高考理科数学试题汇编三角函数大题

2015年高考理科数学试题汇编(含答案)：三角函数大题(江苏)15.(本小题满分14分),在中，已知.,ABCAB,2,AC,3,A,60(1)求的长;BC(2)求的值.sin2C

43【答案】

(1)(2)77

【剖析】

考点:余弦定理，二倍角公式

fxx,,，sin,,,(10)(安徽)已知函数

(，，均为正的常数

)的最小正

,,，，，，

2,fx,,周期为，当

x时，函数获取最小值，则以下结论正确的选项是

()

，，3

fff220,,,fff022,,,(A)(B)，，，，，，，，，，，，fff,,,202fff202,,,(C)(D)，，，，，，，，，，，，【答案】Aoftheaudit.Fireisacombinationofauditinginthefieldoflawenforcementlawenforcementjobrotationandfirepracticeoflawenforcement,inthepromotionchanged,retrainingretiredkeytimingsynchronizationinplaceauditmechanism,andconducivetoself-urgedself-restraint,thefirelawenforcement,toprotectthemselves.1.3preventivefire-fightingurgedthecurrentfirelawenforcementcorruptioninvolvingcasesofviolatinglawenforcementhasanupwardtrend,judgingfromnationalreportsofcomplaintsinrecentyears,reportedcasesreflectthefirelawenforcementisover50%.Inaddition,themultiplecorruptioncases,withahighincidenceofleadingcadres,asmallnumberofleadingcadres'corruption,corrupt,power-for-moneytransaction,especiallyteams,groupreportingincreasingcomplaintsinvolvingthesupervisionoflawenforcementjobs,bringgreatpressuretoforcehonestandseriouschallenges.Meanwhile,teamandBrigadelevelsbearanadministrativelicenseapproval,administrativepunishment,routineinspectionsandmorethan95%oftheamount,somefirestaffhandledayeargothroughlegalinstrumentsandthousandsofcontacthundredsofsocialunits,onceinsidelax,theirqualityisnothigh,extremelyeasytoinducedisciplineandcorruption.Inpracticaloperation,thefirelawenforcementas'athletes'and'umpire'internalinspector,inspection,testandnotenoughproblemtouchesnotfoundtoleadingpolicymakers,deviatefromsystemdesign,systemimplementationhastodealwith.2buildingfirelaw1/6

页考点

:1.三角函数的图象与应用

;2.函数值的大小比较

.

(福建

)19(已知函数的图像是由函数的图像经以下变换获取

:先将

f()xgxx()cos=

图像上所有点的纵坐标伸长到原来的2倍(横坐标不变)，再将所获取的图像向右平

gx()

p移个单位长度

.2

(?)求函数的剖析式，并求其图像的对称轴方程

;f()x

(?)已知对于的方程在内有两个不同样的解

(xf()g()xxm+=[0,2)pab,

(1)求实数

m的取值范围

;

22mcos)1.(ab-=-(2)证明:5

pxk=+p(kZ).【答案】

(?)

，;(?)(1);(2)

详看法

f()2sinxx=(5,5)-2

析(

【剖析】

试题剖析

:(?)纵向伸缩或平移

:gxkgx()(),或

gxgxk()(),

，;横向伸缩或平移

:

1a,0gxgx()(),,(

纵坐标不变，横坐标变为原来的倍)，gxgxa()(),，(时，向,a,0aa左平移f()2sinxx=个单位;时，向右平移个单位);(?)(1)由(?)得，则oftheaudit.Fireisacombinationofauditinginthefieldoflawenforcementlawenforcementjobrotationandfirepracticeoflawenforcement,inthepromotionchanged,retrainingretiredkeytimingsynchronizationinplaceauditmechanism,andconducivetoself-urgedself-restraint,thefirelawenforcement,toprotectthemselves.1.3preventivefire-fightingurgedthecurrentfirelawenforcementcorruptioninvolvingcasesofviolatinglawenforcementhasanupwardtrend,judgingfromnationalreportsofcomplaintsinrecentyears,reportedcasesreflectthefirelawenforcementisover50%.Inaddition,themultiplecorruptioncases,withahighincidenceofleadingcadres,asmallnumberofleadingcadres'corruption,corrupt,power-for-moneytransaction,especiallyteams,groupreportingincreasingcomplaintsinvolvingthesupervisionoflawenforcementjobs,bringgreatpressuretoforcehonestandseriouschallenges.Meanwhile,teamandBrigadelevelsbearanadministrativelicenseapproval,administrativepunishment,routineinspectionsandmorethan95%oftheamount,somefire12)，方程在内有staffhandledayeargothroughlegalinstrumentsandthousandsofcontacthundredsofsocialunits,onceinsidelax,theirqualityisnothigh,extremelyeasytoinducedisciplineandcorruption.Inpracticaloperation,thefirelawenforcementas'athletes'and'umpire'internalinspector,inspection,testandnotenoughproblemtouchesnotfoundtoleadingpolicymakers,deviatefromsystemdesign,systemimplementationhastodealwith.2buildingfirelaw2/6页，利用协助角公式变形为(其中f()g()2sincosxxxx+=+f()g()xx+=+5sin()xj两个不同样的解，等价f()g()xxm+=[0,2)pab,sin,cosjj==55于直线和函数有两个不同样交点，数形联合求实数m的取值范围;ym,yx=+5sin()j

