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高二精选题库3-2. 数学 数学doc北师大版

第3模块 第2节

[知能演练]

一、选择题

1.α是第四象限角,tan α=-512

,则sin α等于 ( )

A.15

B .-15C.513

D .-513解析:?????

sin αcos α=-512,sin 2α+cos 2α=1,

∴??? sin α=

513,cos α=-1213或??? sin α=-513,cos α=1213. ∵α是第四象限角,∴sin α<0,cos α>0. ∴sin α=-

513

.选D. 答案:D 2.已知cos(π2+φ)=32,且|φ|<π2

,则tan φ等于 ( )

A .-33

B.33 C .- 3 D. 3

解析:由cos(π2φ)=32,得sin φ=-32

. 又|φ|<π2,∴cos φ=12

.∴tan φ=- 3. 答案:C

3.若α是第三象限角,且cos(75°+α)=13

,则tan(15°-α)的值为 ( )

A .-

223 B .-24 C.223

D.24

解析:cos(75°+α)=sin(90°-75°-α)=sin(15°-α)=13

,又∵α为第三象限角, ∴-α为第二象限角.

∴-α+15°为第二象限角.

∴cos(15°-α)=-

1-19=-223. ∴tan(15°-α)=-

24. 答案:B

4.若△ABC 的内角A 满足sin2A =23

,则sin A +cos A 等于 ( )

A.153

B .-153 C.53 D .-53

解析:在△ABC 中,2sin A cos A =23

>0, ∴sin A >0,cos A >0. ∴sin A +cos A =(sin A +cos A )2

=sin 2A +cos 2A +2sin A cos A =1+23

=53=153. 答案:A

二、填空题 5.如果cos α=15,且α是第四象限角,那么cos(α+π2

)=________. 解析:由已知?cos(α+π2)=-sin α=-(-1-cos 2α)=265

. 答案:265

6.化简:sin 2(α+π)·cos(π+α)·cos(-α-2π)tan(π+α)·sin 3(π2

+α)·sin(-α-2π)=

________.

解析:sin 2

(α+π)·cos(π+α)·cos(-α-2π)tan(π+α)·sin 3(π2

+α)·sin(-α-2π)

=(-sin α)2·(-cos α)·cos(-α)tan α·cos 3α·sin(-α)

=-sin 2α·cos α·cos αsin αcos α

·cos 3α·(-sin α)=sin 2αcos 2αsin 2αcos 2α=1. 答案:1

三、解答题

7.已知cos(π+α)=-12

,且α是第四象限角,计算: (1)sin(2π-α);

(2)sin[α+(2n +1)π]+sin[α-(2n +1)π]sin(α+2nπ)·cos(α-2nπ)

(n ∈Z). 解:∵cos(π+α)=-12,∴-cos α=-12,cos α=12

, 又∵α是第四象限角,

∴sin α=-1-cos 2α=-

32

. (1)sin(2π-α)=sin[2π+(-α)] =sin(-α)=-sin α=

32

. (2)sin[α+(2n +1)π]+sin[α-(2n +1)π]sin(α+2nπ)·cos(α-2nπ) =sin(2nπ+π+α)+sin(-2nπ-π+α)sin(2nπ+α)·cos(-2nπ+α)

=sin(π+α)+sin(-π+α)sin α·cos α

=-sin α-sin(π-α)sin α·cos α=-2sin αsin αcos α

=-2cos α

=-4. 8.已知sin(π-α)-cos(π+α)=

23(π2<α<π).求下列各式的值: (1)sin α-cos α;

(2)sin 3(π2-α)+cos 3(π2

+α). 解:由sin(π-α)-cos(π+α)=

23

, 得sin α+cos α=23.① 将①式两边平方,得1+2sin α·cos α=29

, 故2sin α·cos α=-79

, 又π2

<α<π,∴sin α>0,cos α<0. ∴sin α-cos α>0.

(1)(sin α-cos α)2=1-2sin α·cos α

=1-(-79)=169

∴sin α-cos α=43

. (2)sin 3(π2-α)+cos 3(π2

+α)=cos 3α-sin 3α =(cos α-sin α)(cos 2α+cos α·sin α+sin 2α)

=(-43)×(1-718=-2227

[高考·模拟·预测]

1.已知tan θ=2,则sin 2θ+sin θcos θ-2cos 2θ=

( )

A .-43

B.54 C .-34 D.45

解析:由于tan θ=2,则sin 2θ+sin θcos θ-2cos 2θ

=sin 2θ+sin θcos θ-2cos 2θsin 2θ+cos 2θ=tan 2θ+tan θ-2tan 2θ+1

=22

+2-222+1=45

,故选D. 答案:D

2.已知△ABC 中,1tan A =-125

,则cos A = ( )

A.1213

B.513 C .-

513 D .-1213

解析:∵1tan A =-125,∴tan A =-512,∴π2

=-1213,选D. 答案:D

3.下列关系式中正确的是 ( )

A .sin11°

B .sin168°

C .sin11°

D .sin168°

解析:注意到sin168°=sin(180°-12°)=sin12°,cos10°=sin80°,且0°<11°<12°<80°<90°,因此sin11°

答案:C

4.若sin θ=-45

,tan θ>0,则cos θ=________. 解析:∵sin θ<0,tan θ>0,θ在第三象限内,cos θ=-1-sin 2θ=-35

.

答案:-35

5.已知cos θ=-

23,θ∈(π2,π),求2sin2θ-cos θsin θ

的值. 解:原式=22sin θcos θ-cos θsin θ=1-cos 2θsin θcos θ=sin θcos θ. 又cos θ=-

23,θ∈(π2,π), ∴sin θ=

1-29=73,2sin2θ-cos θsin θ=-142. [备选精题]

6.已知函数f (x )=1-2sin(2x -π4)cos x

. (1)求f (x )的定义域;

(2)设α是第四象限的角,且tan α=-43

,求f (α)的值. 解:(1)由cos x ≠0得x ≠kπ+π2

(k ∈Z), 故f (x )的定义域为??????x ??

x ≠kπ+π2,k ∈Z . (2)因为tan α=-43

,且α是第四象限的角, 所以sin α=-45cos α=35

, 故f (α)=1-2sin(2α-π4)cos α

=1-2(22sin2α-22cos2α)cos α

=1-sin2α+cos2αcos α=2cos 2α-2sin αcos αcos α

=2(cos α-sin α)=145

.

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