数学4(必修) 第二章 平面向量 [提高训练C组]
一、选择题
1.C AB?(1,a?3),AC?(2,b?3),AB//AC?b?3?2a?6,2a?b?3
?????2.C P1P2?(2?sin??cos?,2?cos??sin?),
????? P1P2?2(2?cos?)?2sin??22????????????????10?8cos??18?32 ??3.C 单位向量仅仅长度相等而已,方向也许不同;当b?0时,a与c可以为任意向量;
|a?b|?|a?b|,即对角线相等,此时为矩形,邻边垂直;还要考虑夹角 ??4.C a?3b???2??2a?6a?b?9b?1?6cos60?9?013 ??a?b21?5.C cos??????,??
423ab????26.D 设b?ka?(2k,k),,而|b|?25,则5k?25,k??,b?(4,2),或(?4,?2)
二、填空题
??1.4 2a?b?(2co?s?3,2?s?in??a?1),b?2?8??8s?in(3?? )164????????2.直角三角形 AB?(1,1)A,C??(????????????????3,3A?)B,AC?0A?,B AC3.(22,22),或(?22,?22)
设所求的向量为(x,y),2x?2y?0,x?y?1,x?y??2222
4. 6 由平行四边形中对角线的平方和等于四边的平方和得
2??2?? a?b?a?b43??2?2a?2b2???a?b2?2?2???2a?2b?a?b2?2?2?4?4?6
?43225.(,?) 设b?(x,y),4x?3y?5,x?y?1,x?,y??
5555三、解答题
????????1.解:(1)若a?b?a?c且a?0,则b?c,这是一个假命题
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?????????? 因为a?b?a?c,a?(b?c)?0,仅得a?(b?c)
???????(2)向量a在b的方向上的投影是一模等于acos?(?是a与b的夹角),方向与a在b相同或相反的一个向量.这是一个假命题
?? 因为向量a在b的方向上的投影是个数量,而非向量。
?????2.证明:设x?(a,b),y?(c,d),则x?y?ac?bd,x??22a?b,y?c?d22
而x?y?xycos?,x?y?xycos??xy ????即x?y?xy,得ac?bd?22222??????????a?b22c?d22
?(ac?bd)?(a?b)(c?d)
?????13?3.解:由a?(3,?1),b?(,)得a?b?0,a?2,b?1
22??????2????2222[a?(t?3)b]?(?ka?tb)?0,?ka?ta?b?k(t?3)a?b?t(t?3)b?0
4????????????????4. 解:?AB?AC,?AB?AC?0.
?4k?t?3t?0,k?31(t?3t),f(t)?314(t?3t)
3?????????????????????????????????AP??AQ,BP?AP?AB,CQ?AQ?AC, ?????????????????????????BP?CQ?(AP?AB)?(AQ?AC)?AP?AQ?AP?AC?AB?AQ?AB?AC??a?AP?AC?AB?AP??a?AP?(AB?AC)22
??a???a?2221212PQ?BCPQ?BC2??a?acos?.故当cos??1,即??0(PQ与BC方向相同)时,BP?CQ最大.其最大值为0.
数学4(必修)第三章 三角恒等变换 [基础训练A组]
一、选择题
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1.D x?(??2,0),cosx?45,sinx??2?1?2?
35,tanx??34,tan2x?2tanx1?tanx2??247
2.D y?5sin(x??)?5,T?3.C cosAcosB?sinAsinB?cos(A?B)?0,?cosC?0,cosC?0,C为钝角 4.D a?2sin59,b?02sin61,c?2202sin60
2?41205.C y??2sin2xcos2x??sin4x,为奇函数,T???22
6.B sin4??cos4??(sin2??cos2?)2?2sin2?cos2??1? ?1?1112(1?cos?2?) 218sin2?
