人教A版 必修1 第5章 三角恒等变换 单元测试卷(解析版)

人教A版 必修1 第5章 三角恒等变换 单元测试卷(解析版)
人教A版 必修1 第5章 三角恒等变换 单元测试卷(解析版)

第15单元三角恒等变换单元测试卷

一、选择题(共12小题).

1.(5分)(sin15°+cos15°)2的值为()

A.B.C.D.

2.(5分)已知cosα=,α∈(0,),则cos(α﹣)=()A.﹣B.C.D.

3.(5分)设tanα,tanβ是方程x2﹣3x+2=0的两个根,则tan(α+β)的值为()A.﹣3B.﹣1C.1D.3

4.(5分)已知α,β均为锐角,,cos(α+β)=()

A.B.C.D.

5.(5分)若sinθ=,且90°<θ<180°,则cos的值为()A.B.C.D.

6.(5分)已知角α,β均为锐角,且cosα=,sinβ=,则α﹣β的值为()A.B.C.D.

7.(5分)已知α∈(0,),2sin2α=cos2α+1,则sinα=()

A.B.C.D.

8.(5分)在△ABC中,若tan A tan B>1,则△ABC是()

A.锐角三角形B.直角三角形C.钝角三角形D.无法确定

9.(5分)使函数y=sin(2x+φ)+cos(2x+φ)为奇函数,且在[0,]上是减函数的φ的一个值为()

A.B.C.D.

10.(5分)中国古代数学家赵爽设计的弦图(如图1)是由四个全等的直角三角形拼成,四个全等的直角三角形也可拼成图2所示的菱形,已知弦图中,大正方形的面积为100,小正方形的面积为4,则图2中菱形的一个锐角的正弦值为()

A.B.C.D.

11.(5分)已知,且cosθ>0,则()

A.tanθ<0B.

C.sin2θ>cos2θD.sin2θ>0

12.(5分)下列四个等式,其中正确的是()

A.tan25°+tan35°+tan 25°tan35°=

B.=

C.cos2﹣sin2=

D.﹣=4

二、填空题(本题共4小题,每小题5分,共20分)

13.(5分)若=﹣,则cosα+sinα的值为.

14.(5分)已知sinα+cosα=,其中α∈(,π),则tan2α=.

15.(5分)已知函数f(x)=sin x﹣cos(x+),则f()=,f(x)的值域为.

16.(5分)已知A是函数f(x)=sin(2020x+)+cos(2020x﹣)的最大值,若存在实数x1,x2,使得对任意实数x,总

有f(x1)≤f(x)≤f(x2)成立,则A?|x1﹣x2|的最小值为.

三、解答题(本题共4小题,共40分.解答应写出文字说明、证明过程或演算步骤)17.(8分)已知角α的顶点与原点O重合,始边与x轴的非负半轴重合,它的终边过点P (﹣,﹣).

(Ⅰ)求sin(α+π)的值;

(Ⅱ)若角β满足sin(α+β)=,求cosβ的值.

18.(10分)已知α,β为锐角,tanα=,cos(α+β)=﹣.

(1)求cos2α的值;

(2)求tan(α﹣β)的值.

19.(10分)在△ABC中,A、B为锐角且B<A,sin A=,sin2B=.(1)求角C的值;

(2)求证:5cos A cos(A+3B)=2sin B.

20.(12分)已知函数f(x)=cos(+x)cos(),g(x)=sin2x﹣.(1)求函数f(x)的最小正周期;

(2)求函数h(x)=f(x)﹣g(x)的最大值,并求使h(x)取得最大值的x的集合.附加题(本大题共2小题,每小题10分,共20分.解答应写出文字说明、证明过程或演算步骤)

21.(10分)某同学在一次研究性学习中发现,以下五个式子的值都等于同一个常数.(1)sin213°+cos217°﹣sin13°cos17°

(2)sin215°+cos215°﹣sin15°cos15°

(3)sin218°+cos212°﹣sin18°cos12°

(4)sin2(﹣18°)+cos248°﹣sin(﹣18°)cos48°

(5)sin2(﹣25°)+cos255°﹣sin(﹣25°)cos55°

(Ⅰ)试从上述五个式子中选择一个,求出这个常数;

(Ⅱ)根据(Ⅰ)的计算结果,将该同学的发现推广为三角恒等式,并证明你的结论.22.(10分)已知函数f(x)=cos(ωx+φ)(ω>0,0<φ≤π)为奇函数,且其图象上相邻的一个最高点与一个最低点之间的距离为.

