三角函数练习及高考题(带答案).pdf

该【三角函数练习及高考题(带答案) 】是由【小屁孩】上传分享，文档一共【16】页，该文档可以免费在线阅读，需要了解更多关于【三角函数练习及高考题(带答案) 】的内容，可以使用淘豆网的站内搜索功能，选择自己适合的文档，以下文字是截取该文章内的部分文字，如需要获得完整电子版，请下载此文档到您的设备，方便您编辑和打印。:..三角函数练****及高考题(带答案)-CAL-FENGHAI-(2020YEAR-YICAI)_JINGBIAN:..三角函数练****及高考题?π??cos?2x??的图像,只需将函数y?sin2x的图像(A)?3??a与函数f(x)?sinx和g(x)?cosx的图像分别交于M,N两点,则MN的最大值为(B).?tanx?cotx?cos2x?(D)(A)tanx(B)sinx(C)cosx(D)???2?,sin??3cos?,则?的取值范围是:(C)?????????4??(A),(B),?(C),(D)???????32??3??33???3???,??32???sinx(x?R)的图象上所有点向左平行移动个单位长度,再把31所得图象上所有点的横坐标缩短到原来的倍(纵坐标不变),得到的图象所2表示的函数是C?x?(A)y?sin(2x?),x?R(B)y?sin(?),x?R326?2?(C)y?sin(2x?),x?R(D)y?sin(2x?),x?R335?2?2??sin,b?cos,c?tan,则D777(A)a?b?c(B)a?c?b(C)b?c?a(D)b?a?c???sin(2x?)的图象按向量?平移后所得的图象关于点(?,0)中心312对称,则向量?的坐标可能为(C)2:..???A.(?,0)B.(?,0)C.(,0)12612?D.(,0)(α-)+sinα=3,则sin(α?)的值是656232344(A)-(B)(C)-(D)5555?9.(湖北)将函数y?3sin(x??)的图象F按向量(,3)平移得到图象F?,若F?3?的一条对称轴是直线x?,则?的一个可能取值是A4551111A.?B.??C.?D.??12121212????(x)?sin2x?3sinxcosx在区间,上的最大值是(C)???42?1?+322sinx?(x)=(0?x?2?)的值域是B3?2cosx?2sinx2(A)[-,0](B)[-1,0](C)[-2,0](D)[-23,0](x)=cosx(x)(x?R)的图象按向量(m,0)平移后,得到函数y=-f′(x)的图象,则m的值可以为A?.??-?D.-2x3?,函数y?cos(?)(x?[0,2?])的图象和直线221y?的交点个数是C2(A)0(B)1(C)2(D)43:..?2sina??5,则tana=B11(A)(B)2(C)?(D)?=2sin(ωx+φ)(ω>0)在区间[0,2π]的图像如下:那么ω=(B)/33?sin70016.=(C)2??(x)=3sinx+sin(+x),b,c为△ABC的三个内角A,B,C的对边,向量m=(3,?1),nπ=(cosA,sinA).若m⊥n,且acosB+bcosA=csinC,则角B=.6?????x??cos?x?的最小正周期为,其中??0,则?=.10???6?(x)?(sinx?cosx)sinx,x?R,则f(x)的最小正周期是.??????????????(x)?sin??x??(??0),f???f??,且f(x)在区间?,?有最?3??6??3??63?14小值,无最大值,则?=△ABC的内角A,B,C所对的边长分别为a,b,c,且3acosB?bcosA?(Ⅰ)求tanAcotB的值;(Ⅱ)求tan(A?B):..3解析:(Ⅰ)在△ABC中,由正弦定理及acosB?bcosA?c53333可得sinAcosB?sinBcosA?sinC?sin(A?B)?sinAcosB?cosAsinB5555即sinAcosB?4cosAsinB,则tanAcotB?4;(Ⅱ)由tanAcotB?4得tanA?4tanB?0tanA?tanB3tanB33tan(A?B)???≤1?tanAtanB1?4tan2BcotB?4tanB41当且仅当4tanB?cotB,tanB?,tanA?2时,等号成立,213故当tanA?2,tanB?