p3p(2)联合图像可得和，

进而利用引诱公式联合已知条件

abj+=2()-abj+=2()-22

求解

(

试题剖析

:解法一

:(1)将的图像上所有点的纵坐标伸长到原来的

2倍(横坐

gxx()cos=

p标不变

)获取的图像，再将的图像向右平移个单位长度后获取

y2cos=xy2cos=x2

py2cos()=-x

的图像，故，进而函数图像的对称轴方程为f()2sinxx=f()2sinxx=2pxk=+p(kZ).221(2)1)f()g()2sincos5(sincos)xxxxxx+=+=+5512(其中)sin,cosjj===+5sin()xj

55mm

依题意，在区间内有两个不同样的解当且仅当，故

msin()=x+j||1[0,2)pab,

55的取值范围是

.(5,5)-

2)因为是方程在区间内有两个不同样的解，

ab,[0,2)p5sin()=mx+j

mm因此，

.sin()=aj+sin()=bj+55p1m?abjabpbj+=2(),2();--=-+3p-5abjabpbj+=2(),32();--=-+当时,

当时，2

22mm222

因此cos)cos2()2sin()12()11.(abbjbj-=-+=+-=-=-55解法二:(1)同解法一.(2)1)同解法一.oftheaudit.Fireisacombinationofauditinginthefieldoflawenforcementlawenforcementjobrotationandfirepracticeoflawenforcement,inthepromotionchanged,retrainingretiredkeytimingsynchronizationinplaceauditmechanism,andconducivetoself-urgedself-restraint,thefirelawenforcement,toprotectthemselves.1.3preventivefire-fightingurgedthecurrentfirelawenforcementcorruptioninvolvingcasesofviolatinglawenforcementhasanupwardtrend,judgingfromnationalreportsofcomplaintsinrecentyears,reportedcasesreflectthefirelawenforcementisover50%.Inaddition,themultiplecorruptioncases,withahighincidenceofleadingcadres,asmallnumberofleadingcadres'corruption,corrupt,power-for-moneytransaction,especiallyteams,groupreportingincreasingcomplaintsinvolvingthesupervisionoflawenforcementjobs,bringgreatpressuretoforcehonestandseriouschallenges.Meanwhile,teamandBrigadelevelsbearanadministrativelicenseapproval,administrativepunishment,routineinspectionsandmorethan95%oftheamount,somefirestaffhandledayeargothroughlegalinstrumentsandthousandsofcontacthundredsofsocialunits,onceinsidelax,theirqualityisnothigh,extremelyeasytoinducedisciplineandcorruption.Inpracticaloperation,thefirelawenforcementas'athletes'and'umpire'internalinspector,inspection,testandnotenoughproblemtouchesnotfoundtoleadingpolicymakers,deviatefromsystemdesign,systemimplementationhastodealwith.2buildingfirelaw3/6页2)因为是方程在区间内有两个不同样的解，

ab,[0,2)p5sin()=mx+j

mm因此，

.sin()=aj+sin()=bj+55p当时，1m?abjajpbj+=2(),+();-=-+即23p当时,-5abjajpbj+=2(),+3();-=-+即2因此cos+)cos()(ajbj=-+于是cos)cos[()()]cos()cos()sin()sin()(abajbjajbjajbj-=+-+=+++++2mmm2222=-++++=--+=-cos()sin()sin()[1()]()1.bjajbj555

