2023-03-25 02:15:16
(资料图片仅供参考)
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!。
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