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

A+B=π-C
cos(A+B)=cos(π-C)=-cosC
cos[(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*(π-C)]=sin(3/2π-3/2*C)=sin(3C)

A+B=180°-C
cos(A+B)=cos(180°-C)=-cosC
cos[(A+B)/2]=cos[(180°-C)/2]=cos(180°/2-C/2)=sin(C/2)
sin(3A+3B)=sin3*(A+B)=sin3*(180°-C)=sin(3*180°-3C)=sin(3C)
sin[(3A+3B)/2]=-cos[(3C)/2]
sin[(3A+3B)/2]=sin[3/2*(A+B)]=sin[3/2*(180°-C)]=sin(270°-3*C/2)=cos[(3C)/2]