时,tan(A?B)△ABC中,cosB??,cosC?.135(Ⅰ)求sinA的值;33(Ⅱ)设△ABC的面积S?,求BC的长.△ABC2解:512(Ⅰ)由cosB??,得sinB?,131343由cosC?,得sinC?.5533所以sinA?sin(B?C)?sinBcosC?cosBsinC?.·········5分6533133(Ⅱ)由S?得?AB?AC?sinA?,△ABC22233由(Ⅰ)知sinA?,65故AB?AC?65,························8分AB?sinB20又AC??AB,sinC132013故AB2?65,AB?.132AB?sinA11所以BC??.····················10分sinC2?π?(x)?sin2?x?3sin?xsin?x?(??0)的最小正周期为???2?:..(Ⅰ)求?的值;?2π?(Ⅱ)求函数f(x)在区间0,上的取值范围.???3?1?cos2?x3311解:(Ⅰ)f(x)??sin2?x?sin2?x?cos2?x?22222?π?1?sin2?x??.???6?2因为函数f(x)的最小正周期为π,且??0,2π所以?π,解得????π?1(Ⅱ)由(Ⅰ)得f(x)?sin?2x???.?6?22π因为0≤x≤,3ππ7π所以?≤2x?≤,6661?π?所以?≤sin2x?≤1,??2?6??π?13?3?因此0≤sin2x??≤,即f(x)的取值范围为0,.?????6?22?2??7?4sinxcosx?4cos2x?4cos4x的最大值与最小值。【解】:y?7?4sinxcosx?4cos2x?4cos4x???7?2sin2x?4cos2x1?cos2x?7?2sin2x?4cos2xsin2x?7?2sin2x?sin22x??1?sin2x?2?6由于函数z??u?1?2?6在??11,?中的最大值为z???1?1?2?6?10max6:..最小值为z??1?1?2?6?6min故当sin2x??1时y取得最大值10,当sin2x?(x)?2cos2?x?2sin?xcos?x?1(x?R,??0)的最小值正周期是?.2(Ⅰ)求?的值;(Ⅱ)求函数f(x)的最大值,并且求使f(x)取得最大值的x的集合.(17)本小题主要考查特殊角三角函数值、两角和的正弦、二倍角的正弦与余弦、函数y?Asin(?x??)的性质等基础知识,.(Ⅰ)解:1?cos2?xf?x??2??sin2?x?12?sin2?x?cos2?x?2?????2?sin2?xcos?cos2?xsin??2?44?????2sin?2?x???2?4??2??由题设,函数f?x?的最小正周期是,可得?,所以???2???(Ⅱ)由(Ⅰ)知,f?x??2sin?4x???2.?4????k????4x???2k?,即x???k?Z?sin4x?取得最大值1,所当时,??42162?4???k??以函数f?x?的最大值是2?2,此时x的集合为x|x??,k?Z.???162????(x)?cos(2x?)?2sin(x?)sin(x?)344(Ⅰ)求函数f(x)的最小正周期和图象的对称轴方程7:..??(Ⅱ)求函数f(x)在区间[?,]上的值域122???解:(1)f(x)?cos(2x?)?2sin(x?)sin(x?)34413?cos2x?sin2x?(sinx?cosx)(sinx?cosx)2213?cos2x?sin2x?sin2x?cos2x2213?cos2x?sin2x?cos2x22??sin(2x?)62?∴周期T???2??k??由2x??k??(k?Z),得x??(k?Z)6223?∴函数图象的对称轴方程为x?k??(k?Z)3????5?(2)x?[?,],?2x??[?,]122636?????因为f(x)?sin(2x?)在区间[?,]上单调递增,在区间[,]上单调612332递减,?所以当x?时,f(x)取最大值13?3?1?又f(?)???f()?,当x??时,f(x)取最小值12222123?2??3所以函数f(x)在区间[?,]上的值域为[?,1](x)=3sin(?x??)?cos(?x??)(0???