≠kπ/2,A+B≠(π/2)+kπ(k∈Z),∴cos(A+B)≠0 ,cosA≠0 ,∴两边同时除以[cosA·cos(A+B)]即得：3tan(A+B)+tanA=0
有时三角函数题目需要用到角度变化.