鲁京津琼专用2020版高考数学大一轮复习第四章三角函数解三角形阶段自测卷三课件

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1、阶段自测卷(三),第四章 三角函数、解三角形,一、选择题(本大题共12小题，每小题5分，共60分) 1.(2019浏阳六校联考)已知点P(4,3)是角终边上的一点，则sin()等于,解析 点P(4,3)是角终边上的一点，,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,故选A.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,故选C.,3.(2019长沙长郡中学调研)cos 210cos 752cos215sin 15等于,1,2,3,4,5,6,7,8,9,10

2、,11,12,13,14,15,16,17,18,19,20,21,22,解析 根据相应公式可得cos 210cos 752cos215sin 15cos 30cos 75sin 30cos 15 (sin 15cos 30cos 15sin 30)sin 45 故选B.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,又tan 2，,故选D.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,

3、16,17,18,19,20,21,22,7.(2019成都七中诊断)设a，b，c分别是ABC的内角A，B，C的对边，已知(bc)sin(AC)(ac)(sin Asin C)，则A的大小为 A.30 B.60 C.120 D.150,解析 (bc)sin(AC)(ac)(sin Asin C)， 由正弦定理可得(bc)b(ac)(ac)， 整理可得b2c2a2bc，,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,由A(0，)，可得A120. 故选C.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,1

4、6,17,18,19,20,21,22,即ysin(2x).,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,故选A.,9.(2019吉林通榆一中期中)函数f(x)cos(x)的部分图象如图所示，则f(x)的单调递减区间为,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,f(x)cos(x)，,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,故选D.,1,2,3,4,5,6,7,8,9,10,11,

5、12,13,14,15,16,17,18,19,20,21,22,故选A.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,解析 如图以OA,2OB为邻边作平行四边形OAED，F为AE中点，根据题意知，P点在以BF，BD为邻边的平行四边形上及其内部， 动点P的轨迹所覆盖图形的面积为2SAOB.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,又O为AB

6、C的内心，,故选A.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,即f(x)2cos x(sin xcos cos xsin )m,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,总能以f(a)，f(b)，f(c)的长为边构成三角形， 则2f(x)minf(x)max 0 ，,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,

7、16,17,18,19,20,21,22,故选D.,二、填空题(本大题共4小题，每小题5分，共20分) 13.(2019南充适应性考试)已知sin 则cos 2_.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,化简得sin Acos Bcos Asin Bsin(AB)0， 0A，0B，AB， AB，ab.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21

8、,22,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,三、解答题(本大题共70分) 17.(10分)(2019武汉示范高中联考)已知函数f(x) (1)求函数f(x)的单调递增区间；,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,1,2

9、,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,(1)求的值；,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,19.(12分)(2019佛山禅城区调研)ABC的对边分别为a，b，c，且满足abcos Ccsin B. (1)求角B；,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,解 已知a

10、bcos Ccsin B，由正弦定理得sin Asin Bcos Csin Csin B， sin(BC)sin Bcos Csin Csin B, sin Bcos Ccos Bsin Csin Bcos Csin Csin B， cos Bsin Csin Csin B， 因为在ABC中sin C0，所以cos Bsin B，,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,20,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,(1)求f(x)的解析式；,21,22,20,1,

11、2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,T，2，,21,22,20,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,(2)在锐角ABC中，角A，B，C的对边分别为a，b，c，且满足(2ca)cos Bbcos A，求f(A)的取值范围.,解 (2ca)cos Bbcos A， 由正弦定理得2sin Ccos Bsin(AB)sin C，,21,22,20,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,21.(12分)已知向量m( sin x,1)

12、，n(cos x，cos2x1)，设函数f(x)mnb. (1)若函数f(x)的图象关于直线x 对称，且当0,3时，求函数f(x)的单调增区间；,21,22,20,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,21,22,解得3k1(kZ)，0,3，1，,20,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,21,22,(2)在(1)的条件下，当x 时，函数f(x)有且只有一个零点，求实数b的取值范围.,20,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18

13、,19,21,22,20,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,21,22,20,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,22.(12分)(2019衡水中学考试)如图，在ABC中，P是BC边上的一点，APC60，AB APPB4. (1)求BP的长；,21,22,在ABP中，由余弦定理，,解 由已知，得APB120，,整理，得BP24BP40.解得BP2.,20,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,21,22,解 由(1)知，AP2， 所以在ACP中，由正弦定理,从而ACPAPC，即ACP是锐角，,