2022年高考数学专题复习测试卷 专题4 三角函数 解三角形（word版含答案）.doc

1112022年高考数学专题复习测试卷专题4三角函数解三角形一单选题1将函数119910sin1199096119909119877图像上所有的点向左平行移动6个单位长度再把图像上各点的横坐标扩大到原来的2倍纵坐标不变则所得到的图像的解析式为A119910sin21199093B119910sin11990923C119910sin1199092D119910cos11990922函数ytan2x4的最小正周期为A4B2CD23将函数119910sin1199091205932cos1199091205932120593数的图象则120593的值是A31205874B1205874C1205874D12058724函数1198911199094tan12058741199091205878的单调递增区间是A411989652411989632119896119885B411989632411989652119896119885C811989652811989632119896119885D8119896328119896521198961198855函数119891119909119860sin120596119909120593其中1198600120593119860sin120596119909的图象则只要将119891119909的图象A向右平移1205876个单位长度B向右平移1205873个单位长度C向左平移1205876个单位长度D向左平移1205873个单位长度6已知点119860120587240在函数119891119909cos21205961199091205931205960且1205961198730上直线1199091205876是函数119891119909的图象的一条对称轴若119891119909在区间12058761205873内单调则120593211A1205876B1205873C21205873D512058767已知函数119910sin12059611990912059312059600标是A21205872B21205874C41205872D412058748用五点法作函数119910119860sin120596119909120593119860012059601205931199091205876212058731205961199091205930120587212058731205872212058711991004040则119860120596120593的值分别为A421205873B4121205873C421205876D41212058769已知函数119891119909sin119909120593的图象关y轴对称则实数120593的取值可能是A1205874B1205873C1205872D12058710下列四个等式中cos360300cos300cos180300cos300cos180300cos300cos360300cos300其中正确的等式有A1个B2个C3个D4个11已知cos1205721205876331205876A33B63C33D6312为得到函数119910cos1199091205876的图象只需将函数ysinx的图象A向左平移1205873个长度单位B向右平移1205873个长度单位311C向左平移21205873个长度单位D向右平移21205873个长度单位13已知1198911199091199041198941198992017119909120587611988811990011990420171199091205873的最大值为119860若存在实数11990911199092使得对任意实数119909总有11989111990911198911199091198911199092成立则11986011990911199092的最小值为A1205872017B21205872017C41205872017D1205874034二填空题14在119860119861119862中角119860119861119862的对边分别为119886119887119888119886151198871011986060则cos11986115已知角120579的终边经过点11987512则tan120579120587416如果cos12057213且120572是第四象限角那么cos120587212057217如图某港口一天6时到18时的水深变化曲线近似满足函数ysin6xk据此函数可知这段时间水深单位m的最大值为18已知sinxm1且xR则m的取值范围是三解答题19已知fsincos2cos112057232cos2120572sin1化简f2若为第三象限角且cos3215求f的值3若313求f的值20已知点1198601199091119891119909111986111990921198911199092是函数119891119909211990411989411989912059611990912059312059601205872任意两点且角120593的终边经过点11987513若119891119909111989111990924时11990911199092的最小值为12058731求函数119891119909的解析式2求函数119891119909的单调递增区间4113求当11990901205873时119891119909的值域211已知函数119891119909119860sin120596119909120593其中11986001205960120593于AB两点AB两点的最小距离为1205872且该函数的图象上的一个最高点的坐标为120587122求函数119891119909的解析式2已知角120572的终边在直线1199103119909上求下列函数的值sin120572cos120572tan12057222已知函数fxcos2x1205876300的最大值为2最小正周期为212058731求函数yfx的解析式2当x01205872时求函数fx的值域23已知函数119891119909119886sin120596119909cos12059611990911988601205960从下列四个条件中选择两个作为已知使函数119891119909存在且唯一确定1求119891119909的解析式2设1198921199091198911199092cos21205961199091求函数119892119909在0120587上的单调递增区间条件11989112058741条件119891119909为偶函数条件119891119909的最大值为1条件119891119909图象的相邻两条对称轴之间的距离为1205872注如果选择的条件不符合要求第1问得0分如果选择多个符合要求的条件分别解答按第一个解答计分24设函数fx3cos312058722xsin2x101若1fx是偶函数求的最小值2若1存在x120587121205873使fx12fx1m30成立求m取值范围3若yfx1在x02015上至少存在2016个最值点求范围25在平面四边形ABCD中AB4AD22对角线AC与BD交于点EE是BD的中点且11986011986421198641198625111若1198601198611198631205874求cosAED的值2若1198601198623求BD的长26已知函数fx2sinxcosxcos2x0的最小正周期为1求的值2求fx的单调递增区间611答案解析部分1答案B2答案B3答案B4答案A5答案B6答案B7答案B8答案A9答案C10答案C11答案B12答案C13答案B14答案6315答案316答案22317答案418答案0219答案解1fsin120572cos120572sin120572sin120572sin120572cos2cos32sin15sin15又为第三象限角cos1sin2120572265f26533136253f313cos313cos6253711cos53cos31220答案1解角120593的终边经过点1198751311990511988611989912059331205872由119891119909111989111990924时11990911199092的最小值为1205873得11987921205873即21205871205962120587312059631198911199092119904119894119899311990912058732解由120587221198