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!
和差化积公式的推导（和差化积公式证明）
免责声明：本站所有文章和图片均来自用户分享和网络收集，文章和图片版权归原作者及原出处所有，仅供学习与参考，请勿用于商业用途，如果损害了您的权利，请联系网站客服处理。
