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

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

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

令A=B得 1=(cosAcosB/2)/cos(A-B/2)+(cosBcosA/2)/cos(B-A/2)= cosAcos(A/2)/cos(A/2)+(cosAcosA/2)/cos(A/2)
=cosA+cosA (注意cosA/2 !=0)
此时cosA+cosB=cosA+cosA=1
如果cosA+cosB的值存在，则A=B时计算的结果1就是这个值，否则co...

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令A=B得 1=(cosAcosB/2)/cos(A-B/2)+(cosBcosA/2)/cos(B-A/2)= cosAcos(A/2)/cos(A/2)+(cosAcosA/2)/cos(A/2)
=cosA+cosA (注意cosA/2 !=0)
此时cosA+cosB=cosA+cosA=1
如果cosA+cosB的值存在，则A=B时计算的结果1就是这个值，否则cosA+cosB的值不存在

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sinα＆＃47；cosα＝tanα＝3sinα＝3cosα带入恒等式sin＆sup2；α＋cos＆sup2；α＝19cos＆sup2；α＋cos＆sup2；α＝1cos＆sup2；α＝1＆＃47；10sin＆sup2；α＝9＆＃47；10sinαcosα＝（3cosα）cosα＝3cos＆sup2；α＝3＆＃47；10所以2sin＆sup2；α－sinαcosα＋cos＆sup2；α＋1＝13＆...

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sinα＆＃47；cosα＝tanα＝3sinα＝3cosα带入恒等式sin＆sup2；α＋cos＆sup2；α＝19cos＆sup2；α＋cos＆sup2；α＝1cos＆sup2；α＝1＆＃47；10sin＆sup2；α＝9＆＃47；10sinαcosα＝（3cosα）cosα＝3cos＆sup2；α＝3＆＃47；10所以2sin＆sup2；α－sinαcosα＋cos＆sup2；α＋1＝13＆＃47；5sinαcosα＝3＆＃47；10

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