Question

一道三角函数题目

已知sin2a=3/5(π/4＜a＜π/2),tan(a-b)=1/2,则tan(a+b)=?
后来得出tan(a+b)=tan(2a-（a-b）)=tan2a-tan(a-b)/1+tan2atan(a-b)，为什么我算出来等于2/11啊？，正确答案是-2

Answer 1

π/4＜a＜π/2
π/2＜2a＜π
2a是钝角
sin2a=3/5
cos2a=-4/5
tan2a=-3/4
tan(a+b+a-b)=-3/4
[tan(a+b)+tan(a-b)]/[1-tan(a+b)tan(a-b)]=-3/4
令tan(a+b)=t
(t+1/2)/(1-1/2t)=-3/4
4t+2=-3+3t/2
8t+4=-6+3t
5t=-10
t=-2
tan(a+b)=-2

Answer 2

π/4＜a＜π/2
π/2＜2a＜π
2a是钝角
sin2a=37cos2a=-4/5
[tan(a+b)+tan(a-b)]/[1-tan(a+b)tan(a-b)]=-3/4
令tan(a+b)=t
(t+1/2)/(1-1/2t)=-3/4
4t+2=-3+3t/2
8t+4=-6+3t
5t=-10
t=-2
tan(a+b)=-2