5sin(A B) 3sinA,求tan(A B/2)*cotB/2

题目

5sin(A B) 3sinA,求tan(A B/2)*cotB/2
5sin(A+B)+3sinA=0,求tan(A+B/2)*cotB/2

答案

利用凑角的思想解此题
由5sin(A+B)+3sinA=0
得5sin[(A+B/2)+B/2]+3sin[(A+B/2)-B/2]=0
即5[sin(A+B/2)cosB/2+cos(A+B/2)sinB/2]+3[sin(A+B/2)cosB/2-cos(A+B/2)sinB/2]=0
整理得8sin(A+B/2)cosB/2+2cos(A+B/2)sinB/2=0
即8sin(A+B/2)cosB/2=-2cos(A+B/2)sinB/2
所以sin(A+B/2)cosB/2除以cos(A+B/2)sinB/2等于-1/4
即tan(A+B/2)*cotB/2=-1/4