发布时间 : 星期一 文章高中数学人教A版高一必修四课下能力提升:(七)更新完毕开始阅读7e76423141323968011ca300a6c30c225901f0d0
课下能力提升(七) [学业水平达标练]
题组1 化简求值
1.下列与sin??θ-π
2??的值相等的式子为( )
A.sin?ππ
?2+θ??B.cos??2+θ??
C.cos?
3π3?2-θ??D.sin?π
?2+θ??
2.化简:sin(-α-7π)·cos?3π?α-2??=________.
3.化简:1
1
tan2(-α)+.
sin?π?2-α??·cos?3π?α-2??·tan(π+α)
题组2 条件求值问题
sin?π
+θ?-cos(π4.已知tan θ=2,则?2?
-θ)sin?π等于( )
?2-θ??-sin(π-θ)A.2 B.-2 C.0 D.2
3
5.若sin(π+α)+cos?π?2+α?3π
?=-m,则cos??2-α??+2sin(2π-α)的值为( )
A.-23mB.2
3m
C.-332mD.2
m
6.已知cos(60°+α)=1
3,且-180°<α<-90°,则cos(30°-α)的值为( )
A.-223B.223
C.-
23D.23
7.已知α是第三象限角,且cos(85°+α)=4
5
,则sin(α-95°)=________.
8.已知sin α是方程3x2-10x-8=0的根,且α为第三象限角,sin??α+3π2??·sin?3π
?2-α??
·
tan2(2π-α)·tan(π-α)cos?π?2-α??·cos?π?2+α?的值.
?
题组3 三角恒等式的证明
求3π
tan(2π-α)cos?-α?cos(6π-α)
?2?
9.求证:=1.
3π3π
tan(π-α)sin?α+?cos?α+?2??2??cos(π-θ)
10.求证:+
3π
cos θ?sin?-θ?-1???2??cos(2π-θ)2
=2.
π3πsinθ
cos(π+θ)sin?+θ?-sin?+θ??2??2?
[能力提升综合练]
π1
1.如果cos(π+A)=-,那么sin?+A?等于( )
2?2?11
A.-B.
22C.-
33D. 22
ππ1
2.已知sin?α+?=,α∈?-,0?,则tan α的值为( )
2?3??2?A.-22B.22 C.-
22
D. 44
1
3.已知sin(75°+α)=,则cos(15°-α)的值为( )
311A.-B.
332222C.-D.
33
4.在△ABC中,下列各表达式为常数的是( ) A.sin(A+B)+sin C B.cos(B+C)-cos A A+BA+BCC
C.sin2+sin2 D.sinsin
2222
5.sin21°+sin22°+sin23°+…+sin289°=________. 6.已知tan(3π+α)=2,
ππ
sin(α-3π)+cos(π-α)+sin?-α?-2cos?+α??2??2?则=________.
-sin(-α)+cos(π+α)
3π3π
sin?-α-?cos?-α?2??2??2
7.已知sin α是方程5x2-7x-6=0的根,且α是第三象限角,求·tan(π-
ππ????cos-αsin+α?2??2?α)的值.
πππ
8.是否存在角α,β,α∈?-,?,β∈(0,π),使等式sin(3π-α)=2cos?-β?,3cos(-α)
?22??2?=-2cos(π+β)同时成立?若存在,求出α,β的值;若不存在,请说明理由.
答 案
[学业水平达标练]
ππ
1. 解析:选D 因为sin?θ-?=-sin?-θ?=-cos θ,
2???2?对于A,sin?
π?
?2+θ?=cos θ;
π
对于B,cos?+θ?=-sin θ;
?2?对于C,cos?=-cos?
3π?π
-θ=cos?π+?-θ?? ?2???2??
π?
?2-θ?=-sin θ;
3π?π
+θ=sin?π+?+θ?? ?2???2??
对于D,sin?
π
=-sin?+θ?=-cos θ.
?2?2. 解析:原式=-sin(7π+α)·cos?
3π??-cos?π-α??=sin α·(-sin α)=-sin2
-α=-sin(π+α)·?2???2??
α.
答案:-sin2α
π
3. 解:∵tan(-α)=-tan α,sin?-α?=cos α,
?2?3π3π
cos?α-?=cos?-α?=-sin α,
2???2?tan(π+α)=tan α,
cos2α-1sin2α1111
∴原式=2+=+==-2=-1.
tanαcos α·(-sin α)·tan αsin2α-sin2αsin2αsinαcos2αcos θ+cos θ2cos θ2
4. 解析:选B 原式====-2.
cos θ-sin θcos θ-sin θ1-tan θπm
5. 解析:选C ∵sin(π+α)+cos?+α?=-sin α-sin α=-m,∴sin α=.
2?2?∴cos?
3πm3
-α?+2sin(2π-α)=-sin α-2sin α=-3sin α=-3×2=-2m. ?2?
1
6. 解析:选A 由-180°<α<-90°,得-120°<60°+α<-30°,又cos(60°+α)=>0,所
3以-90°<60°+α<-30°,即-150°<α<-90°,所以120°<30°-α<180°,cos(30°-α)<0,
所以cos(30°-α)=sin(60°+α)=-1-cos2(60°+α)= -
1?221-?=-. ?3?3
4
7. 解析:由α是第三象限角,cos(85°+α)=>0,
5知85°+α是第四象限角, 3
∴sin(85°+α)=-,
5
3
sin(α-95°)=sin[(85°+α)-180°]=-sin[180°-(85°+α)]=-sin(85°+α)=. 53答案: 5
2
8. 解:∵方程3x2-10x-8=0的两根为x1=4或x2=-,
32
又∵-1≤sin α≤1,∴sin α=-.
3又∵α为第三象限角, ∴cos α=-1-sin2α=-
525,tan α=. 35
2
(-cos α)·(-cos α)·tan2α·(-tan α)
∴原式= sin α·(-sin α)25=tan α=.
5
π
-α??cos(-α)
??2??
9. 证明:左边= π???π?????(-tan α)-sin+α-cos+α
??2????2??
tan(-α)?-cos?
=
(-tan α)(-sin α)cos α=1=右边.∴原式成立.
(-tan α)(-cos α)sin α-cos θcos θ11
+=+=
cos θ(-cos θ-1)-cos θcos θ+cos θ1+cos θ1-cos θ10. 证明:左边=
1-cos θ+1+cos θ22
==右边.∴原式成立. 2=(1+cos θ)(1-cos θ)1-cosθsin2θ[能力提升综合练]
1
1. 解析:选B cos(π+A)=-cos A=-,
21
∴cos A=,
2∴sin?
π1+A?=cos A=2. ?2?
12. 解析:选A 由已知得,cos α=,
3