![](https://img.fx361.cc/images/2023/0205/c9c4d4d89436955c3c7034f62ccb9b82f7f34598.webp)
![](https://img.fx361.cc/images/2023/0205/dd9e140382db1183d2c879abe746cb32d60f87dc.webp)
(2)题目的结论是求cosC的值.
(3)沟通题目的条件与结论的联系.由cosC=cos[π-(A+B)]=-cos(A+B),得
cosC=sinAsinB-cosAcosB.(1)
可见,只需由sinA,cosB求出cosA,sinB,便可求出cosC的值.给出以下2种解法:
![](https://img.fx361.cc/images/2023/0205/1bd1c89fb9a7c25c90d57a9500415133f5ad15d9.webp)
![](https://img.fx361.cc/images/2023/0205/837e71ce6729ba0e9fbc8b61e8b02a18d356e257.webp)
从而
![](https://img.fx361.cc/images/2023/0205/ef4d0f49932eec31ee79026e2fcc649d23ac0f16.webp)
![](https://img.fx361.cc/images/2023/0205/0f7d9147b98403ee5962db68cdaefbce32136ca2.webp)
![](https://img.fx361.cc/images/2023/0205/0b91c94f15f33ea8720caba1ba64088271c9089d.webp)
sinA=sin[π-(B+C)]=sin(B+C)=sinBcosC+cosBsinC,
![](https://img.fx361.cc/images/2023/0205/199199ac9b3a8db8e4c48ebb58778d47f61af081.webp)
![](https://img.fx361.cc/images/2023/0205/d05640015a17fe581179da1b0e390f4d72154eea.webp)
(2)
将式(2)移项,2边平方,整理得关于cosC的二次方程
解得
![](https://img.fx361.cc/images/2023/0205/5da3aff86824b2360f9ef6e11500ae65d3a28d34.webp)
![](https://img.fx361.cc/images/2023/0205/89d6ba3a435ad0e42dea840aa800ebc7d20ed5ca.webp)
![](https://img.fx361.cc/images/2023/0205/fe259ddf14fcbde53e5287d7b2d06c12dd05ac8e.webp)
图1
说明如图1所示,cosC的2个解分别可在△AB1C,△AB2C中求得,由余弦定理可验证其正确.
![](https://img.fx361.cc/images/2023/0205/1201a02283bc3be38fbe6e6be68b53785eba719e.webp)
![](https://img.fx361.cc/images/2023/0205/b9053945071f68b4a4ed0feafaa963ba4b91e489.webp)
![](https://img.fx361.cc/images/2023/0205/ef8c8d1e271c386d14bd21d8ac2cbda651c2033d.webp)
![](https://img.fx361.cc/images/2023/0205/8ced991c3c5cb59ac78227f96c18ac71953f4ac6.webp)
![](https://img.fx361.cc/images/2023/0205/8889e7e3d436165d4f6c353daf0c5bbfac0e4d42.webp)
sinA=sin[π-(B+C)]=sin(B+C)=sinBcosC+cosBsinC,
![](https://img.fx361.cc/images/2023/0205/3f593f881de84d1b845185429902eb9485161818.webp)
![](https://img.fx361.cc/images/2023/0205/036ec48fd95bd8ac56bd257bcca6e694a1ef63ad.webp)
(3)
![](https://img.fx361.cc/images/2023/0205/3554aec1aaf37da94240b5731dbec190af5ab025.webp)
![](https://img.fx361.cc/images/2023/0205/4c0cb386c2f3525fbd22985681174e5ea9e57fee.webp)
图2
说明如图2所示,例2也可以在△ABC中用余弦定理求解,在此不再赘述.
从上述求解过程可以看到,由sinA到cosA有2个可能的取值,但这2个值能不能都取到还需要进一步讨论.更一般地,当sinA=m,cosB=n时,cosC是不是有解?有几个解?具体数值是什么?就更加需要抽象的讨论了,这是一个很有探究价值的问题.
2 一般情况的探究
例3在△ABC中,sinA=m,cosB=n(0下面给出的2种思路都是数形结合并分类讨论,但思路1重在几何,思路2重在代数.
思路1数形结合讨论已知条件中的角.
(1)思路分析.
![](https://img.fx361.cc/images/2023/0205/2db92e645575974e67637d23637a4310fd41295d.webp)
关键点1确定cosA.
由于A为三角形的内角,对每一个m∈(0,1],有m=sinA=sin(π-A),故内角A最多有2个取值,记为A1,A2(如图3,其中B1=B2),满足
![](https://img.fx361.cc/images/2023/0205/e58f985de75cc0ac69fcf9142e7e1a006c574284.webp)
关键点2保证A,B能在同一个三角形内.
由A+B+C=π知,A,B在同一个三角形内的充要条件是0下面只需讨论A1+B,A2+B与π(平角)的关系,便可确定Ai(其中i=1,2)与B是否在同一个三角形内,以及有几个Ai(其中i=1,2)与B在同一个三角形内.
(2)讨论A1=A2的情况.
