已知0＜a＜π/4,0＜b＜π/4,5sinb=sin(2a+b),4tan(a/2)=1-tan^2(a/2)求a+b
问题描述：
已知0＜a＜π/4,0＜b＜π/4,5sinb=sin(2a+b),4tan(a/2)=1-tan^2(a/2)求a+b
答
4tan(a/2)=1-tan^2(a/2)
tana=2tan(a/2)/[1-tan^2(a/2)]=1/2
5sinb=sin(2a+b)
5sin[(a+b)-a]=sin[(a+b)+a]
5sin(a+b)cosa-5cos(a+b)sina=sin(a+b)cosa+cos(a+b)sina
2sin(a+b)cosa=3cos(a+b)sina
tan(a+b)=sin(a+b)/cos(a+b)=(3/2)sina/cosa=(3/2)tana=3/4
所以a+b=arctan(3/4)