三角形ABC,求证cos(A+B)=-cosC,cos[(A+B)/2]=sin(C/2)和sin(3A+3B)=sin(3C),sin[(3A+3B)/2]=-cos[(3C)/2]

题目

三角形ABC,求证cos(A+B)=-cosC,cos[(A+B)/2]=sin(C/2)和sin(3A+3B)=sin(3C),sin[(3A+3B)/2]=-cos[(3C)/2]

答案

A+B=π-Ccos(A+B)=cos(π-C)=-cosCcos[(A+B)/2]=cos[(π-C)/2]=cos(π/2-C/2)=sin(C/2)sin(3A+3B)=sin3*(A+B)=sin3*(π-C)=sin(3π-3C)=sin(3C)sin[(3A+3B)/2]=-cos[(3C)/2]sin[(3A+3B)/2]=sin[3/2*(A+B)]=sin[3/2*(...