三角函数,求解!(cosAcosB/2)/cos(A-B/2)+(cosBcosA/2)/cos(B-A/2)=1,求c

来源：学生作业帮编辑：搜搜做题作业网作业帮分类：数学作业时间：2024/05/03 05:38:18

三角函数,求解!
(cosAcosB/2)/cos(A-B/2)+(cosBcosA/2)/cos(B-A/2)=1,求cosA+cosB

该题目比较繁琐,不过还是帮帮你吧
　　(cosAcosB/2)/cos(A-B/2)+(cosBcosA/2)/cos(B-A/2)=1
=>
(cosAcosB/2)/cos(A-B/2)=1-(cosBcosA/2)/cos(B-A/2)
=>
(cosAcosB/2)/cos(A-B/2)=[cos(B-A/2)-(cosBcosA/2)]/cos(B-A/2)
=>
(cosAcosB/2)/cos(A-B/2)=(sinBsinA/2)/cos(B-A/2)
=>
(cosAcosB/2)cos(B-A/2)=(sinBsinA/2)cos(A-B/2)
=>
(cosAcosB/2)(cosBcosA/2+sinBsinA/2)-(sinBsinA/2)(cosAcosB/2+sinAsinB/2)=0
=>
(cosAcosB/2)(cosBcosA/2+sinBsinA/2)-(sinBsinA/2)(cosAcosB/2+sinAsinB/2)=0
=>
cosAcosBcos(A/2)cos(B/2)-sinAsinBsin(A/2)sin(B/2)=0
=>
cosAcosBcos(A/2)cos(B/2)-4sin²(A/2)sin²(B/2)cos(A/2)cos(B/2)=0
=>
cos(A/2)cos(B/2)[cosAcosB-4sin²(A/2)sin²(B/2)]=0
由(cosAcosB/2)/cos(A-B/2)+(cosBcosA/2)/cos(B-A/2)=1知cos(A/2)=0、cos(B/2)=0不能同时成立,故只需进行如下讨论：
①若cos(A/2)=0,cos(B/2)≠0则 (cosAcosB/2)/cos(A-B/2)=1且 A/2=kπ+π/2
即[cos(2kπ+π)cosB/2)/cos(2kπ+π-B/2)=1
[-cos(B/2)]/[-cos(B/2)]=1
[-cos(B/2)]/[-cos(B/2)]=1
此时,cosA+cosB=2cos²(A/2)-1+cosB=cosB-1不确定
②若cos(A/2)≠0,cos(B/2)=0,同上cosA+cosB=cosA-1不确定
③若cos(A/2)cos(B/2)≠0则
cosAcosB-4sin²(A/2)sin²(B/2)=0

cosAcosB-[(1-cosA)(1-cosB)]=0

cosAcosB-(1-cosA-cosB+cosAcosB)=0

1-cosA-cosB=0
即
cosA+cosB=1
再问：谢谢，和我做的一致，我做这题时对结果不唯一不是很自信，因为不大见这类题。
再答：呵，是，稍微复杂点，应该加点条件简化问题，即便这样也不太好做，谢谢采纳，好运！

三角函数,求解!(cosAcosB/2)/cos(A-B/2)+(cosBcosA/2)/cos(B-A/2)=1,求c cosacosb=1/2[cos(a+b)+cos(a-b)]如何证明 cosacosb=1 cos(a-b)=? 三角函数证明题证明cos(A+B)=cosAcosB-sinAsinB 设p=cosacosb,q=cos平方（a+b）/2,比较q,p大小 p=cosacosb,q=cos[(a+b)/2],且90° 证明cos(A+B)=cosAcosB-sinAsinB 证cos(A-B)=cosAcosB+sinAsinB cosacosb+sinasinb=cos(a-b) 1设p=cosAcosB,q=cos²（A+B)/2,比较p与q的大小 Cos(a+b)*cos(a-b)=1/5 求cos ^2-sin^2 求证:a^2(cos^2b-cos^2c)+b^2(cos^c-cos^2a)+c^2(cos^2a-cos^2b)=0