十年真题(2010-2019)高考数学(文)分类汇编专题05 三角函数与解三角形(新课标Ⅰ卷)(解析版) 下载本文

QA,2???????A均为锐角,可得:?A?,??A??,?a?c?(6,43]. 362636故答案为: (6,43].

16.在?ABC中,已知AB边上的中线CM?1,且________. 【答案】【解析】

111,,成等差数列,则AB的长为tanAtanCtanB23 3111

,,成等差数列, tanAtanCtanB2112cosCcosAcosBsin(A?B)sinC??????所以,即, tanCtanAtanBsinCsinAsinBsinAsinBsinAsinB因为

sin2Cc2所以2cosC?,由正弦定理可得cosC?,

sinAsinB2aba2?b2?c2a2?b2?c2c2又由余弦定理可得cosC?,所以,故a2?b2?2c2, ?2ab2ab2abuuuuvuuuuv1uuuvuuuvCA?CB, 又因为AB边上的中线CM?1,所以CM?1,因为CM?2uuuur2uuur2uuur2uuuruuuruuur2uuur2uuuruuur所以4CM?CA?CB?2CA?CB?CA?CB?2CACBcosC,

??c223. 即4?b?a?2ab??3c2,解c?32ab22即AB的长为

23. 3故答案为23 3cosB?3. 317.在?ABC中,A,B,C的对边分别a,b,c,A?60?,(Ⅰ)若D是BC上的点,AD平分?BAC,求

DC的值; BD(Ⅱ)若 ccos B?bcosC?2,求?ABC的面积.

【答案】(Ⅰ)26?4;(Ⅱ)【解析】

(Ⅰ)因为cosB?62+43 936,∴sinB?, 33sinC?sin?A?B??sinAcosB?cosAsinB?由正弦定理得

33163?6, ????23236ADBDADDC?,?, sinBsin?BADsinCsin?CAD因为AD平分?BAC,

6DCsinB??3?26?4. 所以

BDsinC3?66a2?c2?b2a2?b2?c2(Ⅱ)由ccosB?bcosC?2,即ccosB?bcosC?c??b??a?2,

2ac2ab所以

abasinB42?,∴b?, ?sinAsinBsinA311423?662+43. absinC??2???22369故SVABC?18.在?ABC中,角A,B,C所对的边分别a,b,c,f?x??2sin?x?A?cosx?sin?B?C??x?R?,函数

f?x?的图象关于点?(1)当x??0,???,0?对称. ?6?????时,求f?x?的值域; ?2?133,求?ABC的面积. 14(2)若a?7且sinB?sinC?【答案】(1)?????3?,1?(2)103 2?【解析】

(1)f?x??2sin?x?A?cosx?sin?B?C? ?2sin?x?A?cosx?sinA

=2sin(x?A)cosx?sin(x?(x?A))=2sin(x?A)cosx?sinxcos(x?A)?sin(x?A)cosx =sin(x?A)cosx?sinxcos(x?A)?sin?2x?A?

∵函数f?x?的图像关于点??π?,0?对称, 6??∴f??π???0 6??∴A?π 3??π?? 3?∴f?x??sin?2x?∵f?x?在区间?0,??5π??5ππ?上是增函数,?,?上是减函数, ?12??122?且f?0????5π?3,f???1,

12??23?π?。 f????2?2?3?∴f?x?的值域为???2,1?

??(2)∵sinB?sinC?133 14?sinB?sinC?∴b?c?13

1313sinA?b?c??a?13 77由余弦定理,a2?b2?c2?2bccosA ∴bc?40 ∴SVABC?1bcsinA?103 219.在?ABC中,已知AB?2,cosB?(1)求BC的长; (2)求sin(2A??2,C=.

410?3)的值.

【答案】(1)BC?【解析】

8224?73(2) 550解:(1)因为cosB?2,0?B??, 102?2?72所以sinB?1?cos2B?1??. ???10?10??在?ABC中,A?B?C??,所以A???(B?C), 于是sinA?sin(??(B?C))?sin(B?C)

?sinBcosC?cosBsinC?722224????. 1021025在?ABC中,由正弦定理知

BCAB?, sinAsinC所以

BC?AB2482?sinA???sinC5. 252(2)在?ABC中,A?B?C??,所以A???(B?C), 于是cosA?cos(??(B?C))??cos(B?C)

?22722?3??(cosBcosC?sinBsinC)????10?2?10?2???5,

??于是sin2A?2sinAcosA?2??24324?, 552527?3??4?cos2A?cos2A?sin2A????????.

25?5??5?因此,sin?2A???????sin2Acos?cos2Asin 3?33???241?7?324?73. ???????252?25?2503,BD?6.

20.如图,在四边形ABCD中,?A?60?,?ABC?90?.已知AD?