2020_2021學(xué)年新教材高中數(shù)學(xué)第八章向量的數(shù)量積與三角恒等變換8.2.1兩角和與差的余弦課時(shí)素養(yǎng)檢測(cè)含解析新人教B版必修第三冊(cè)202103091137

1、課時(shí)素養(yǎng)檢測(cè)十七兩角和與差的余弦(30分鐘60分)一、選擇題(每小題4分,共24分,多選題全部選對(duì)得4分,選對(duì)但不全對(duì)的得2分,有選錯(cuò)的得0分)1.計(jì)算cos 8cos 38+sin 8sin 38等于()A.B.C.D.-【解析】選C.逆用兩角差的余弦公式,得cos 8cos 38+sin 8sin 38=cos(8-38)=cos(-30)=cos 30=.2.式子coscos-sinsin的值為()A.B.0C.1D.-1【解析】選D.根據(jù)題意,coscos-sinsin=cos(+)=cos =-1.3.sin 40sin 10+cos 40sin 80=()A.B.-C.cos 50

2、D.【解析】選D.原式=sin 40sin 10+cos 40cos 10=cos(40-10)=cos 30=.4.已知sin =,是第二象限角,則cos(-60)為()A.B.C.D.【解析】選B.因?yàn)閟in =,是第二象限角,所以cos =-,故cos(-60)=cos cos 60+sin sin 60=+=.5.已知cos=,0,則cos 等于()A.B.C.D.【解析】選A.因?yàn)?所以+,所以sin =.故cos =cos=coscos +sinsin=+=.6.(多選題)下列關(guān)于函數(shù)f(x)=coscos(x)sinsin x的性質(zhì)敘述中正確的是()A.最小正周期為B.函數(shù)圖像關(guān)

3、于直線x=對(duì)稱C.函數(shù)圖像關(guān)于直線x=對(duì)稱D.函數(shù)圖像關(guān)于點(diǎn)對(duì)稱【解析】選ABC.函數(shù)f(x)=coscos(x)sinsin x=coscos(x)+sinsin (x)=cos=cos,所以函數(shù)的最小正周期是,由2x+=k,kZ,得x=,kZ,所以函數(shù)圖像關(guān)于直線x=,kZ,對(duì)稱,故選項(xiàng)B,C都正確.由2x+=k+,kZ,得x=+,kZ,所以函數(shù)圖像關(guān)于點(diǎn)對(duì)稱,其中,kZ,故選項(xiàng)D不正確.二、填空題(每小題4分,共8分)7.計(jì)算cos(+120)cos sin(+120)sin ()=.【解析】方法一:cos(+120)cos sin(+120)sin()=cos(+120)cos()si

4、n(+120)sin()=cos(+120)+(-)=cos 120=-.方法二:cos(+120)cos sin(+120)sin()=cos(+120)cos +sin(+120)sin =cos(+120)=cos 120=.答案:8.已知向量a=(cos ,sin ),b=(cos ,sin ),若a與b的夾角為,則cos(-)=.【解析】因?yàn)閍=(cos ,sin ),b=(cos ,sin ),所以=1,又因?yàn)閍與b的夾角為,所以ab=cos =11=.又ab=(cos ,sin )(cos ,sin )=cos cos +sin sin =cos(-),所以cos(-)=.答案:

5、三、解答題(每小題14分,共28分)9.,為銳角,cos(+)=,cos(2+)=,求cos 的值.【解析】因?yàn)?為銳角,所以0+0,所以0+,又因?yàn)閏os(2+)=,所以02+,所以sin(+)=,sin(2+)=,所以cos =cos (2+)-(+)=cos(2+)cos(+)+sin(2+)sin(+)=+=.10.已知sin +sin =,cos +cos =,0,求-的值.【解析】因?yàn)?sin +sin )2=,(cos +cos )2=,以上兩式展開(kāi)兩邊分別相加得2+2cos(-)=1.所以cos(-)=-.因?yàn)?,所以-0,所以-=-.(35分鐘70分)一、選擇題(每小題4分,

