各地市高考数学最新联考试题分类汇编三角函数(20220319144122).pdf
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1、学习好资料欢迎下载湖南省各地市 20XX年高考数学最新联考试题分类汇编(5)三角函数一、选择题:3(湖南省 十二校 20XX届高三第二次联考理)函数,xxxfcos)2sin()(的最小正周期 是A2BC2D4【解析】由三个向量)2cos,(Aam,)2cos,(Bbn,)2cos,(Ccp共线及正弦定理可得:sincos,sincos,sincos,222ABCABC由sin2sincoscos222AAAA,因为cos02A,所以1sin22A,因为0A,所以022A,所以26A,即3A.同理可得,33BC,5.(湖南师大附中20XX届高三第六次月考理)函数)2|,0,0)(sin()(A
2、xAxf的部分图象如图示,将()yf x的图象向右平移6个单位后得到函数)(xgy的图像,则)(xg的单调递增区间为()A.32,62kk B.652,32kk C.3,6kk D.65,3kk精品资料-欢迎下载-欢迎下载 名师归纳-第 1 页,共 8 页 -学习好资料欢迎下载【答案】C【解析】由图象知1A,T,262,2,234)61211(6,),62sin()(xxf将)(xf的图象平移6个单位后的解析式为).62sin(6)6(2sinxxy则由:36226222kxkkxk,Zk.6(湖 南 省 五 市 十 校 20XX届 高 三 第 一 次 联 合 检 测 理)在 斜 三 角 形
3、ABC中,s i n=2 c o sc o sABC,且tantan=12BC,则A的值为(A)A4 B3 C2 D34二、填空题:10 (湖 南 省 十 二 校20XX届 高 三 第 二次联考 文)已知 向量2t a n,/),1,2(),cos,(sin则且baba.【答案】4312(湖南省长沙市20XX 年高考模拟试卷一文科)已知x(0,2)时,sinxxtanx,若精品资料-欢迎下载-欢迎下载 名师归纳-第 2 页,共 8 页 -文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K
4、2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7
5、I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1
6、O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q
7、4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9
8、Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10
9、M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T
10、2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3学习好资料欢迎下载p=23sin18+21cos18、ooq10tan110tan22,oor20tan3120tan3,那么 p、q、r 的大小关系为;【答案】qpr 13(湖南省五市十校20XX届高三第一次联合检测理)已知)2,0(且tan()34,则lg(sin2cos)lg(3sincos)0 三、解答题:20、(湖南师大附中20XX 届高三第六次月考理)(满分13 分)随着私家车的逐渐增多,居民小区“停车难”
11、问题日益突出本市某居民小区为缓解“停车难”问题,拟建造地下停车库,建 筑设计师提供了该地下停车库的入口和进入后的直角转弯处的平面设计示意图.(1)按规定,地下停车库坡道口上方要张贴限高标志,以便告知停车人车辆能否安全驶入,为标明限高,请你根据该图所示数据计算限定高度CD的值(精确到 0.1m)(下列数据提供参考:sin20 0.3420,cos20 0.9397,tan20 0.3640)(2)在车库内有一条直角拐弯车道,车道的平面图如图所示,设(rad)PAB,车道宽为 3 米,现有一辆转动灵活的小汽车,其水平截面图为矩形,它的宽为1.8 米,长为4.5 米,问此车是否能顺利通过此直角拐弯车
12、道?解:(1)在ABE中,ABE=90,BAE=20 ,tan BAE=BEAB,又AB=10,BE=AB?tan BAE=10tan20 3.6m,BC=0.6CE=BE-BC=3m,在CED中,CD AE,ECD=BAE=20 ,cosECD=CDCE,CD=CE?cos ECD=3cos2 030.942.8m 故答案为 2.8m 5 分(2)延长CD与直角走廊的边相交于,E F,如下图.33cossinEFOEOF,其中02.容易得到1.8tantanDADE,tan1.8tanCFBC.又()ABDCEFDECF,A 3 米3 米1.8 米P B C D E O F 精品资料-欢迎下
13、载-欢迎下载 名师归纳-第 3 页,共 8 页 -文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6
14、 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文
15、档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5
16、N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4
17、X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D
18、3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6
19、N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3学习好资料欢迎下载于是331()1.8(tan)cossintanf3(sincos)1.8sincos,其中02.8 分设sincost,则2sin()4t,于是12t.又21sincos2t,因此263.6()
20、()1tfg tt 11 分因为22222267.266(0.6)3.84()(1)(1)tttg ttt,又12t,所以()0g t恒成立,因此函数263.6()1tg tt在(1,2t是减函数,所以min()(2)6 23.64.5g tg,故能顺利通过此直角拐弯车道 13 分17(湖 南省长 沙市20XX年高 考模拟试卷一文科)设函数)0(sinsincos2cossin22xxxxf在x处取最小值.(1)求的值;(2)在ABC中,cba,分别是角A,B,C 的对边,已知,2,1 ba22)(Bf,求值)cos()sin()3sin(2CCC.17.解:解:(1)f(x)=2xxxsin
21、sincos2cossin2=sinx(2cos22-1)+cosxsin =sinxcos+cosxsin=sin(x+),依题意,sin(+)=-1,0,=2;4分(2)由(1)f(x)=sin(x+)=sin(x+2)=cosx,22)(Bf,cosB=-22,0B,B=43;,2,1 ba由正弦定理,BAsinsin=ba=21 sinA=21,ab,AB,0A2,A=6C=-A-B=12;9 分)cos()sin()3sin(2CCC=)15cos()15sin()45sin(20oo精品资料-欢迎下载-欢迎下载 名师归纳-第 4 页,共 8 页 -文档编码:CO9Q7I6N5N6
22、HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6
23、ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档
24、编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N
25、6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X6 ZE7T2Q4K2D3文档编码:CO9Q7I6N5N6 HN10M1O4X4X
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