961205873119909120587312058722119896120587得120587182119896120587311990951205871821198961205873119896119885函数119891119909的单调递增区间为1205871821198961205873512058718211989612058731198961198853解当11990901205873时即01199091205873则0311990912058712058733119909120587321205873由函数单调性可得3211990411989411989931199091205873131198911199092函数119891119909的值域为3221答案1解由题意得119860211987921205872则1198791205872120587120596解得1205962根据最高点120587122得2sin2120587121205932所以21205871212059321198961205871205872即12059321198961205871205873119896119885因为120593所以1198911199092sin211990912058732解由题意得tan1205723则120572为第一或第三象限角当120572为第一象限角时由tan1205723得sin1205723cos120572代入sin2120572cos21205721得cos212057214又cos1205720所以cos12057212则sin120572cos120572tan12057232所以sin12057232cos12057212tan1205723当120572为第三象限角时同理可得sin12057232cos12057212tan120572322答案1解fxcos2x1205876300811111988811990011990421205961199091205873231205822cos2x120587312058223又函数fx的最大值为2可得1205822120582232解得5最小正周期为2120587321205872120596解得32fx52cos3x1205873122解119909012058723119909120587312058731112058761198881199001199043119909120587313211989111990935324所以fx的值域是3532423答案1解因为119891119909119886sin120596119909cos12059611990911988601205960所以11989111990912119886sin2120596119909显然当1198860时119891119909为奇函数故不能选若选择即11989111990912119886sin2120596119909最大值为1所以121198861解得1198862所以119891119909sin2120596119909又11989112058741所以1198911205874sin212059612058741即120587212059612058722119896120587119896119885解得12059614119896119896119885故119891119909不能唯一确定故舍去若选择即119891119909图象的相邻两条对称轴之间的距离为1205872所以21205872120596120587解得1205961所以11989111990912119886sin2119909又119891120587412119886sin212058741所以121198861解得1198862所以119891119909sin2119909若选择即119891119909图象的相邻两条对称轴之间的距离为1205872所以21205872120596120587解得1205961所以11989111990912119886sin2119909又119891119909的最大值为1所以121198861解得1198862所以119891119909sin21199092解由1可得1198921199091198911199092cos21205961199091sin21199092cos21199091sin2119909cos2119909222sin211990922cos21199092sin21199091205874令211989612058712058722119909120587421198961205871205872119896119885解得119896120587120587811990911989612058731205878119896119885所以函数的单调递增区间为119896120587120587811989612058731205878119896119885又1199090120587所以119892119909在0120587上的单调递增区间有71205878120587和91103120587824答案1解函数fx3cos312058722xsin2x13sin2xsin2x12sin2x10若1fx2sin2x1是偶函数则2k1205872即11989612058721205874kZ故的最小值为12058742解若1fx12sin2x112sin2x当x120587121205873时2x120587621205873fx12sin2x12存在x120587121205873使fx12fx1m30成立mfx13119891119909123当且仅当fx13时取等号故要求的m取值范围为233解若yfx12sin2x在x02015上至少存在2016个最值点则在02015上至少包含100734个周期100734212058721205962015求得4031806025答案1由题意可知在119860119861119863中1198601198611198631205874119860119861411986011986322由正弦定理可知119860119861sin119860119863119861119860119863sin119860119861119863即4sin1198601198631198612222解得sin1198601198631198611所以1198601198631198611205872所以119860119861119863为直角三角形由勾股定理可知1198631198611198601198612119860119863222又因为E是BD的中点所以1198631198642则在直角119860119863119864中11986011986410cos1198601198641198631198631198641198601198642105510112因为1198601198623所以1198601198642又因为E是BD的中点所以11986011986412119860119861119860119863所以1198601198642141198601198611198601198632即119860119864214119860119861119860119863214119860119861211986011986322119860119861119860119863所以16422222422cos119863119860119861解得cos11986311986011986124在119860119861119863中利用余弦定理得1198611198632119860119861211986011986322119860119861119860119863cos1198631198601198614222224222432所以1198611198634226答案1解fx2sinxcosxcos2xsin2xcos2x222sin212059611990922cos21205961199092sin21205961199091205874由T21205872120596120587得12解由1得fx2sin21205961199091205874再由120587221198961205872119909120587412058722119896120587得3120587811989612058711990912058781198961205871198961199111111fx的单调递增区间为312058781198961205871205878119896120587kZ