高中数学湘教版(2019)必修 第一册第5章 三角函数5.3 三角函数的图象与性质练习题
展开1.已知sineq \b\lc\(\rc\)(\a\vs4\al\c1(\f(5π,2)+α))=eq \f(1,5),那么cs α等于( )
A.-eq \f(2,5) B.-eq \f(1,5)
C.eq \f(1,5) D.eq \f(2,5)
解析:选C sineq \b\lc\(\rc\)(\a\vs4\al\c1(\f(5π,2)+α))=sineq \b\lc\(\rc\)(\a\vs4\al\c1(2π+\f(π,2)+α))
=sineq \b\lc\(\rc\)(\a\vs4\al\c1(\f(π,2)+α))=cs α=eq \f(1,5).
2.化简sineq \b\lc\(\rc\)(\a\vs4\al\c1(α+\f(π,2)))·cseq \b\lc\(\rc\)(\a\vs4\al\c1(α-\f(3π,2)))·taneq \b\lc\(\rc\)(\a\vs4\al\c1(\f(π,2)-α))的结果是( )
A.1 B.sin2α
C.-cs2α D.-1
解析:选C 因为sineq \b\lc\(\rc\)(\a\vs4\al\c1(α+\f(π,2)))=cs α,cseq \b\lc\(\rc\)(\a\vs4\al\c1(α-\f(3π,2)))=cseq \b\lc\[\rc\](\a\vs4\al\c1(π+\b\lc\(\rc\)(\a\vs4\al\c1(\f(π,2)-α))))=-sin α,taneq \b\lc\(\rc\)(\a\vs4\al\c1(\f(π,2)-α))=eq \f(sin\b\lc\(\rc\)(\a\vs4\al\c1(\f(π,2)-α)),cs\b\lc\(\rc\)(\a\vs4\al\c1(\f(π,2)-α)))=eq \f(cs α,sin α),所以原式=cs α(-sin α)eq \f(cs α,sin α)=-cs2α,故选C.
3.已知tan θ=2,则eq \f(sin\b\lc\(\rc\)(\a\vs4\al\c1(\f(π,2)+θ))-cs\b\lc\(\rc\)(\a\vs4\al\c1(π-θ)),cs\b\lc\(\rc\)(\a\vs4\al\c1(\f(π,2)-θ))+sin\b\lc\(\rc\)(\a\vs4\al\c1(π-θ)))=( )
A.2 B.-2
C.0 D.eq \f(1,2)
解析:选D ∵eq \f(sin\b\lc\(\rc\)(\a\vs4\al\c1(\f(π,2)+θ))-cs(π-θ),cs\b\lc\(\rc\)(\a\vs4\al\c1(\f(π,2)-θ))+sin(π-θ))=eq \f(cs θ+cs θ,sin θ+sin θ)=eq \f(1,tan θ),又∵tan θ=2,∴原式=eq \f(1,2),故选D.
4.已知α是第四象限角,且3sin2α=8cs α,则cseq \b\lc\(\rc\)(\a\vs4\al\c1(α+\f(2 021π,2)))=( )
A.-eq \f(2\r(2),3) B.-eq \f(1,3)
C.eq \f(2\r(2),3) D.eq \f(1,3)
解析:选C ∵3sin2α=8cs α,∴sin2α+eq \b\lc\(\rc\)(\a\vs4\al\c1(\f(3sin2α,8)))eq \s\up12(2)=
1,整理可得9sin4α+64sin2α-64=0,
解得sin2α=eq \f(8,9)或sin2α=-8(舍去).
又∵α是第四象限角,
∴sin α=-eq \f(2\r(2),3),
∴cseq \b\lc\(\rc\)(\a\vs4\al\c1(α+\f(2 021π,2)))=cseq \b\lc\(\rc\)(\a\vs4\al\c1(α+1 010π+\f(π,2)))
=cseq \b\lc\(\rc\)(\a\vs4\al\c1(α+\f(π,2)))=-sin α=eq \f(2\r(2),3).
5.若角A,B,C是△ABC的三个内角,则下列等式中一定成立的是( )
A.cs(A+B)=cs C B.sin(A+B)=-sin C
C.cseq \f(A+C,2)=sin B D.sineq \f(B+C,2)=cseq \f(A,2)
解析:选D ∵A+B+C=π,∴A+B=π-C,
∴cs(A+B)=-cs C,sin(A+B)=sin C,故A、B错.
∵A+C=π-B,∴eq \f(A+C,2)=eq \f(π-B,2),
∴cseq \f(A+C,2)=cseq \b\lc\(\rc\)(\a\vs4\al\c1(\f(π,2)-\f(B,2)))=sineq \f(B,2),故C错.
