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??|