![](https://img.fx361.cc/images/2023/0205/372551f1a730e0acc983b9f497ca383388764d29.webp)
![](https://img.fx361.cc/images/2023/0205/af027db806ec33820cd3dccae290a9c47ae0d465.webp)
图3 图4
![](https://img.fx361.cc/images/2023/0205/1ef74893eed9d42fdc9088ddd6fec892e668022d.webp)
![](https://img.fx361.cc/images/2023/0205/d96502700c94c13b1fb6a4718bf7f7357b599e86.webp)
(3)讨论A1≠A2的情况.
![](https://img.fx361.cc/images/2023/0205/8eafc32a5925e0a67f2e6383800df7857919e978.webp)
情况3若A2+B<π,则0情况4若0情况5若π≤A1+B,则π≤A1+B(4)相关结论.
分别把情况1和情况4合并,把情况2和情况5合并,可得:
![](https://img.fx361.cc/images/2023/0205/07a8099e07572c7997d5444425865461f461f7a0.webp)
![](https://img.fx361.cc/images/2023/0205/4461d27a606a37746bbf74758aa683224ddaf43e.webp)
结论3当A1思路2代数方法讨论二次方程中的正根.
(1)思路分析.
![](https://img.fx361.cc/images/2023/0205/d4a08fb0a2a411427240814f636f288b46391f8b.webp)
第1步:构造一个二次方程,求出它的2个实根.
![](https://img.fx361.cc/images/2023/0205/2e21e7577bcc1cc38f35547ce2f0d55836e48dde.webp)
sinA=sin[π-(B+C)]=sin(B+C)=sinBcosC+cosBsinC,
![](https://img.fx361.cc/images/2023/0205/e07341b3a23b0e9103955238e9ca5aeadf764f3f.webp)
(4)
![](https://img.fx361.cc/images/2023/0205/a53069275bada82ce54098605dddb3eeee02a86a.webp)
(5)
![](https://img.fx361.cc/images/2023/0205/6e0f1531048d8d1b28bc1e9ddb728715f61f3632.webp)
把式(5),式(6)代入式(4),得关于t的二次方程
(7)
![](https://img.fx361.cc/images/2023/0205/72bba82bd3612669c9d7ef082063ce8cd3411201.webp)
(8)
所以方程(7)恒有实根(包括等根),解得
(9)
其中t1≤t2.接下来只需讨论t1,t2有无正根,正根能取到几个,并把正根代入式(5)便可求出cosC.
第2步:对t1,t2取正值的情况分类讨论.
基本思路是对二次方程(7)中t1,t2的表达式作二级分类:先把t1,t2分为相等的根和不等的根,然后再分为正根与非正根,结合题目所给的m,n(可以先讨论m后讨论n)进行讨论,其逻辑结构如下:
(2)讨论t1=t2的情况.
![](https://img.fx361.cc/images/2023/0205/fc935864936b51e39beae40d3ec317a60144586f.webp)
![](https://img.fx361.cc/images/2023/0205/911a8ee4495338a6fdb685680094584bb93cae12.webp)
![](https://img.fx361.cc/images/2023/0205/a6c13b5024d4a94ea20b0c58cd3b353d1bddbc85.webp)
(3)讨论t1≠t2的情况.
若t1≠t2,则由式(8)或式(9)知m≠1,得00,或t1≤0![](https://img.fx361.cc/images/2023/0205/3ce14cf77d2ed3868ea35565dad765a7ef1d9f54.webp)
![](https://img.fx361.cc/images/2023/0205/b34f192c30fc55fc151101900eedc9fc79714acf.webp)
![](https://img.fx361.cc/images/2023/0205/61ae6596cb723cc8d282c5a2b81ffa46f7d5d1a3.webp)
![](https://img.fx361.cc/images/2023/0205/63f5309c14c7faecdaae44031d1513d20f4f217c.webp)
这时,方程(7)只有1个正根t2>0,对应方程(4)的cosC只有1个解.把式(9)代入式(5)可求得
![](https://img.fx361.cc/images/2023/0205/fae686ff7d6b2f20add2c8239a609cbcc5d79ae6.webp)
![](https://img.fx361.cc/images/2023/0205/ca8471c4cf1bb08a1ecd6a216b3e227d7b4b28a1.webp)
![](https://img.fx361.cc/images/2023/0205/072b8bc8c86aba4712f56aa0fa92b8e167324033.webp)
(4)相关结论.
分别把情况1和情况4合并,把情况2和情况5合并,可得:
![](https://img.fx361.cc/images/2023/0205/61ae6596cb723cc8d282c5a2b81ffa46f7d5d1a3.webp)
![](https://img.fx361.cc/images/2023/0205/3ed4f671f59e850f668a85789b73967bd230d2a6.webp)
![](https://img.fx361.cc/images/2023/0205/7cb3614605500f3839bb52648d3adf9cf4169d65.webp)
图5 图6 图7
![](https://img.fx361.cc/images/2023/0205/6eecb2ed7cd2de00b82186580b9a7bbc300f46af.webp)
如果把已知条件{(m,n)|0由上面的一般性结论,还可以编拟出各种题目用于不同的场合(略).