考点:1、三角函数图像变换和性质

;2、协助角公式和引诱公式((湖南)17.设的内角A，B，C的对边分别为a，b，c，，且B为钝,ABCabA,tan角》,(1)证明:,,BA2sinsinAC，(2)求的取值范围29【答案】(1)详见剖析;(2)(，].28【解析】,试题剖析:(1)利用正弦定理，将条件中的式子等价变形为inB=sin(+A)，进而得证;2sinA，sinC(2)利用(1)中的结论，以及三角恒等变形，将转变为只与相关的表达A式，再利用三角函数的性质即可求解.sinsinAbB试题解析:(1)由a=btanA及正弦定理，得,,,因此sinB=cosA，即coscosAaB

,sinB=sin(+A).2

,,,,,又

B为钝角，因此+A(，A)，故

B=+A，即B-A=;(2)

由(I)知，C=-,2222,,,,,,,(A+B)=-(2A+)=-2A>0，因此A，于是sinA+sinC=sinA+sin(-2A)=0,,,,2224,,19,222sinsinA+cos2A=-2A+sinA+1=-2(sinA-)+，因为0,因此0,因此4428oftheaudit.Fireisacombinationofauditinginthefieldoflawenforcementlawenforcementjobrotationandfirepracticeoflawenforcement,inthepromotionchanged,retrainingretiredkeytimingsynchronizationinplaceauditmechanism,andconducivetoself-urgedself-restraint,thefirelawenforcement,toprotectthemselves.1.3preventivefire-fightingurgedthecurrentfirelawenforcementcorruptioninvolvingcasesofviolatinglawenforcementhasanupwardtrend,judgingfromnationalreportsofcomplaintsinrecentyears,reportedcasesreflectthefirelawenforcementisover50%.Inaddition,themultiplecorruptioncases,withahighincidenceofleadingcadres,asmallnumberofleadingcadres'corruption,corrupt,power-for-moneytransaction,especiallyteams,groupreportingincreasingcomplaintsinvolvingthesupervisionoflawenforcementjobs,bringgreatpressuretoforcehonestandseriouschallenges.Meanwhile,teamandBrigadelevelsbearanadministrativelicenseapproval,administrativepunishment,routineinspectionsandmorethan95%oftheamount,somefirestaffhandledayeargothroughlegalinstrumentsandthousandsofcontacthundredsofsocialunits,onceinsidelax,theirqualityisnothigh,extremelyeasytoinducedisciplineandcorruption.Inpracticaloperation,thefirelawenforcementas'athletes'and'umpire'internalinspector,inspection,testandnotenoughproblemtouchesnotfoundtoleadingpolicymakers,deviatefromsystemdesign,systemimplementationhastodealwith.2buildingfirelaw4/6页22199,,sinA,，,,,2488,,29由此可知sinA+sinC

的取值范围是(，].28

考点

:1.正弦定理

;2.三角恒等变形

;3.三角函数的性质

.

(四川)19.如图，

A,B,C,D

为平面四边形

ABCD的四个内角

.

AA1cos,(1)

证明:tan;,2sinAoACABBCCDAD，,,,,,180,6,3,4,5,(2)若求ABCDtantantantan

，，，的值

.2222

410【答案】

(1)详见剖析;(2).3

【剖析】

AsinA2tan,

试题剖析

:(1)第全部化弦得，为了将半角变为单角，可在分子分母同时A2cos2A2sin

乘以，尔后逆用正弦与余弦的二倍角公式即可

.(2)由题设知，该四边形的两对角

2

22ABCDtantantantan

，，，,，互补.再联合

(1)的结果，有，因此只要

2222sinsinABcoscosCA,,coscosDB,,sin,sinAB求出即可.因为已知四边，且，，故考虑用余cos,cosABsin,sinAB弦定理列方程组求，进而求出.AA2sin2sinAA1cos,22tan,,,试题剖析:(1).AAA2sinAcos2sincos222,,,AC，,180(2)由CADB,,,,180,180，得.oftheaudit.Fireisacombinationofauditinginthefieldoflawenforcementlawenforcementjobrotationandfirepracticeoflawenforcement,inthepromotionchanged,retrainingretiredkeytimingsynchronizationinplaceauditmechanism,andconducivetoself-urgedself-restraint,thefirelawenforcement,toprotectthemselves.1.3preventivefire-fightingurgedthecurrentfirelawenforcementcorruptioninvolvingcasesofviolatinglawenforcementh