b2?c2?a21A???A,?102 02.120 cos2bc203.6?2 A?150,abbsinA6?2 ?,a??4sinA?4sin150?4?sinAsinBsinB44. 1200
a∶b∶c?sinA∶sinB∶sinC?7∶8∶13， 令a?7k,b?8k,c?13k cosC?a2?b2?c22ab??12,C?1200 5. 4
ACsinB?BCsinA?ABAC?BCABsinC,sinB?sinA?sinC,AC?BC ?2(6?2)(sinA?sinB)?4(6?2)sinA?BA?B2cos2
?4cosA?B2?4,(AC?BC)max?4
三、解答题
1. 解：acosA?bcosB?ccosC,sinAcosA?sinBcosB?sinCcosC
sin2A?sin2B?sin2C,2sin(A?B)cos(A?B)?2sinCcosC cos(A?B)??cos(A?B),2cosAcosB?0
cosA?0或cosB?0，得A??2或B??2
所以△ABC是直角三角形。
a2?c2?b2b22. 证明：将cosB?2ac，cosA??c2?a22bc代入右边
a2?2222 得右边?c(c2?b22abc?b?c?a22abc)?2a?2b2ab
?a2?b2ab?ab?ba?左边，
∴
abcosBcob?a?c(b?sAa) 3．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?∴sinA?sinB?sinC?cosA?cosB?cosC
coAs 4.解：∵a?c?2b,∴sinA?sinC?2sinB，即2sinA?CA?CBBcos?4sincos， 2222∴sinB?B1A?C3B13，而0??,∴cos?， ?cos?22222424∴sinB?2sinBB313cos?2???224439 8参考答案（数学5必修）第一章 [综合训练B组]
一、选择题
1.C A??6,B??3,C??2,a:b:c?sinA:sinB:sinC?132::?1:3:2 2222.A A?B??,A???B，且A,??B都是锐角，sinA?sin(??B)?sinB 3.D sinA?sin2B?2sinBcosB,a?2bcosB 4.D lgsinAsinA?lg2,?2,sinA?2cosBsinC
cosBsinCcosBsinCsin(B?C)?2cosBsinC,sinBcosC?cosBsinC?0, sin(B?C)?0,B?C，等腰三角形
5.B (a?b?c)(b?c?a)?3bc,(b?c)?a?3bc,
22b2?c2?a21s??A,? b?c?a?3bc,coA2bc222206 0B??6.C c?a?b?2acbosC?9,c?，3B为最大角，cos2221 7A?BA?BsinA?Ba?bsinA?sinB22， 7.D tan???2a?bsinA?sinB2sinA?BcosA?B22A?BtanA?B2,tanA?B?0，或tanA?B?1 tan?A?B222tan2?所以A?B或A?B?
22cos二、填空题
1.
2393 S113?ABC?2bcsinA?2c?2?3c,?a42,?a1?3, 13
a?b?casinA?siBn?sCi?sA?1323nin3? 932s?2.? A?B????in(?B)2,A?2?B，即tanA?ta2n(?B?)2 cos?2(?B)?cosBsinB?1taBn，tanA?1tanB,taAntBa?n 13. 2 tanB?taCn?sinBcosB?siCncoCs
?sinBcoCs?cBo?sCsicosBcoCs?nBs?inC(1?)sAin A2sin2sinA4. 锐角三角形 C为最大角，cosC?0C,为锐角
2222?8?435. 600 cosA?b?c?a?32bc?422?6?2?3?112?2?(3?1)?2 2?a2?b2?c2?13?c26．(5,13) ??a2?c2?b2,??4?c2?9,5?c?213,5?c?13 ??c2?b2?a2??c?29?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?∴sinAsinBsinC?cosAcosBcosC,sinAsinBsinCcosAcosBcosC?1
∴tanA?tanB?tanC?1
coAs 3. 证明：∵sinA?sinB?sinC?2sinA?2A? ?2sin2C ?2cos?2A ?4cos2 ?2sinA?Bcos?2BA?B(cos?2AB2cosco s22BCcosco s22BA?BA?Bcos?sin(A?B) 22A?BA?B2sincos
22A?Bcos )2∴sinA?sinB?sinC?4cosABCcoscos 222a2?ac?b2?bcab??1，只要证?1， 4．证明：要证2b?ca?cab?bc?ac?c即a2?b2?c2?ab
而∵A?B?1200,∴C?60
0a2?b2?c22cosC?,a?b2?c2?2abcos600?ab
2ab∴原式成立。
CA3b?ccos2? 2221?coCs?1coAs3BsinA??siCn?? ∴sin
222ncCo?sCsi?nCsinA?cos B 即sinA?siA 5．证明：∵acos2 ∴sinA?siCn?A?siCn?即sinsiAn?(C?) 3Bs2sBi，∴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
A?coAs?1.C sin而0?A??,2.B
2sAin?(
4?),?4?A??4?5?2????sin(A?)?1 424a?bsinA?sinB??sinA?siBn csinCA?BA?BA?Bcos?2cos ?2sin2223.D cosA?1,A?600,S2ABC1?bcsinA?63 24.D A?B?900则sinA?cosB,sinB?cosA，00?A?450,
A?coAs，450?B?900,sinB?cosB sin222222,cb?c?a??,bcocs5.C a?c?b?b1A??,20A?1 20sinAcoBs??6.B
cosAsiBn sinA2?2siAn,2siBncBos?cAosAsin,siAnBsincoAs?sBinB cossinB2A,?2或B2A?2B??2
二、填空题
1. 对 sinA?sinB,则2. 直角三角形
ab??a?b?A?B 2R2R)1,1(1?cosA2??1coBs2?)2cAo?sB(? 21(cos2A?cos2B)?cos2(A?B)?0, 2cos(A?B)cos(A?B)?cos2(A?B)?0
cosAcosBcosC?0
??n?3. x?y?z A?B?,A??B,siA22 c?a?b,sinC?sinA?cBosB,s?inAyco?sz , y?siBnx?,yx?,zA?CA?CA?CA?C,2sinco?s4sin cos2222A?CA?CACACcos?2cos,coscos?3sinsin
2222221C2Asin2 则sinAsinC?4sin3221cosA?cosC?cosAcosC?sinAsinC
3AC??(1?cosA)(1?cosC)?1?4sin2sin2
22ACAC??2sin2?2sin2?4sin2sin2?1?1
2222??tanA?tanC25. [,) tanB?tanAtanC,tanB??tan(A?C)?
32tanAtanC?1tanA?taCnB??taAn?(C?) tan 2tanB?1A?siCn?4．1 sin2sBin