积化和差与和差化积三角形ABC中,求证sinA+sinB+sinC=4cos(A/2)cos(B/2)cos(C/2)要详细!

Question

积化和差与和差化积
三角形ABC中,求证sinA+sinB+sinC=4cos(A/2)cos(B/2)cos(C/2)要详细!

Answer 1

sinA+sinB+sinC=sin(B+C)+sinB+sinC (A=π-(B+C) )=sinBcosc+sinB+cosBsinC+sinC=sinB(cosC+1)+sinC(1+sinB) =sinB*2cos^2(C/2)+sinC*2cos^2(B/2) （cos2a=2cos^2a-1变形　1+cos2a=2cos^2a)=4sin(B/2)cos(B/2)cos^2(C/2)+4sin(C/2)cos(C/2)cos^2(B/2)=4cos(B/2)cos(C/2)[sin(B/2)cos(C/2)+sin(C/2)cos(B/2)]=4cos(B/2)cos(C/2)sin[(B+C)]/2 ((B+C)/2=π-A/2)=4cos(A/2)cos(B/2)cos(C/2)