(1)求f(x)的解析式;

(2)若f(α+)=﹣(﹣<α<0),求sin(2α﹣)的值.

参考答案

一、选择题(共12小题).

1.(5分)(sin15°+cos15°)2的值为()

A.B.C.D.

【分析】利用诱导公式、二倍角公式进行化简三角函数式,可得结果.

解:(sin15°+cos15°)2=1+2sin15°cos15°=2+sin30°=1+=,

故选:C.

2.(5分)已知cosα=,α∈(0,),则cos(α﹣)=()A.﹣B.C.D.

【分析】由已知求得sinα,然后展开两角差的余弦求值.

解:∵cosα=,α∈(0,),

∴sinα=,

故选:D.

3.(5分)设tanα,tanβ是方程x2﹣3x+2=0的两个根,则tan(α+β)的值为()A.﹣3B.﹣1C.1D.3

【分析】由tanα,tanβ是方程x2﹣3x+2=0的两个根,利用根与系数的关系分别求出tanα+tanβ及tanαtanβ的值,然后将tan(α+β)利用两角和与差的正切函数公式化简后,将tanα+tanβ及tanαtanβ的值代入即可求出值.

解:∵tanα,tanβ是方程x2﹣3x+2=7的两个根,

∴tanα+tanβ=3,tanαtanβ=2,

故选:A.

4.(5分)已知α,β均为锐角,,cos(α+β)=()

A.B.C.D.

【分析】由同角的三角函数关系与两角和的余弦公式,计算即可.

解:α,β均为锐角,则α﹣∈(﹣,),β+∈(,),

sin(β+)==,

=cos(α﹣)cos(β+)﹣sin(α﹣)sin(β+)

=﹣.

故选:A.

5.(5分)若sinθ=,且90°<θ<180°,则cos的值为()A.B.C.D.

【分析】由已知求得cosθ,再由二倍角的余弦列式求解cos的值.

解:由已知得cosθ=,

又,45°<<90°,

故选:C.

6.(5分)已知角α,β均为锐角,且cosα=,sinβ=,则α﹣β的值为()A.B.C.D.

【分析】利用同角三角函数的基本关系求得sinα和cosβ的值,再利用两角差的正弦公式求得sin(α﹣β)的值,结合α﹣β∈(﹣,),可得α﹣β的值.

解:∵角α,β均为锐角,且cosα=,sinβ=,

∴sinα==,cosβ==,

再根据α﹣β∈(﹣,),可得α﹣β=﹣,

故选:C.

7.(5分)已知α∈(0,),2sin2α=cos2α+1,则sinα=()

A.B.C.D.

【分析】由二倍角的三角函数公式化简已知可得4sinαcosα=2cos2α,结合角的范围可求sinα>0,cosα>0,可得cosα=2sinα,根据同角三角函数基本关系式即可解得sinα的值.

解:∵2sin2α=cos2α+1,

∴可得:8sinαcosα=2cos2α,

∴cosα=2sinα,

∴解得:sinα=.

故选:B.

8.(5分)在△ABC中,若tan A tan B>1,则△ABC是()

A.锐角三角形B.直角三角形C.钝角三角形D.无法确定

【分析】利用两角和的正切函数公式表示出tan(A+B),根据A与B的范围以及tan A tan B >1,得到tan A和tan B都大于0,即可得到A与B都为锐角,然后判断出tan(A+B)小于0,得到A+B为钝角即C为锐角,所以得到此三角形为锐角三角形.