π,??0)为偶函数,且函π数y=f(x)(Ⅰ)美洲f()的值;88:..π(Ⅱ)将函数y=f(x)的图象向右平移个单位后,再将得到的图象上各点的6横坐标舒畅长到原来的4倍,纵坐标不变,得到函数y=g(x)的图象,求g(x):(Ⅰ)f(x)=3sin(?x??)?cos(?x??)?31?=2?sin(?x??)?cos(?x??)?22??π=2sin(?x??-)6因为f(x)为偶函数,所以对x∈R,f(-x)=f(x)恒成立,ππ因此sin(-?x??-)=sin(?x??-).66ππππ即-sin?xcos(?-)+cos?xsin(?-)=sin?xcos(?-)+cos?xsin(?-),6666ππ整理得sin?xcos(?-)=?>0,且x∈R,所以cos(?-)=<?<π,故?-=.所以f(x)=2sin(?x+)=2cos????2?, 所以?=?2故f(x)=2cos2x.??因为f()?2cos???(Ⅱ)将f(x)的图象向右平移个个单位后,得到f(x?)的图象,再将所得66??图象横坐标伸长到原来的4倍,纵坐标不变,得到f(?)????????所以 g(x)?f(?)?2cos2(?)?2cosf(?).??46?46?23??当2kπ≤?≤2kπ+π(k∈Z),239:..2?8?即4kπ+≤≤x≤4kπ+(k∈Z)时,g(x)?2?8??因此g(x)的单调递减区间为4k??,4k??(k∈Z)???33?,在平面直角坐标系xoy中,以ox轴为始边做两个锐角?,?,它们的225终边分别与单位圆相交于A,B两点,已知A,B的横坐标分别为,.105(Ⅰ)求tan(???)的值;(Ⅱ)求??2???,cos??,因为?,?为锐角,所以105725sin?=,sin??1051因此tan??7,tan??2tan??tan?(Ⅰ)tan(???)=??31?tan?tan?2tan?4tan??tan2?(Ⅱ)tan2???,所以tan???2?????11?tan2?31?tan?tan2?3?3?∵?,?为锐角,∴0???2??,∴??2?=?ABC中,角A,B,C所对应的边分别为a,b,c,a?23,A?BCtan?tan?4,222sinBcosC?sinA,求A,B及b,cA?解:由tan?tan?4得cot?tan?42222CCcossin221∴??4∴?4sincossincos22221∴sinC?,又C?(0,?)210:..?5?∴C?,或C?66由2sinBcosC?sinA得2sinBcosB?sin(B?C)即sin(B?C)?0∴B?C?B?C?62?A???(B?C)?3abc由正弦定理??得sinAsinBsinC1sinB2b?c?a?23??2sinA321?t17?(t)?,g(x)?cosx?f(sinx)?sinx?f(cosx),x?(?,).1?t12(Ⅰ)将函数g(x)化简成Asin(?x??)?B(A?0,??0,??[0,2?))的形式;(Ⅱ)求函数g(x)、值域和三角函数的性质等基本知识,考查三角恒等变换、代数式的化简变形和运算能力.(满分12分)1?sinx1?cosx解:(Ⅰ)g(x)?cosx?sinx1?sinx1?cosx(1?sinx)2(1?cosx)2?cosx?sinxcos2xsin2x1?sinx1?cosx?cosx??17??x???,,?cosx??cosx,sinx??sinx,??12?1?sinx1?cosx?g(x)?cosx?sinx?cosx?sinx?sinx?cosx?211:..???=2sin?x???2.?4?17?5??5?(Ⅱ)由?<x?,得<x??.12443?5?3???3?5??sint在?,上为减函数,在?,上为增函数,???42??23?5?5?3??5??17??又sin<sin,?sin?sin(x?)<sin(当x???,),?34244?2??2?即?1?sin(x?)<?,??2?2?2sin(x?)?2<?3,424?故g(x)的值域为??2?2,?3.?(x)?2sincos?23sin2?(Ⅰ)求函数f(x)的最小正周期及最值;?π?(Ⅱ)令g(x)?f?x??,判断函数g(x)的奇偶性,并说明理由.?3?xxxx?xπ?