6、共16分,多選題全部選對(duì)得4分,選對(duì)但不全對(duì)的得2分,有選錯(cuò)的得0分)1.計(jì)算cos 28sin 62+sin 28sin 152等于()A.1B.1C.0B.【解析】選A.cos 28sin 62+sin 28sin 152=cos 28cos 28+sin 28sin(18028)=cos 28cos 28+sin 28sin 28=cos (2828)=cos 0=1.2.滿足cos cos =-sin sin 的一組,的值是()A.=,=B.=,=C.=,=D.=,=【解析】選B.由條件cos cos =-sin sin 得cos cos +sin sin =,即cos(-)=,=,=

7、滿足題意.【補(bǔ)償訓(xùn)練】若sin sin =1,則cos(-)的值為()A.0B.1C.1D.-1【解析】選B.因?yàn)閟in sin =1,-1sin 1,-1sin 1,所以或者解得于是cos(-)=cos cos +sin sin =1.3.已知tan =4,cos(+)=-,均為銳角,則的值是()A.B.C.D.【解析】選B.因?yàn)閠an =4,sin2+cos2=1,為銳角,所以sin =,cos =,因?yàn)閏os(+)=-,且,均為銳角,所以sin(+)=,所以cos =cos (+)-=cos(+)cos +sin(+)sin =-+=,又因?yàn)闉殇J角,則=.4.(多選題)若,為兩個(gè)銳角,則

8、下列大小關(guān)系正確的是()A.cos(+)cos +cos C.cos(-)cos cos D.cos(-)sin sin 【解析】選AC.方法一:不妨取=,=,則cos(+)=cos=cos=0,cos +cos =+,所以cos(+)cos cos ,cos(-)sin sin 所以C正確,D錯(cuò)誤.方法二:作差得cos-(cos +cos )=cos cos -sin sin -cos -cos =cos (cos -1)-sin sin -cos ,因?yàn)?是銳角,所以cos -10,cos (cos -1)0,-sin sin 0,-cos 0,故cos-(-)-(cos +cos )0,

9、即cos(+)0,sin sin 0,所以cos(-)=cos cos +sin sin cos cos ,同理cos(-)sin sin ,故C正確,D錯(cuò)誤.二、填空題(每小題4分,共16分)5.計(jì)算cos 1cos 361+sin 1sin 361=.【解析】由兩角差的余弦公式,得cos 1cos 361+sin 1sin 361=cos(1-361)=cos(-360)=1.答案:16.已知sin =-,cos =-,則cos(-)=.【解析】因?yàn)閟in =-,所以cos =-=-,又cos =-,所以sin =,所以cos(-)=cos cos +sin sin =-+=.答案:7.已

10、知cos=,則cos +sin 的值為.【解析】因?yàn)閏os=coscos +sinsin =cos +sin =,所以cos +sin =.答案:8.已知ABC中,sin A=,cos B=-,則cos(A-B)=.【解析】因?yàn)閏os B=-,且0B,所以B,所以sin B=,且0A,所以cos A=,所以cos(A-B)=cos Acos B+sin Asin B=+=-.答案:-三、解答題(共38分)9.(12分)已知cos(-)=-,sin(+)=-,-,+2,求的值.【解析】因?yàn)?,cos(-)=-,所以sin(-)=.因?yàn)?2,sin(+)=-,所以cos(+)=.所以cos 2=c

11、os(+)-(-)=cos(+)cos(-)+sin(+)sin(-)=+=-1.因?yàn)?,+2,所以2,2=,所以=.10.(12分)已知cos +cos =,sin +sin =,計(jì)算cos(-)的值.【解析】由cos +cos =(cos +cos )2=,即cos2+2cos cos +cos2=,由sin +sin =(sin +sin )2=,即sin2+2sin sin +sin2=,所以+可得2+2(cos cos +sin sin )=,即2cos(-)=-,即cos(-)=-.11.(14分)(1)把向量=繞原點(diǎn)順時(shí)針?lè)较蛐D(zhuǎn)角,得到向量=,用x,y及角的三角函數(shù)表示x.(2)利用(1)的結(jié)論解答下面的問(wèn)題:如圖點(diǎn)B(2,0),半圓上動(dòng)點(diǎn)A,求等邊三角形ABC(逆時(shí)針?lè)较蚺帕?的頂點(diǎn)C的橫坐標(biāo)的取值范圍.【解析】(1)設(shè)的模為r,在角的終邊上,則x=rcos ,y=rsin ,由題意可得在角-的終邊上,且的模也是r,由三角函數(shù)的定義可得x=rcos=rcos cos +rsin sin =xcos +