www.890555.com:人教A版高中数学必修四 3.2 简单的三角恒等变换(2)测试(教师版)

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3.2 简单的三角恒等变换(2) (检测教师版)
时间:40 分钟 班级: 一、 选择题(共 6 小题,每题 5 分,共 30 分) π? π? 2? 1.函数 y=cos2? ?x-12?+sin ?x+12?-1 是( A.奇函数 C.奇函数且是偶函数 ) B.偶函数 D.既不是奇函数,又不是偶函数 总分:60 分 姓名:

π? π? ? 1+cos? ?2x-6? 1-cos?2x+6? π π?? 1 1 ? 解析:y= + -1= ? cos?2x-6? ?-cos?2x+6??=2sin2x,是 2 2 2? ? 奇函数. 答案:A π 1 x+ ?,若 a=f(lg5),b=f?lg ?,则( 2.已知 f(x)=sin2? ? 4? ? 5? A.a+b=0 C.a+b=1
2

)

B.a-b=0 D.a-b=1

π 2x+ ? 1-cos? 2 ? ? 1+sin2x π 1 ? 解析:因为 f(x)=sin ? = ,不妨令 lg5=t,则 lg =-t, ?x+4?= 2 2 5 1+sin2t 1? 1-sin2t 所以 a=f(lg5)=f(t)= ,b=f? ?lg5?=f(-t)= 2 .所以 a+b=1.故选 C. 2 答案:C π 2α- ? 1+ 2cos? 4? ? 3 3.已知角 α 在第一象限,且 cosα= ,则 等于( 5 π? ? α + sin? 2? 2 A. 5 7 B. 5 14 C. 5

)

2 D.- 5

π π? 1+ 2? ?cos2αcos4+sin2αsin4? 1+cos2α+sin2α 2cos2α+2sinαcosα 解析:原式= = = cosα cosα cosα

?3+4?=14. =2× (cosα+sinα)=2× ?5 5? 5
答案:C π 0, ?上的最小值为-4,那么 a 的 4.设函数 f(x)=2cos2x+ 3sin2x+a(a 为实常数)在区间? ? 2? 值等于( )

A.4

B.-6

C.-4

D.-3

π? 解 析 : f(x) = 2cos2x + 3 sin2x + a = 1 + cos2x + 3sin2x + a = 2sin ? ?2x+6? + a + 1. 当 π? x∈? ?0,2?时, π π 7π? ?-1?+a+1=-4.∴a=-4. , ,∴f(x)min=2· 2x+ ∈? ? 2? 6 ?6 6 ? 答案:C π? 5.使 f(x)=sin(2x+θ)+ 3cos(2x+θ)为奇函数,且在区间? ?0,4?上是减函数的 θ 的一个值是 ( ) π A.- 3 π B. 3 2 C. π 3 4 D. π 3

π? π π 4 ? π? 解析:f(x)=2sin? ?2x+θ+3?,当 θ 取-3时,为奇函数,但在?0,4?上递增;θ 取3和3π 时为非奇非偶函数;当 θ 取 答案:C 6.设△ABC 的三个内角为 A,B,C,向量 m=( 3sinA,sinB),n=(cosB, 3cosA),若 m· n =1+cos(A+B),则 C=( π A. 6 π B. 3 ) 2π C. 3 5π D. 6 2π 时,f(x)=-2sin2x 符合题意. 3

解析:依题意得 3sinAcosB+ 3cosAsinB=1+cos(A+B), π? ? π? 1 3sin(A+B)=1+cos(A+B), 3sinC+cosC=1,2sin? ?C+6?=1,sin?C+6?=2. π π 7π π 5π 2π 又 <C+ < ,因此 C+ = ,C= ,故选 C. 6 6 6 6 6 3 答案:C 二、填空题(共 2 小题,每题 5 分,共 10 分) π π? 7.函数 f(x)=sin2x+ 3sinxcosx 在区间? ?4,2?上的最大值是 解析:由已知得 f(x) = π 5π? ∈? ?3 , 6 ?, π? ?1 ? 1 3 sin? ?2x-6?∈?2,1?,因此 f(x)的最大值等于2+1=2. 3 答案: 2 π 8.若函数 f(x)=(1+ 3tan x)cos x,0≤x< ,则 f(x)的最大值是 2 1-cos2x π? 3 1 π ?π π? + sin2x = + sin ? ?2x-6? ,当 x∈ ?4,2? 时, 2x - 6 2 2 2

sin x ? ? π? 解析: f(x)=(1+ 3tan x)cos x=? ?1+ 3cos x?cos x= 3sin x+cos x=2sin?x+6?. π π π 2 π π ∵0≤x< ,∴ ≤x+ < π,∴当 x+ = 时,f(x)取到最大值 2. 2 6 6 3 6 2 答案: 2 三、解答题(共 2 小题,每题 10 分,共 20 分) x x x 1 9.已知函数 f(x)=cos2 -sin cos - . 2 2 2 2 (1)求函数 f(x)的最小正周期和值域; 3 2 (2)若 f(α)= ,求 sin2α 的值. 10 π x x x 1 1 1 1 2 x+ ?. 解析:(1)f(x)=cos2 -sin cos - = (1+cosx)- sinx- = cos? 4? ? 2 2 2 2 2 2 2 2 所以 f(x)的最小正周期为 2π,值域为?-

?

2 2? . , 2 2?

(2)由(1)知 f(α)=

π 3 2 2 ? ? π? 3 cos?α+4? ?= 10 ,所以 cos?α+4?=5. 2

π π π 18 7 +2α?=-cos?2?α+ ??=1-2cos2?α+ ?=1- = . 所以 sin2α=-cos? 2 4 4 ? ? ? ? ?? ? ? 25 25 10.已知函数 f(x)=sin(π-ωx)cosωx+cos2ωx(ω>0)的最小正周期为 π. (1)求 ω 的值; 1 (2)将函数 y=f(x)的图象上各点的横坐标缩短到原来的 , 纵坐标不变, 得到函数 y=g(x) 2 的图象, π? 求函数 g(x)在区间? ?0,16?上的最小值. 解析:(1)因为 f(x)=sin(π-ωx)cosωx+cos2ωx. 1+cos2ωx 1 π? 1 1 1 2 所以 f(x)=sinωxcosωx+ = sin2ωx+ cos2ωx+ = sin? ?2ωx+4?+2. 2 2 2 2 2 2π 由于 ω>0,依题意得 =π.所以 ω=1. 2ω (2)由(1)知 f(x)= π 1 π? 1 2 ? 2 ? sin?2x+4? ?+2.所以 g(x)=f(2x)= 2 sin?4x+4?+2. 2

π 1+ 2 π π π π 2 4x+ ?≤1.因此 1≤g(x)≤ 当 0≤x≤ 时, ≤4x+ ≤ .所以 ≤sin? . 4 ? ? 16 4 4 2 2 2 π 0, ?上的最小值为 1. 故 g(x)在区间? ? 16?


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