解:因为A和B都为三角形中的内角,

由tan A tan B>1,得到1﹣tan A tan B<0,

所以tan(A+B)=<0,

所以△ABC是锐角三角形.

故选:A.

9.(5分)使函数y=sin(2x+φ)+cos(2x+φ)为奇函数,且在[0,]上是减函数的φ的一个值为()

A.B.C.D.

【分析】利用两角和正弦公式化简函数的解析式为2sin(2x+θ+),由于它是奇函数,故θ+=kπ,k∈z,由此排除A,D;再逐一检验其它3个选项,可得结论

解:∵函数f(x)=sin(2x+θ)+cos(2x+θ)=2sin(2x+θ+)是奇函数,故θ+=kπ,k∈Z,θ=kπ﹣,故排除A,D.

在[0,]上,2x∈[2,],

若θ=,f(x)=2sin(2x+π)=﹣2sin4x是奇函数;

满足f(x)在[0,]上是减函数,故C满足条件.

故选:C.

10.(5分)中国古代数学家赵爽设计的弦图(如图1)是由四个全等的直角三角形拼成,四个全等的直角三角形也可拼成图2所示的菱形,已知弦图中,大正方形的面积为100,小正方形的面积为4,则图2中菱形的一个锐角的正弦值为()

A.B.C.D.

【分析】由题意,图2是四个全等的直角三角形拼成,只需求出图1中一个直角三角形的小锐角的正余弦值,利用二倍角即可求出图2中菱形的一个锐角的正弦值.

解:由题意,大正方形的面积为100,其边长为10,小正方形的面积为4,其边长为2.每个直角三角形的面积为.

可得:,

设小边所对的角为θ,则,,

故选:A.

11.(5分)已知,且cosθ>0,则()

A.tanθ<0B.

C.sin2θ>cos2θD.sin2θ>0

【分析】由同角三角函数的基本关系,求出cosθ及tanθ,进而得解.

解:∵,且cosθ>0,

∴,

故选:AB.

12.(5分)下列四个等式,其中正确的是()

A.tan25°+tan35°+tan 25°tan35°=

B.=

C.cos2﹣sin2=

D.﹣=4

【分析】由题意利用二倍角公式、两角和差的三角公式逐一判断各个选项是否正确,从而得出结论.

解:∵tan25°+tan35°+tan 25°tan35°=tan60°(1﹣tan 25°tan35°)+tan 25°tan35°

=(1﹣tan 25°tan35°)+tan 25°tan35°=,故A正确;

∵cos2﹣sin2=cos=,故C错误;

故选:ABD.

二、填空题(本题共4小题,每小题5分,共20分)

13.(5分)若=﹣,则cosα+sinα的值为.

【分析】由题意利用二倍角的余弦公式、两角和差的正弦公式,求得cosα+sinα的值.解:若==﹣(sinα+cosα)=﹣,

则cosα+sinα=,

故答案为:.

14.(5分)已知sinα+cosα=,其中α∈(,π),则tan2α=.【分析】由题意利用同角三角函数的基本关系求得sinα﹣cosα的值,可得sinα和cosα、tanα的值,再利用二倍角公式求得tan2α的值.

解:∵sinα+cosα=,∴1+2sinαcosα=,

∴sinαcosα=﹣<2,α为钝角.

求得sinα=﹣,cosα=,∴tan4α==,

故答案为:.

15.(5分)已知函数f(x)=sin x﹣cos(x+),则f()=,f(x)的值域为[﹣,].

【分析】把x=代入,展开两角和的余弦,则f()可求;把函数解析式展开两角和的余弦,整理后再由辅助角公式化积,则函数的值域可求.

解:∵f(x)=sin x﹣cos(x+),

∴f()=sin﹣cos()=﹣cos cos+sin sin

∵f(x)=sin x﹣cos x cos+sin x sin=

∵,∴,

∴f(x)的值域为[﹣,].