二、填空题
1.3 tan60?tan(20?40)?000tan20?tan4000001?tan20tan400?3 2.2008
3?13tan20?tan?2?20tan?40?co?s210ta?n20 tan40sin?2?1s?in2? c?os2?cos20cos?2 ?(cos??sin?)22cos??sin??cos??sin?cos??sin??1?tan?1?tan??2008
?2?2cxo?s(,2T?)??
32??24171724., (sin?cos)?1?sin??,sin??,cos2??1?2sin??
22339392?3.? f(x)?cosx3sixn?25.60,032 cosA?2B?C2cos?2A212?2sinA2cAo?sA2A2?sin?2?12)?21322A2?sin22
A2sin ??2sinA2?1??2(sin
?)x32 当sin?三、解答题
0s?,即A?60时,得(coAB?C2cos2ma
1.解:sin??sin???sin?,cos??cos???cos?,
(sin??sin?)?(cos??cos?)?1, 2?2cos(???)?1,cos(???)??1222。
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2.解:令cos??cos??t,则(sin??sin?)2?(cos??cos?)2?t2?2?2cos(???)?t?212,
122,2cos(???)?t?232
?2?t?232?2,?120?t?72,?142?t?142
3.解:原式?2cos10024sin10cos10cos10000?sin10(0cos5sin5000?sin5cos500)
?2sin100?2cos10?0cos10?2sin202sin10000
?cos10?2sin(30?10)2sin1032x200?cos10?2sin30cos10?2cos30sin102sin10000000
?cos300?4.解:y?sin (1)当
x2x2?? ?2sin(x2?3cos?3)
?3?2k???2,即x?4k???3,k?Z时,y取得最大值
??x|x?4k?????,k?Z?为所求 3?右移(2)y?2sin(x2??3?3)??????y?2sin个单位x2????????y?2sinx
横坐标缩小到原来的2倍????????y?sinx纵坐标缩小到原来的2倍
数学4(必修)第三章 三角恒等变换 [综合训练B组]
一、选择题
1.C a?sin30cos6?cos30sin6?sin24,b?sin26,c?sin25,
1?tan2x1?tan2x?0?0?00022.B y?2?cos4x,T?2?4??2
?????03.B sin17(?sin43)?(?sin73)(?sin47)?cos17cos43?sin17sin43?cos60 4.D sin2x?cos(?2?2x)?cos2(???4?x)?1?2sin(2?4?x)?725
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5.A (cos??sin?)2?19,sin?cos???49,而sin??0,cos??0
cos??sin???(cos??sin?)?4sin?cos???2173
cos2??cos??sin??(cos??sin?)(cos??sin?)??221334?(?173)
6.B y?(sin2x)2?cos2x?(sin2x)2?sin2x?1?(sin2x? ?14cosx2?212)?2
34?18?(13coxs?4 )4二、填空题 ?1. (3sinA?4cosB)2?(4sinB?3cosA)2?37,25?24sin(A?B)?37
6 sin(A?B)?sin(8?00012,sinC?012,事实上A为钝角,?C?000?6
02.2?3?2sin15sin10sin80cos153 ??0?02?00000sin(15?10)?cos15cos80sin15cos10sin151?5)cos153
3. y?sin2x2x?cos33cos?6?2x?sins?in362x?2xcos?cos363s in6?sin ?cos(2x32??6),T?2?23?3?,相邻两对称轴的距离是周期的一半
4.
34 f(x)??cosx??21coxs?当,2T2?1cxo?s时2,T?2?3?3fm,x(?) ax42?5.f(x)?2sin(3x?三、解答题
) A?2,?3?,??3,sin???1,可取????2
1.解:(1)原式?sin6cos12cos24cos48?1?20000000sin6cos6cos12cos24cos48cos60000000
sin12cos12cos24cos48cos6sin48cos48cos600001?41sin24cos24cos48cos600000
1?81
?160cos60sin960?160cos6cos6?1160(2)原式?1?cos402120?1?cos10020?12(sin70?sin30)
140?1?(cos100?cos40)?012sin70?0
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