怎样证明三角函数的和差化积公式sinA+sinB=2*sin[(A+B)/2]*cos[(A-B)/2] sinA-si

第一个公式的证明:
右边=2*sin[(A+B)/2]*cos[(A-B)/2]
=2*[sin(A/2)*cos(B/2)+cos(A/2)sin(B/2)]*[cos(A/2)cos(B/2)+sin(A/2)sin(B/2)]
=2*sin(A/2)*cos(A/2)*cos(B/2)*cos(B/2)+2*cos(A/2)*cos(A/2)*sin(B/2)*cos(B/2)+2*sin(A/2)*sin(A/2)*cos(B/2)*sin(B/2)+2*sin(A/2)*cos(A/2)*sin(B/2)*sin(B/2)
=sinA*[cos(B/2)*cos(B/2)+sin(B/2)*sin(B/2)]+sin(B/2)*[cos(B/2)*cos(B/2)+sin(B/2)*sin(B/2)]
=sinA+sinB=左边
证毕
其中用到公式:
sinA=2*sin(A/2)*cos(A/2),sinB=2*cos(B/2)*sin(B/2)
cos(B/2)*cos(B/2)+sin(B/2)*sin(B/2)=1
其他的公式依此类推,自己推推看吧!

第一个公式的证明:设:½(A B) =C,½(A-B)=D,,,右边=2*sin[(A B)/2]*cos[(A-B)/2]=2sinC*cosD=sin(C＋D) sin(C-D)（注：这是积化和差公式，反过来展开很好推）接下来把C,D换回去就OK了=sinA sinB 证毕