∵B+C=π-A,∴sineq \f(B+C,2)=sineq \b\lc\(\rc\)(\a\vs4\al\c1(\f(π,2)-\f(A,2)))=cseq \f(A,2),故D正确.
6.已知α∈eq \b\lc\(\rc\)(\a\vs4\al\c1(0,\f(3π,2))),cseq \b\lc\(\rc\)(\a\vs4\al\c1(\f(3π,2)-α))=eq \f(\r(3),2),则tan(2 020π-α)=________.
解析:由cseq \b\lc\(\rc\)(\a\vs4\al\c1(\f(3π,2)-α))=eq \f(\r(3),2)得sin α=-eq \f(\r(3),2),
又0<α
所以tan(2 020π-α)=tan(-α)=-tan α=-eq \r(3).
答案:-eq \r(3)
7.sin2eq \b\lc\(\rc\)(\a\vs4\al\c1(\f(π,3)-x))+sin2eq \b\lc\(\rc\)(\a\vs4\al\c1(\f(π,6)+x))=________.
解析:sin2eq \b\lc\(\rc\)(\a\vs4\al\c1(\f(π,3)-x))+sin2eq \b\lc\(\rc\)(\a\vs4\al\c1(\f(π,6)+x))=sin2eq \b\lc\(\rc\)(\a\vs4\al\c1(\f(π,3)-x))+sin2eq \b\lc\[\rc\](\a\vs4\al\c1(\f(π,2)-\b\lc\(\rc\)(\a\vs4\al\c1(\f(π,3)-x))))=sin2eq \b\lc\(\rc\)(\a\vs4\al\c1(\f(π,3)-x))+cs2eq \b\lc\(\rc\)(\a\vs4\al\c1(\f(π,3)-x))=1.
答案:1
8.化简eq \f(cs\b\lc\(\rc\)(\a\vs4\al\c1(α-\f(π,2))),sin\b\lc\(\rc\)(\a\vs4\al\c1(\f(5π,2)+α)))·sin(α-π)·taneq \b\lc\(\rc\)(\a\vs4\al\c1(\f(3π,2)-α))的结果为________.
解析:原式=eq \f(cs\b\lc\(\rc\)(\a\vs4\al\c1(\f(π,2)-α)),sin\b\lc\(\rc\)(\a\vs4\al\c1(2π+\f(π,2)+α)))·(-sin α)·eq \f(1,tan α)=eq \f(sin α,sin\b\lc\(\rc\)(\a\vs4\al\c1(\f(π,2)+α)))·(-sin α)·eq \f(cs α,sin α)=eq \f(sin α,cs α)·(-sin α)·eq \f(cs α,sin α)=-sin α.
答案:-sin α
9.在①角α的终边经过点P(4m,-3m)(m≠0);②taneq \b\lc\(\rc\)(\a\vs4\al\c1(\f(π,2)-α))=eq \f(3,4);③3sin α+4cs α=0,这三个条件中任选一个,求sin2α-sin αcs α-2cs2α的值.
注:如果选择多个条件分别解答,按第一个解答计分.
解:sin2α-sin αcs α-2cs2α
=eq \f(sin2α-sin αcs α-2cs2α,sin2α+cs2α)
=eq \f(tan2α-tan α-2,tan2α+1),
若选①:角α的终边经过点P(4m,-3m)(m≠0);
可得tan α=eq \f(-3m,4m)=-eq \f(3,4),
原式=eq \f(\b\lc\(\rc\)(\a\vs4\al\c1(-\f(3,4)))\s\up12(2)+\f(3,4)-2,\b\lc\(\rc\)(\a\vs4\al\c1(-\f(3,4)))\s\up12(2)+1)=-eq \f(11,25).
若选②:taneq \b\lc\(\rc\)(\a\vs4\al\c1(\f(π,2)-α))=eq \f(3,4),可得tan α=eq \f(4,3),
原式=eq \f(\b\lc\(\rc\)(\a\vs4\al\c1(\f(4,3)))\s\up12(2)-\f(4,3)-2,\b\lc\(\rc\)(\a\vs4\al\c1(\f(4,3)))\s\up12(2)+1)=-eq \f(14,25).
若选③:3sin α+4cs α=0,tan α=-eq \f(4,3),
原式=eq \f(\b\lc\(\rc\)(\a\vs4\al\c1(-\f(4,3)))\s\up12(2)+\f(4,3)-2,\b\lc\(\rc\)(\a\vs4\al\c1(-\f(4,3)))\s\up12(2)+1)=eq \f(2,5).
