课时作业(十七) 第17讲 同角三角函数的基本关系式与诱导公式

时间 / 30分钟 分值 / 80分

基础热身

1.sin(-750°)的值为 ( ) A.-√3 B.√32

2

C.-1

D.122 2.已知α是第四象限角,sin α=-17,则tan α= ( ) A.-√2 B.-√344 C.-√2√312 D.-12 3.已知θ是第三象限角,tan θ=3,则cos3π2

+θ= A.-√10√1010

B.-

310

C.-√10 D.-√255

4.√1?2sin(π+2)cos(π-2)= ( ) A.sin 2-cos 2 B.sin 2+cos 2 C.±(sin 2-cos 2) D.cos 2-sin 2

5.已知α∈[0,π],sin α+√3cos α=0,则α= . 能力提升

6.已知sin α=√5,则sin45α-cos4α的值为

( )

A.-1 B.-355 C.1 D.35

5

7.已知sin??+cos??

sin??-cos??=2,则

sin θcos θ= ( )

A.34 B.±3

10

( )

C. D.-

8.已知函数f(x)=asin(πx+α)+bcos(πx+β),且f(4)=6,则f(2020)的值为 ( ) A.6 B.-6 C.1 D.-1

9.已知2sin θ=1+cos θ,则tan θ= A.-或0 B.或0 C.- D. 10.已知

12

1+sin??1cos??

=-,则的值是 cos??2sin??-11243434343310

310

( )

( )

A. B.- C.2 D.-2

11.已知角θ的顶点在坐标原点,始边与x轴正半轴重合,终边在直线y=2x上,则

sin(2+??)+cos(π-??)sin(2-??)?sin(π-??)

π3π

= .

π-α4

12.[2018·兰州一诊] 若sin13.化简:

=-,则cos

25π+α4

= .

sin2(??+π)cos(π+??)cos(-??-2π)

= πtan(??+π)sin3(2+??)sin(???-2π) .

1

= tan2??

14.已知α为第二象限角,则cos α√1+tan2??+sin α√1+ .

难点突破

15.(5分)在△ABC中,√3sinA. B. C. D.

23π

6

π2

2π3π3

π4

π-A2

=3sin(π-A),且cos A=-√3cos(π-B),则C等于 ( )

16.(5分)设函数f(x)满足f(x+π)=f(x)+sin x(x∈R),当0≤x<π时,f(x)=0,则f

= .

课时作业(十七)

1.C [解析] sin(-750°)=sin(-720°-30°)=sin(-30°)=-sin 30°=-.故选C. 2.D [解析] 因为α是第四象限角,sin α=-,所以cos α=√1?sin2??=

3π

+θ2

π2

π2174√3,故712tan α=

sin??√3=-.故选cos??12D.

3.B [解析] cos=cos2π-+θ=cos-+θ=sin θ.由tan θ=3,得

910

3√10

,因为10

sin??sin2??

=3,所以2=9,即cos??cos??

3√10

.故选10

sin2θ=9-9sin2θ,即sin2θ=,得sin θ=±θ是第三象限角,所以sin θ<0,所以sin θ=-

B.

4.A [解析] √1?2sin(π+2)cos(π-2)=√1?2sin2cos2=√(sin2-cos2)2=|sin 2-cos 2|=sin 2-cos 2.故选A.

5. [解析] 由sin α+√3cos α=0,得cos α≠0,则tan α=-√3,因为α∈[0,π],所以α=. 6.B [解析] sin4α-cos4α=sin2α-cos2α=2sin2α-1=-.故选B.

7.C [解析] 由条件,得sin θ+cos θ=2sin θ-2cos θ,即3cos θ=sin θ,所以tan θ=3,所以sin θcos θ=

sin??cos??tan??3

=2=.故选22sin??+cos??tan??+1103

5

2π32π3C.

8.A [解析] 因为f(4)=asin(4π+α)+bcos(4π+β)=asin α+bcos β=6,所以f(2020)=asin(2020π+α)+bcos(2020π+β)=asin α+bcos β=6.故选A.

9.B [解析] 将2sin θ=1+cos θ两边平方并整理,得5cos2θ+2cos θ-3=0,解得cos θ=-1或cos θ=.当cos θ=-1时,θ=2kπ+π,k∈Z,得tan θ=0;当cos θ=时,sin θ=(1+cos θ)=,得tan θ=.故选B. 10.A [解析] 因为1-sin2α=cos2α,cos α≠0,1-sin α≠0,所以(1+sin α)(1-sin α)=cos αcos α,所以

1+sin??cos??cos??1cos??1

=,所以=-,即=.故选cos??1?sin??1?sin??2sin??-123

5

12

45

43

35

A.

-cos??-cos??-2

==2.

cos??-sin??1?tan??π-α4

25

11.2 [解析] 由题意可得tan θ=2,所以原式=12.- [解析] cos

25

π+α4

ππ-+α24

=sin=sin=-.

13.1 [解析] 原式=

sin2??(-cos??)cos??sin2??cos2??

==1.

tan??cos3??(-sin??)sin2??cos2??

2??+cos2??

sin

14.0 [解析] 原式=cos α√

cos2??

+sin α√

sin2??+cos2??cos??sin??

=+,因为

sin2??|cos??||sin??|

α是第二象限角,所以sin

α>0,cos α<0,所以

cos??sin??

+=-1+1=0. |cos??||sin??|