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
原式=(-√3a+4√3)(a-6+3a/2)
=(-√3a+4√3)(5a/2-6)
=-5√3a²/2+10√3a+6√3a-24√3
=-5√3a²/2+16√3a-24√3
两边乘x(x+3)
7(x+3)=5x
7x+21=5x
x=-21/2
分式方程要检验
经检验，x=-21/2是方程的解
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)