2018-2019学年人教B版高中数学-必修4-课时跟踪检测(十二)已知三角函数值求角(Word)

2018-2019学年人教B版高中数学-必修4-课时跟踪检测(十二)已知三角函数值求角(Word)
2018-2019学年人教B版高中数学-必修4-课时跟踪检测(十二)已知三角函数值求角(Word)

课时跟踪检测(十二) 已知三角函数值求角

层级一 学业水平达标

1.使arcsin(1-x )有意义的x 的取值范围是( )

A .[1-π,1]

B .[0,2]

C .(-∞,1]

D .[-1,1]

解析:选B 由题知应有-1≤1-x ≤1,∴0≤x ≤2.

2.cos ?

???arcsin 12的值为( ) A. 12

B. 32 C .-12 D .-

32 解析:选B ∵在????-π2,π2上,arcsin 12=π6

, ∴cos ????arcsin 12=cos π6=32

. 3.方程cos x +

22=0,x ∈[0,2π]的解集是( ) A. ??????3π4,5π4 B. ????

??-π4,3π4 C. ????

??-π4,-3π4 D. ??????5π4,7π4 解析:选A 在[0,2π]内,cos

3π4=cos 5π4=-cos π4=-22. 4.若tan α=

33,且α∈????π2,3π2,则α=( ) A. π6

B. 5π6

C. 7π6

D. 11π6

解析:选C ∵tan π6=33

,又α∈????π2,3π2, ∴α=π+π6=7π6

. 5.已知sin x =-1213

,x ∈????π,3π2,则x 等于( ) A .arcsin ???

?-1213 B .π-arcsin 1213 C .π+arcsin 1213 D. 3π2-arcsin 1213

解析:选C ∵x ∈????π,3π2,∴x =π+arcsin 1213

. 6.若sin(x -π)=-22

,且-2π<x ≤0,则角x =________. 解析:∵sin(x -π)=-sin(π-x )=-sin x =-

22, ∴sin x =22.∴x =2k π+π4或2k π+34

π(k ∈Z). 又-2π

π. 答案:-74π或-54

π 7.若α∈(0,2π),tan α=1,cos α=-22

,则α=________. 解析:由已知,得α是第三象限的角.又α∈(0,2π),tan

5π4=1,cos 5π4=-22, ∴α=5π4

. 答案:5π4

8.已知等腰三角形的顶角为arccos ???

?-12,则底角的正切值是________. 解析:∵arccos ????-12=2π3

, ∴底角为π-2π32=π6.∴tan π6=33

. 答案:33

9.求方程tan x =-3,x ∈(-π,π)的解集.

解:∵tan ????-π3=-tan π3

=-3, tan ????π-π3=-tan π3

=-3, -π3,π-π3=2π3

都在(-π,π)内, ∴方程tan x =-3,x ∈(-π,π)的解集为????

??-π3,2π3. 10.已知cos ????2x +π3=-12,x ∈[0,2π],求x 的集合.

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