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

【解析】

函数f?x??2sin??x???的图像过点0,3 ?2sin??3,即:sin????3 2Q0????2 ????3

又函数图象关于点??2,0?对称 ?2sin??2????????0,即:?2???k?,k?Z 3?3?1?????k??,k?Z

26Q0???1 ???????? ?f?x??2sin?x??,

3?6?6??????f??1??2sin?????2sin?1

6?63?本题正确结果:1

10.若实数x,y满足2cos2?x?y?1?x?1???y?1???x?y?122?2xy.则xy的最小值为____________

【答案】. 【解析】

∵2cos2?x?y?1??14?x?1???y?1??2xyx?y?122>0, ,∴x?y?1?x?1???y?1?Qx?y?1??x?y?1??22?2xy?x?y?1??2?1x?y?1??x?y?1??1

x?y?11?2x?y?1?x?y?1??1?2,

x?y?1当且仅当x?y?1?1时即x=y时取等号

Q2cos2?x?y?1??2,当且仅当x?y?1?k??k?Z?时取等号

?x?1???y?1??2xyx?y?122?2cos2?x?y?1??2,即x?y?1?1且

x?y?1?k??k?Z?,

即x?y?1?k??k?Z?, 22?1?k??1(当且仅当

因此xy??, k?0时取等号)???2?4从而xy的最小值为. 11.设函数f(x)?sin(2x?【答案】(【解析】

不妨设x1?0?x2,则x2?x1?x2?x1,由图可知x2?x1?0?(?14?3),若x1x2?0,且f(x1)?f(x2)?0,则x2?x1的取值范围是_______.

?,??) 3?3)??3.

故答案为:(

?,??) 312.已知角?为第一象限角,【答案】(1,2] 【解析】

a?sin??3,则实数a的取值范围为__________.

cos?由题得a?sin??3cos??2sin(???3),

因为2k????2k???2,k?Z,

所以2k?+?3??+?3?2k??5?,k?Z, 6所以

1???sin(??)?1,?1?2sin(??)?2. 233故实数a的取值范围为(1,2]. 故答案为:(1,2]

13.已知函数f(x)?sin(x??)?2cos(x??)(0????)的图象关于直线x??对称,则cos2??___.

【答案】 【解析】

因为函数f(x)?sin(x??)?2cos(x??)(0????)的图象关于直线x??对称,

35????3??f???f??2??2??, ?即cos??2sin???cos??2sin?,即cos???2sin?, 即tan???1, 22221cos??sin?1?tan?34cos2?????, 则2221cos??sin?1?tan?1?541?故答案为.

14.如图,四边形ABCD中,AB?4,BC?5,CD?3,?ABC?90?,?BCD?120°,则AD的长为______

35

【答案】65?123 【解析】

连接AC,设?ACB??,则?ACD?120o??,如图:

故在Rt?ABC中,sin??45,cos?? , 414113153443?5, Qcos?120o?????cos??sin???????22224141241又在?ACD中由余弦定理有cos120o???????2?3?241?32?AD241?43?5,解得AD2?65?123,即

241AD?65?123,

故答案为:65?123. 15.在锐角?ABC中,角A,B,C的对边分别为a,b,c.且

cosAcosB23sinC??,b?23.则ab3aa?c的取值范围为_____.

【答案】(6,43] 【解析】

QcosAcosB23sinC23???bcosA?acosB?bsinC ab3a323sinBsinC, 3由正弦定理可得: sinBcosA?sinAcosB?可得:sin(A?B)?sinC?233, sinBsinC,?sinB?32又?ABC为锐角三角形,?B??,可得: 3a?c?bsinAbsinC???2?????4(sinA?sinC)?4sinA?4sin??A??43cos?A?? sinBsinB3??3??