题目

三角函数题cot.
已知A、B、C是三角形的三个内角
求证cot(A/2)+cot(B/2)+cot(C/2)=cot(A/2)*cot(B/2)*cot(C/2)

提问时间：2020-10-25

答案

cot(A/2)+cot(B/2)+cot(C/2)=cos(A/2)/sin(A/2)+cos(B/2)/sin(B/2)+cos(C/2)/sin(C/2)=[sin(B/2)cos(A/2)+cos(B/2)sin(A/2)]/sin(A/2)sin(B/2)+cos(C/2)/sin(C/2)=sin(A/2+B/2)/sin(A/2)sin(B/2)+cos(C/2)/sin(C/2)=sin(π/2-C/2)/sin(A/2)sin(B/2)+cos(C/2)/sin(C/2)=cos(C/2)/sin(A/2)sin(B/2)+cos(C/2)/sin(C/2)=[cos(C/2)sin(C/2)+sin(A/2)sin(B/2)cos(C/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)={cos(C/2)[cos(B/2)cos(A/2)-sin(A/2)sin(B/2)+sin(A/2)sin(B/2)]}/sin(A/2)sin(B/2)sin(C/2)=cos(A/2)cos(B/2)cos(C/2)/sin(A/2)sin(B/2)sin(C/2)=cot(A/2)*cot(B/2)*cot(C/2)