16.(5分)已知A是函数f(x)=sin(2020x+)+cos(2020x﹣)的最大值,若存在实数x1,x2,使得对任意实数x,总

有f(x1)≤f(x)≤f(x2)成立,则A?|x1﹣x2|的最小值为.

【分析】先根据正余弦的和差角公式及辅助角公式将函数解析式化简为f(x)=

,易得A=2,若存在实数x1,x2,使得对任意实数x,总有f(x1)≤f(x)≤f(x2)成立,则f(x1)和f(x_2)分别是函数f(x)的最小、最大值,进而求出|x1﹣x2|的最小值,从而得答案.

解:∵

∴,

∴f(x2)=f(x)max=2,f(x1)=f(x)min=﹣2,

故答案为:.

三、解答题(本题共4小题,共40分.解答应写出文字说明、证明过程或演算步骤)17.(8分)已知角α的顶点与原点O重合,始边与x轴的非负半轴重合,它的终边过点P (﹣,﹣).

(Ⅰ)求sin(α+π)的值;

(Ⅱ)若角β满足sin(α+β)=,求cosβ的值.

【分析】(Ⅰ)由已知条件即可求r,则sin(α+π)的值可得;

(Ⅱ)由已知条件即可求sinα,cosα,cos(α+β),再由cosβ=cos[(α+β)﹣α]=cos (α+β)cosα+sin(α+β)sinα代值计算得答案.

解:(Ⅰ)∵角α的顶点与原点O重合,始边与x轴非负半轴重合,终边过点P(﹣,﹣).

∴x=﹣,y=,r=|OP|=,

(Ⅱ)由x=﹣,y=,r=|OP|=1,

又由sin(α+β)=,

则cosβ=cos[(α+β)﹣α]=cos(α+β)cosα+sin(α+β)sinα=

∴cosβ的值为或.

18.(10分)已知α,β为锐角,tanα=,cos(α+β)=﹣.

(1)求cos2α的值;

(2)求tan(α﹣β)的值.

【分析】(1)由已知结合平方关系求得sinα,cosα的值,再由倍角公式得cos2α的值;

(2)由(1)求得tan2α,再由cos(α+β)=﹣求得tan(α+β),利用tan(α﹣β)=tan[2α﹣(α+β)],展开两角差的正切求解.

解:(1)由,解得,

∴cos6α=;

∵α,β∈(0,),∴α+β∈(7,π),

则tan(α+β)=.

∴tan(α﹣β)=tan[2α﹣(α+β)]==.

19.(10分)在△ABC中,A、B为锐角且B<A,sin A=,sin2B=.(1)求角C的值;

(2)求证:5cos A cos(A+3B)=2sin B.

【分析】(1)由已知sin A,sin2B可求cos A,cos2B,利用半角公式可求cos B,从而可得cos C=﹣cos(A+B),

(2)根据(1)的结论代入证明左边等于右边即可.

解:(1)∵A为锐角,sin A=

∴cos A==﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣(2分)

∴B<45°﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣(3分)

∴cos2B==

cos C=﹣cos(A+B)=﹣cos A cos B+sin A sin B=﹣=

(2)证明:左边=4cos A cos(π﹣C+2B)=﹣5cos A cos(C﹣2B)=﹣5cos A[cos C cos2B+sin C sin8B]=﹣5××(﹣×+)==5×=2sin B=右边

从而得证.

20.(12分)已知函数f(x)=cos(+x)cos(),g(x)=sin2x﹣.(1)求函数f(x)的最小正周期;

(2)求函数h(x)=f(x)﹣g(x)的最大值,并求使h(x)取得最大值的x的集合.【分析】(1)利用两角和与差的余弦公式将f(x)展开,化简得f(x)=cos2x﹣sin2x,再根据二倍角的余弦公式化简整理,即可得到f(x)=cos2x﹣,结合三角函数的周期公式即可算出函数f(x)的最小正周期;

(2)根据(1)中化简的结果,得h(x)=f(x)﹣g(x)=sin2x﹣cos2x,利用辅助角公式合并得h(x)=sin(2x﹣),再由三角函数的图象与性质,即可得到使h(x)取得最大值的x的集合.

