（浙江專用）2020版高考數(shù)學(xué)大一輪復(fù)習(xí) 課時(shí)17 4.2 同角三角函數(shù)的基本關(guān)系和誘導(dǎo)公式夯基提能作業(yè).docx

4.2同角三角函數(shù)的基本關(guān)系和誘導(dǎo)公式A組基礎(chǔ)題組1.(2017浙江臺(tái)州質(zhì)量評(píng)估)已知cos =1,則sin-6= () A.12B.32C.-12D.-32答案C由題意知,=2k(kZ),所以sin-6=sin2k-6=-sin 6=-12,故選C.2.(2019鎮(zhèn)海中學(xué)月考)已知cos+2<0,cos(-)>0,則下列不等式中必成立的是() A.tan2>0B.sin2>cos2C.tan2<0D.sin2<cos2答案A由cos+2<0得sin >0,由cos(-)>0得cos <0,2k+2<<2k+(kZ),則k+4<2<k+2(kZ),選項(xiàng)A必成立,故選A.3.已知sin +cos =430<<4,則sin -cos 的值為()A.23B.-23C.13D.-13答案B將sin +cos =43兩邊平方得1+2sin cos =169,2sin cos =79,(sin -cos )2=1-2sin cos =1-79=29,又0<<4,sin <cos ,sin -cos =-23,故選B.4.當(dāng)kZ時(shí),sin(k-)cos(k+)sin(k+1)+cos(k+1)-=()A.-1B.1C.-1或1D.0答案A若k為偶數(shù),則原式=sin(-)cossin(+)cos(-)=-sincos(-sin)(-cos)=-1;若k為奇數(shù),則原式=sin(-)cos(+)sincos(-)=sin(-cos)sincos=-1.故選A. 5.(2016課標(biāo)全國文,6,5分)若tan =-13,則cos 2=()A.-45B.-15C.15D.45答案D解法一:cos 2=cos2-sin2=cos2-sin2cos2+sin2=1-tan21+tan2,tan =-13,cos 2=45.故選D.解法二:由tan =-13,可得sin =110,因而cos 2=1-2sin2=45.6.(2019鎮(zhèn)海中學(xué)月考)若2,且2cos 2=sin4-,則sin 2=()A.14B.-14C.34D.-34答案D由2cos 2=sin4-,得2(cos2-sin2)=22(cos -sin ),又2,則cos -sin 0,得cos +sin =12,兩邊平方得cos2+sin2+2sin cos =14,即sin 2=-34.7.已知為鈍角,且sin +cos =15,則tan 2=()A.-247B.247C.-724D.724答案B由sin +cos =15得(sin +cos )2=125,即2sin cos =-2425,亦即sin 2=-2425.因?yàn)闉殁g角,所以2,所以2(,2),cos 2=-725,所以tan 2=247,故選B.8.(2019效實(shí)中學(xué)月考)已知2sin(+)-cos2-sin(-)-cos(+)=4,則tan =.答案4解析2sin(+)-cos2-sin(-)-cos(+)=-2sin-sin-sin+cos=-3sincos-sin=4,即4cos -4sin =-3sin ,4cos =sin ,tan =sincos=4.9.已知sin +2cos =0,則2sin cos -cos2的值是.答案-1解析由sin +2cos =0得tan =-2.2sin cos -cos2=2sincos-cos2sin2+cos2=2tan-1tan2+1=2(-2)-1(-2)2+1=-55=-1.10.若0,2,且sin2+cos 2=14,則cos =,tan =.答案12;3解析由sin2+cos 2=14,得sin2+1-2sin2=1-sin2=cos2=14,因?yàn)?,2,所以cos =12,所以=3,故tan =3.11.1-sin6x-cos6x1-sin4x-cos4x=.答案32解析sin6x+cos6x=(sin2x+cos2x)(sin4x-sin2xcos2x+cos4x)=(sin2x+cos2x)2-3sin2xcos2x=1-3sin2xcos2x.sin4x+cos4x=(sin2x+cos2x)2-2sin2xcos2x=1-2sin2xcos2x.原式=1-(1-3sin2xcos2x)1-(1-2sin2xcos2x)=32.12.(2018寧波調(diào)研)已知cos(+)=-12,求sin+(2n+1)+sin(+)sin(-)cos(+2n)(nZ).解析因?yàn)閏os(+)=-12,所以-cos =-12,cos =12.sin+(2n+1)+sin(+)sin(-)cos(+2n)=sin(+2n+)-sinsincos=sin(+)-sinsincos=-2sinsincos=-2cos=-4.B組提升題組1.已知2sin tan =3,則sin4-cos4的值是() A.-34B.-12C.34D.12答案D由2sin tan =3,得2sin2=3cos ,即有2cos2+3cos -2=0,得cos =12(cos =-2舍去),則sin4-cos4=(sin2+cos2)(sin2-cos2)=sin2-cos2=1-2cos2=12.2.若1sin+1cos=3,則sin cos =()A.-13B.13C.-13或1D.13或-1答案A1sin+1cos=sin+cossincos=3,sin +cos =3sin cos ,兩邊平方得1+2sin cos =3(sin cos )2,(sin cos -1)(3sin cos +1)=0,因?yàn)閟in cos =12sin 212,所以sin cos =-13,故選A.3.(2017浙江鎮(zhèn)海中學(xué)模擬)已知f(x)=x2+sin+cos-sincosx為偶函數(shù),則sin 2的值為()A.2-22B.33-6C.32-5D.1-3答案A因?yàn)閒(x)為偶函數(shù),即f(-x)=f(x)恒成立,所以sin +cos =sin cos .設(shè)sin +cos =sin cos =t,則t2-1=2t,故t=1-2或t=1+2(舍).所以sin 2=2t=2-22,故選A.4.若sin3+=512,則cos6-=.答案512解析因?yàn)?-+3+=2,所以cos6-=sin3+=512.5.若sin +2cos =-25(0<<),則tan =;cos2+4=.答案-43;17250解析由sin +2cos =-25(0<<)可知,為鈍角,結(jié)合sin2+cos2=1,可得sin =45,cos =-35,所以tan =-43,sin 2=2sin cos =-2425,cos 2=cos2-sin2=-725,所以cos2+4=cos 2cos4-sin 2sin4=17250.6.(2019浙江鎮(zhèn)海中學(xué)月考)已知x,y0,2,且2sin x=6sin y,tan x=3tan y,則cos x=.答案12解析由2sinx=6siny,tanx=3tany得2sinx=6siny,sinxcosx=3sinycosy,即2sinx=6siny,cosy=2cosx,所以sin2y+cos2y=23sin2x+2cos2x=23+43cos2x=1,則cos2x=14,又x0,2,故cos x=12.7.在ABC中,若sin(2+A)=-2sin(2-B),3cos A=-2cos(-B),求角A,B,C.解析由題意得sin A=2sin B,3cos A=2cos B,2+2得sin2A+3cos2A=2,cos2A=12,由可知,cos A與cos B同號(hào),又A,B均為ABC的內(nèi)角,A必為銳角,A=4,cos B=32,B=6,C=712.