已知a、b为锐角,a+b≠π/2,且满足3sinb=sin(2a+b)1.求证tan(a+b)=2tana2.求证tanb≤根号2_数学解答

已知a、b为锐角,a+b≠π/2,且满足3sinb=sin(2a+b)
1.求证tan(a+b)=2tana
2.求证tanb≤根号2/4,并求等号成立时tana和tanb的值

人气：335 ℃　时间：2020-10-02 02:41:21

解答

1 3sinb=sin(2a+b)3sin(a+b-a)=sin(a+b+a)3sin(a+b)cosa-3cos(a+b)sina=sin(a+b)cosa+sinbcos(b+a)2sin(a+b)cosa=4cos(a+b)sinatan(a+b)=2tana2 tanb=tana/(1+2tana^2)当tana=√2/2时有最小值=√2/4所以tanb≤√2/4 ...