在角ABC中,角A、B、C对应边分别为a,b,c,试证明下列恒等式;cotA/2+cotB/2+cotC/2=cotA/2*cotB/2*cotC/2

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有个挺有用的公式 ctgA + ctgB = sin(A+B) / sinAsinB
ctg (A/2) + ctg (B/2) + ctg (C/2)
= sin(A/2+B/2) / sin(A/2)sin(B/2) + ctg(C/2)
= cos(C/2) / sin(A/2)sin(B/2) + cos(C/2) / sin(C/2)
= [ cos(C/2)sin(C/2) + cos (C/2)sin(A/2)sin(B/2) ] / sin(A/2)sin(B/2)sin(C/2)
= cos(C/2)[ sin(C/2) + sin(A/2)sin(B/2) ] / sin(A/2)sin(B/2)sin(C/2)
由sin(C/2) = sin[(pi-A-B)/2] = cos(A/2+B/2) = cos(A/2)cos(B/2) - sin(A/2)sin(B/2) 代入上式
原式= cos(C/2)cos(A/2)cos(B/2) / sin(A/2)sin(B/2)sin(C/2)
= ctg(A/2) * ctg(B/2) * ctg(C/2)

cotA/2+cotB/2+cotC/2=cotA/2*cotB/2*cotC/2
等价于：tanA/2tanB/2+tabB/2tanC/2+tanC/2tanA/2=1
证明：
tanC/2=tan(180-(A+B))/2
=cot(A/2+B/2)
=1/tan(A/2+B/2)
=(1-tanA/2tanB/2)/(tanA/2+tanB/2)
故：tanC/2*(tanA/2+tanB/2)=1-tanA/2tanB/2
tanA/2tanB/2+tabB/2tanC/2+tanC/2tanA/2=1
故原式成立