sin(π+a)=-3/5,a属于〔π/2,π〕则tg(a/2)＝

Answer 1

a属于〔π/2,π〕
a/2属于〔π/4,π/2〕
tan(a/2)>0
cosa<0
sin(π+a)=-sina=-3/5
sina=3/5
(cosa)^2=1-(sina)^2
cosa<0
cosa=-4/5
tana=sina/cosa=-3/4
tana=2tan(a/2)/[1-(tana/2)^2]
令tana=x
则2x/(1-a^2)=-3/4
3x^2-3=8x
3x^2+8x-3=0
(3x-1)(x+3)=0
tan(a/2)>0
所以tan(a/2)=x=1/3