10.化简:(1)eq \f(sin\b\lc\(\rc\)(\a\vs4\al\c1(\f(π,2)+α))cs\b\lc\(\rc\)(\a\vs4\al\c1(\f(π,2)-α)),cs(π+α))+eq \f(sin(π-α)cs\b\lc\(\rc\)(\a\vs4\al\c1(\f(π,2)+α)),sin(π+α));
(2)eq \f(tan(3π-α),sin(π-α)sin\b\lc\(\rc\)(\a\vs4\al\c1(\f(3π,2)-α)))+eq \f(sin(2π-α)cs\b\lc\(\rc\)(\a\vs4\al\c1(α-\f(7π,2))),sin\b\lc\(\rc\)(\a\vs4\al\c1(\f(3π,2)+α))cs(2π+α)).
解:(1)∵sineq \b\lc\(\rc\)(\a\vs4\al\c1(\f(π,2)+α))=cs α,cseq \b\lc\(\rc\)(\a\vs4\al\c1(\f(π,2)-α))=sin α,
cs(π+α)=-cs α,sin(π-α)=sin α,
cseq \b\lc\(\rc\)(\a\vs4\al\c1(\f(π,2)+α))=-sin α,sin(π+α)=-sin α,
∴原式=eq \f(cs α·sin α,-cs α)+eq \f(sin α·(-sin α),-sin α)=-sin α+sin α=0.
(2)∵tan(3π-α)=-tan α,
sin(π-α)=sin α,
sineq \b\lc\(\rc\)(\a\vs4\al\c1(\f(3π,2)-α))=-cs α,sin(2π-α)=-sin α,
cseq \b\lc\(\rc\)(\a\vs4\al\c1(α-\f(7π,2)))=cseq \b\lc\(\rc\)(\a\vs4\al\c1(\f(7π,2)-α))
=cseq \b\lc\(\rc\)(\a\vs4\al\c1(4π-\f(π,2)-α))=cseq \b\lc\(\rc\)(\a\vs4\al\c1(\f(π,2)+α))=-sin α,
sineq \b\lc\(\rc\)(\a\vs4\al\c1(\f(3π,2)+α))=-cs α,cs(2π+α)=cs α,
∴原式=eq \f(-tan α,sin α(-cs α))+eq \f((-sin α)(-sin α),(-cs α)·cs α)
=eq \f(1,cs2α)-eq \f(sin2α,cs2α)=eq \f(1-sin2α,cs2α)
=eq \f(cs2α,cs2α)=1.
[B级 综合运用]
11.如果f(sin x)=cs 2x,那么f(cs x)的值为( )
A.-sin 2x B.sin 2x
C.-cs 2x D.cs 2x
解析:选C f(cs x)=feq \b\lc\[\rc\](\a\vs4\al\c1(sin\b\lc\(\rc\)(\a\vs4\al\c1(\f(π,2)-x))))=cs 2eq \b\lc\(\rc\)(\a\vs4\al\c1(\f(π,2)-x))=cs(π-2x)=-cs 2x.
12.计算sin21°+sin22°+sin23°+…+sin289°=( )
A.89 B.90
C.eq \f(89,2) D.45
解析:选C ∵sin21°+sin289°=sin21°+cs21°=1,sin22°+sin288°=sin22°+cs22°=1,…,∴sin21°+sin22°+sin23°+…+sin289°=sin21°+sin22°+sin23°+…+sin244°+sin245°+cs244°+cs243°+…+cs23°+cs22°+cs21°=44+eq \f(1,2)=eq \f(89,2).
13.已知sineq \b\lc\(\rc\)(\a\vs4\al\c1(α+\f(π,6)))=eq \f(1,3),则sineq \b\lc\(\rc\)(\a\vs4\al\c1(\f(5π,6)-α))+sin2eq \b\lc\(\rc\)(\a\vs4\al\c1(\f(π,3)-α))的值为________.
