在三角形中，已知(sinA/2)^2＋2(sinB/2)^2＋(sinC/2)^2＝1,求tanA/2tanC/2

为了方便书写,设a=A/2,b=B/2,c=C/2
那么题目化为:
已知a+b+c=90度,(sina)^2+2(sinb)^2+(sinc)^2=1
求tanatanc

(sina)^2+2(sinb)^2+(sinc)^2=1
(sina)^2+(sinb)^2+(sinc)^2=1-(sinb)^2=(cosb)^2
=[sin(a+c)]^2
(sina)^2+(sinb)^2=[sin(a+c)]^2-(sinc)^2
=[sin(a+c)-sinc][sin(a+c)+sinc]
=2cos(a/2+c)sin(a/2)*2sin(a/2+c)cos(a/2)
=sin(a+2c)sina
所以再把左边的(sina)^2移到右边,得到:
(sinb)^2=sin(a+2c)sina-(sina)^2
=sina*[sin(a+2c)-sina]
=sina*2cos(a+c)sinc
=2sinasinbsinc
所以左右约去sinb得到:
sinb=2sinasinc
又:
cos(a+c)=sinb=2sinasinc=cosacosc-sinasinc
也就是:
3sinasinc=cosacosc
左右都除以cosacosc就得到:
tanatanc=1/3
所以就得到结果:
tanA/2tanC/2=tanatanc=1/3

要求解的结果表示不很清楚，请多加些括号把问题说清楚。

●要证明这个题目很简单，不需要三角公式,令等式左边为EL,则
EL＝[sin(A/2)－sin(B/2)]^2＋[sin(B/2)－sin(C/2)]^2＋2sin(A/2)sin(B/2)＋2sin(B/2)sin(C/2)
≥2sin(A/2)sin(B/2)＋2sin(B/2)sin(C/2)

当且仅当sin(A/2)＝sin(B/2)＝sin(C/2)，即A/2＝B/2＝C/2＝π/6时
EL(min)＝2sin(π/6)s