Question

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

Answer 1

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!