解:(Ⅰ)f(x)?sin?3(1?2sin2)?sin?3cos?2sin?.??2422?23?2π?f(x)的最小正周期T???xπ??xπ?当sin?????1时,f(x)取得最小值?2;当sin????1时,f(x)取得最?23??23?大值2.?xπ??π?(Ⅱ)由(Ⅰ)知f(x)?2sin???.又g(x)?f?x??.?23??3??1?π?π??xπ?x?g(x)?2sin?x????2sin????2cos.???2?3?3??22?2?x?xg(?x)?2cos????2cos?g(x).?2?2?函数g(x):..?ABC的内角A,B,C的对边分别为a,b,c,且A=60,c=:a(Ⅰ)的值;c(Ⅱ)cotB+:(Ⅰ)由余弦定理得a2?b2?c2?2bcosA1117=(c)2?c2?2cc?c2,3329a7故?.c3(Ⅱ)解法一:cotB?cotCcosBsinC?cosCsinB=sinBsinCsin(B?C)sinA=?,sinBsinCsinBsinC由正弦定理和(Ⅰ)的结论得7c2sinA1a22141439?·?·??.sinBsinCsinAbc31339c·c3143故cotB?cotC?.9解法二:由余弦定理及(Ⅰ)的结论有71c2?c2?(c)2a2?c2?b293cosB??2ac72cc35=.27253故sinB?1?cos2B?1??.2827同理可得13:..71c2?c2?c2a2?b2?c2199cosC????,2ab71272cc33133sinC?1?cos2C?1??.2827cosBcosC51143从而cotB?cotC???3?3?.=(sinA,cosA),n=(3,?1),m·n=1,且A为锐角.(Ⅰ)求角A的大小;(Ⅱ)求函数f(x)?cos2x?4cosAsinx(x?R)、三角函数的基本公式、三角恒等变换、一元二次函数的最值等基本知识,:(Ⅰ)由题意得mn?3sinA?cosA?1,??12sin(A?)?1,sin(A?)?.662???由A为锐角得A??,A?.6631(Ⅱ)由(Ⅰ)知cosA?,2所以13f(x)?cos2x?2sinx?1?2sin2x?2sins??2(sinx?)2?.221因为x∈R,所以sinx???1,1?,因此,当sinx?时,f(x)=-1时,f(x)有最小值-3,所以所求函数f(x)的值域是?3??3,.???2?14:..(x)?Asin(x??)(A?0,0???π),x?R的最大值是1,其图像经?π1??π?过点M?,?.(1)求f(x)的解析式;(2)已知?,???0,?,且?32??2?312f(?)?,f(?)?,求f(???)?1(1)依题意有A?1,则f(x)?sin(x??),将点M(,)代入得32?1?5?sin(??)?,而0????,?????,???,故32362?f(x)?sin(x?)?cosx;2312?(2)依题意有cos??,cos??,而?,??(0,),513234125?sin??1?()2?,sin??1?()2?,55**********f(???)?cos(???)?cos?cos??sin?sin??????。△ABC中,内角A,B,C对边的边长分别是a,b,c,已知c?2,?C?.3(Ⅰ)若△ABC的面积等于3,求a,b;(Ⅱ)若sinC?sin(B?A)?2sin2A,求△,三角函数公式等基础知识,:(Ⅰ)由余弦定理及已知条件得,a2?b2?ab?4,1又因为△ABC的面积等于3,所以absinC?3,得ab?4.·······4分2?a2?b2?ab?4,联立方程组?解得a?2,b?2.··············6分?ab?4,(Ⅱ)由题意得sin(B?A)?sin(B?A)?4sinAcosA,即sinBcosA?2sinAcosA,·······················8分??4323当cosA?0时,A?,B?,a?,b?,2633当cosA?0时,得sinB?2sinA,由正弦定理得b?2a,15:..?a2?b2?ab?4,2343联立方程组?解得a?,b?.b?2a,33?123所以△ABC的面积S?absinC?.·················12分2316