解析:因为eq \b\lc\(\rc\)(\a\vs4\al\c1(α+\f(π,6)))+eq \b\lc\(\rc\)(\a\vs4\al\c1(\f(5π,6)-α))=π,eq \b\lc\(\rc\)(\a\vs4\al\c1(α+\f(π,6)))+eq \b\lc\(\rc\)(\a\vs4\al\c1(\f(π,3)-α))=eq \f(π,2),所以sineq \b\lc\(\rc\)(\a\vs4\al\c1(\f(5π,6)-α))=sineq \b\lc\[\rc\](\a\vs4\al\c1(π-\b\lc\(\rc\)(\a\vs4\al\c1(α+\f(π,6)))))=sineq \b\lc\(\rc\)(\a\vs4\al\c1(α+\f(π,6)))=eq \f(1,3),cseq \b\lc\(\rc\)(\a\vs4\al\c1(\f(π,3)-α))=cseq \b\lc\[\rc\](\a\vs4\al\c1(\f(π,2)-\b\lc\(\rc\)(\a\vs4\al\c1(α+\f(π,6)))))=sineq \b\lc\(\rc\)(\a\vs4\al\c1(α+\f(π,6)))=eq \f(1,3),
所以sineq \b\lc\(\rc\)(\a\vs4\al\c1(\f(5π,6)-α))+sin2eq \b\lc\(\rc\)(\a\vs4\al\c1(\f(π,3)-α))
=eq \f(1,3)+1-cs2eq \b\lc\(\rc\)(\a\vs4\al\c1(\f(π,3)-α))=eq \f(4,3)-eq \b\lc\(\rc\)(\a\vs4\al\c1(\f(1,3)))eq \s\up12(2)=eq \f(11,9).
答案: eq \f(11,9)
14.是否存在角α,β,α∈eq \b\lc\(\rc\)(\a\vs4\al\c1(-\f(π,2),\f(π,2))),β∈(0,π),使等式sin(3π-α)=eq \r(2)cseq \b\lc\(\rc\)(\a\vs4\al\c1(\f(π,2)-β)),eq \r(3)cs(-α)=-eq \r(2)cs(π+β)同时成立?若存在,求出α,β的值;若不存在,请说明理由.
解:假设存在角α,β满足条件,
则由题可得eq \b\lc\{(\a\vs4\al\c1(sin α=\r(2)sin β, ①,\r(3)cs α=\r(2)cs β, ②))
①2+②2,得sin2α+3cs2α=2.
∴cs2α=eq \f(1,2),∴cs α=±eq \f(\r(2),2).
∵α∈eq \b\lc\(\rc\)(\a\vs4\al\c1(-\f(π,2),\f(π,2))),∴cs α=eq \f(\r(2),2).
由cs α=eq \f(\r(2),2),eq \r(3)cs α=eq \r(2)cs β,得cs β=eq \f(\r(3),2).
∵β∈(0,π),∴β=eq \f(π,6).
∴sin β=eq \f(1,2),结合①可知sin α=eq \f(\r(2),2),则α=eq \f(π,4).
故存在α=eq \f(π,4),β=eq \f(π,6)满足条件.
[C级 拓展探究]
15.已知A,B,C为△ABC的内角.
(1)求证:cs2eq \f(A+B,2)+cs2eq \f(C,2)=1;
(2)若cseq \b\lc\(\rc\)(\a\vs4\al\c1(\f(π,2)+A))sineq \b\lc\(\rc\)(\a\vs4\al\c1(\f(3π,2)+B))tan(C-π)<0,求证:△ABC为钝角三角形.
证明:(1)∵在△ABC中,A+B=π-C,∴eq \f(A+B,2)=eq \f(π,2)-eq \f(C,2),
∴cseq \f(A+B,2)=cseq \b\lc\(\rc\)(\a\vs4\al\c1(\f(π,2)-\f(C,2)))=sineq \f(C,2),
∴cs2eq \f(A+B,2)+cs2eq \f(C,2)=sin2eq \f(C,2)+cs2eq \f(C,2)=1.
(2)∵cseq \b\lc\(\rc\)(\a\vs4\al\c1(\f(π,2)+A))sineq \b\lc\(\rc\)(\a\vs4\al\c1(\f(3π,2)+B))tan(C-π)<0,
∴-sin A·(-cs B)·tan C<0,即sin Acs Btan C<0.
又A,B,C∈(0,π),∴sin A>0,∴cs Btan C<0,
即cs B<0,tan C>0或tan C<0,cs B>0,
∴B为钝角或C为钝角,∴△ABC为钝角三角形.
数学5.3 诱导公式随堂练习题: 这是一份数学5.3 诱导公式随堂练习题,共14页。试卷主要包含了单选题,多选题,填空题,计算题,解答题等内容,欢迎下载使用。
人教A版 (2019)必修 第一册5.3 诱导公式课后练习题: 这是一份人教A版 (2019)必修 第一册5.3 诱导公式课后练习题,共4页。
数学必修 第一册5.3 三角函数的图象与性质练习题: 这是一份数学必修 第一册5.3 三角函数的图象与性质练习题,共6页。