二、填空题
1.
2393 S?ABC?12bcsinA?12c?32?3c,?a42,?a1?3, 13
a?b?ca133sinA?siBn?sCi?nsA?in?2
9332?2.? A?B??sin(?B2,A??2?B，即tanA?ta?2) 2n(?B?)cos?2(?B)?cosBsinB?1taBn，tanA?1tanB,taAntBa?n
13. 2 tanB?taCn?sinB?siCncosBcoCs
?sinBcoCs?cBo?ssCiBs?inC(ncosBcoCs?n1?)2sinAsAin
A2si4. 锐角三角形 C为最大角，cosC?0C,为锐角 32c2?a22?8?4?35. 600 cosA?b?3?112bc?4
22?6?2?2?2?(3?1)?22?a2?b2?c2?13?c26．(5,13) ??a2?c2?b2,??4?c2?9,5?c?213,5?c?13 ??c2?b2?a2?2?c?9?4三、解答题
1.解：S1?ABC?2bcsinA?3,bc?4,
a2?b2?c2?2bcosA,b?c?，而5c?b
所以b?1,c?4
2. 证明：∵△ABC是锐角三角形，∴A?B??2,即
?2?A??2?B?0
∴sinA?si?2n(?B，即)sinA?coBs；同理sinB?coCs；sinC? 21
coAs
∴sinAsinBsinC?cosAcosBcosC,∴tanA?tanB?tanC?1
3. 证明：∵sinA?sinB?sinC?2sin ?2sinA?B22A?BC2A?Bcos?2B?B sco2A?B2A?2sin2sinAsinBsinCcosAcosBcosCA?B2?1
cosB?sin(A?B)
A?Bcos 2) ?2sin ?2cos? ?4cosA2A?(cos2A2cos2A?Bcos 2Bcos2Cco s2A22∴sinA?sinB?sinC?4cosab?cba?ccosB2cos2C2
4．证明：要证??1，只要证
a?ac?b?bcab?bc?ac?c2?1，
即a2?b2?c2?ab
而∵A?B?1200,∴C?600
a?b?c2ab222cosC?,a?b?c?2abcos60?ab
2220∴原式成立。
5．证明：∵acos2C ∴sinA?21?coCs2?ccos2A2?3b2
?12coAs?3Bsin2?siCn?
即sinA? ∴sinA?即sinA?siAncCo?ssCi?nCsinA?cos BsiCn?siCn?siAn?(C?)3 Bs2sBi，∴na?c?2b
参考答案（数学5必修）第一章 [提高训练C组] 一、选择题
1.C sinA?coAs?2sAin?(4?
5?4),而0?A??,?4?A??4???22?sin(A??4)?1
22
2.B
a?bc?sinA?sinBsinC2?sinA?sinB
A?B2cos
212bcsinA?63 ?2sin3.D cosA?A?BA?Bcos?2012,A?60,S?ABC?4.D A?B?900则sinA? sinA?coBs,sBi?n，cAo0?A?45,
00coAs，450?B?900,sinB?cosB
c?225.C a2?c2?b2?b,cb?a??,bcocs21A??,2A?1 2006.B
sinAcosA?coBssiBn2?siAn,2siBncBos?cAosAsin,siAnBsincoAs?sBin cosB sinA2?sinB2A,?2或B2A?2B??2
二、填空题
1. 对 sinA?sinB,则2. 直角三角形
12a2R?b2R?a?b?A?B ?1coBs2?)22(1?cosA2?12cAo?sB(? )1,(cos2A?cos2B)?cos(A?B)?0,
2cos(A?B)cos(A?B)?cos(A?B)?0
cosAcosBcosC?0
??n?3. x?y?z A?B?,A??B,siA22cBosB,s?inAyco?sz , c?a?b,sinC?sinA?4．1 sinA?siCn?2sBinsiBnx?,yx?, y?zA?2A?CA?CA?C,2sinco?s4sin222A?CA?CACACcos?2cos,coscos?3sinsin
22222212 cosC则sinAsinC?4sin3A2sin2C2
cosA?cosC?cosAcosC?13sinAsinC
2??(1?cosA)(1?cosC)?1?4sin??2sin2A22sin2C2
A2?2sin2C2?4sin2A2sinC2?1?1
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5. [??3,2) tanB?tanAtanC,tanB??tan(A?C)?tanA?taCn taAn?(C?)2tanB?12tanA?tanCtanAtanC?1
tanB??3tanB?tanB?tanA?tanC?2tanAtanC?2tanB tanB?3tanB,tanB?0?tanB?33?B??3
6．1 b2?ac,sin2B?sinAsinC,cosA(?C)?cosB?cos2B
?cosAcosC?sinAsinC?cosB?1?2sinB ?cosAcosC?sinAsinC?cosB?1?2sinAsinC ?cosAcosC?sinAsinC?cosB?1
2?cos(A?C)?cosB?1?1
三、解答题
sin(A?B)asinAcosBsinA1. 解：2 ?,??222a?bsin(A?B)bcosAsinBsinBa?b2222
cosBcosA?siAnsiBn,sinA2?siBn2A?,2B或22A?B??2
∴等腰或直角三角形
2. 解：2RsinA?sinA?2RsinC?sinC?(2a?b)sinB,
asinA?csinC?(2a?b)sinB,a?c?222ab?b,
2a?b?c?csinC22222ab,cosC?a?b?c2ab22222?222,C?450
?2R,c?2RsinC?2R,a?b?2R?2ab,
2R?2ab?a?b?2ab,ab?222R2?222 S?12absinC?24ab?24?2R2?2,Smax?2?12R
2另法：S?12absinC?24ab?24?2RsinA?2RsinB
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?24?2RsinA?2RsinB?222RsinAsinB
?2R?12?[cos(A?B)?cos(A?B)]
??2R?2R22212?[cos(A?B)?22)22]
?(1??Smax?2?12R 此时A?B取得等号
A?C2A?C2A?C2A?C223. 解：sinA?sinC?2sinB,2sinB212A?C224cos?4sincos
sin?cos?,cosB23?4?144B2,sinB?2sinB2cosB2?74
A?C??2,A?C???B,A??,C??4?B2
7?14sinA?sin(3?4?B)?sin3?4cosB?cos3?4sinB?
sinC?sin(?4?B)?sin?4cosB?cos?4sinB?7?147)
a:b:c?sinA:sinB:sinC?(7?227):7:(7?24. 解：(a?b?c)(a?b?c)?3ac,a?c?b?ac,cosB?tanA?1?tanA?212,B?60
0(?C?) tanAtaCntCan?,?3?3?1taCn? ,AtanCtan?33 tanAtaCn???tanA?2? 得??tanC?1?03tanA?，联合 33??tanA?1或??tanC?2??00??A?75，即?或03?C?45?0??A?45 ?0?C?75? 当A?75,C?45时，b?43sinA43sinA?4(32?6),c?8(3?1),a?8
当A?45,C?75时，b?00?46,c?4(3?1),a?8
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