解:(1)f(x)=cos(+x)cos()

=(cos cos x﹣sin sin x)(cos cos x+sin sin x)

∵cos2x=,sin2x=

因此,函数f(x)的最小正周期T==π;

∴h(x)=f(x)﹣g(x)=cos2x﹣﹣(sin2x﹣)=sin2x﹣cos2x

∴当2x﹣=+2kπ,即x=+kπ(k∈Z)时,sin2x﹣cos2x取得最大值为由此可得使h(x)取得最大值的x的集合为{x|x=+kπ,k∈Z}

附加题(本大题共2小题,每小题10分,共20分.解答应写出文字说明、证明过程或演算步骤)

21.(10分)某同学在一次研究性学习中发现,以下五个式子的值都等于同一个常数.(1)sin213°+cos217°﹣sin13°cos17°

(2)sin215°+cos215°﹣sin15°cos15°

(3)sin218°+cos212°﹣sin18°cos12°

(4)sin2(﹣18°)+cos248°﹣sin(﹣18°)cos48°

(5)sin2(﹣25°)+cos255°﹣sin(﹣25°)cos55°

(Ⅰ)试从上述五个式子中选择一个,求出这个常数;

(Ⅱ)根据(Ⅰ)的计算结果,将该同学的发现推广为三角恒等式,并证明你的结论.【分析】(Ⅰ)选择(2),由sin215°+cos215°﹣sin15°cos15°=1﹣sin30°=,可得这个常数的值.

(Ⅱ)推广,得到三角恒等式sin2α+cos2(30°﹣α)﹣sinαcos(30°﹣α)=.证明方法一:直接利用两角差的余弦公式代入等式的左边,化简可得结果.

证明方法二:利用半角公式及两角差的余弦公式把要求的式子化为+﹣sinα(cos30°cosα+sin30°sinα),即1﹣+cos2α+sin2α

﹣sin2α﹣,化简可得结果.

解:选择(2),计算如下:

sin215°+cos215°﹣sin15°cos15°=1﹣sin30°=,故这个常数为.

证明:(方法一)sin2α+cos3(30°﹣α)﹣sinαcos(30°﹣α)=

sin2α+﹣sinα(cos30°cosα+sin30°sinα)

(方法二)sin7α+cos2(30°﹣α)﹣sinαcos(30°﹣α)=+

﹣sinα(cos30°cosα+sin30°sinα)

=1﹣+cos5α+sin2α﹣sin5α﹣=1﹣﹣+

=.

22.(10分)已知函数f(x)=cos(ωx+φ)(ω>0,0<φ≤π)为奇函数,且其图象上相邻的一个最高点与一个最低点之间的距离为.

(1)求f(x)的解析式;

(2)若f(α+)=﹣(﹣<α<0),求sin(2α﹣)的值.

【分析】(1)由条件利用余弦函数的奇偶性求得φ,再利用余弦函数的图象特征求得ω的值,可得函数的解析式.

(2)由f(α+)=﹣,求得sin(α+)的值,可得cos(α﹣)的值,进而求得sin(α﹣)的值,从而利用二倍角公式求得sin(2α﹣)的值.

解:(1)由函数f(x)=cos(ωx+φ)(ω>0,0<φ≤π)为奇函数,可得φ=,f (x)=cos(ωx+)=﹣sinωx.

又其图象上相邻的一个最高点与一个最低点之间的距离为,可得

=,∴ω=4,f(x)=﹣sin x.

即=cos[﹣(α+)]=cos(﹣α)=cos(α﹣),∴sin(α﹣)=﹣

=﹣,

∴sin(2α﹣)=2 sin(α﹣)?cos(α﹣)=﹣.

2020年高一数学必修一集合单元测试题

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