浙江专用2020版高考数学复习同角三角函数的基本关系和诱导公式夯基提能作业.docx

4.2同角三角函数的基本关系和诱导公式A组基础题组1.(2017浙江台州质量评估)已知cos =1,则sin-6= () A.12B.32C.-12D.-32答案C由题意知,=2k(kZ),所以sin-6=sin2k-6=-sin 6=-12,故选C.2.(2019镇海中学月考)已知cos+20,则下列不等式中必成立的是() A.tan20B.sin2cos2C.tan20D.sin2cos2答案A由cos+20,由cos(-)0得cos 0,2k+22k+(kZ),则k+42k+2(kZ),选项A必成立,故选A.3.已知sin +cos =4304,则sin -cos 的值为()A.23B.-23C.13D.-13答案B将sin +cos =43两边平方得1+2sin cos =169,2sin cos =79,(sin -cos )2=1-2sin cos =1-79=29,又04,sin cos ,sin -cos =-23,故选B.4.当kZ时,sin(k-)cos(k+)sin(k+1)+cos(k+1)-=()A.-1B.1C.-1或1D.0答案A若k为偶数,则原式=sin(-)cossin(+)cos(-)=-sincos(-sin)(-cos)=-1;若k为奇数,则原式=sin(-)cos(+)sincos(-)=sin(-cos)sincos=-1.故选A. 5.(2016课标全国文,6,5分)若tan =-13,则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.故选D.解法二:由tan =-13,可得sin =110,因而cos 2=1-2sin2=45.6.(2019镇海中学月考)若2,且2cos 2=sin4-,则sin 2=()A.14B.-14C.34D.-34答案D由2cos 2=sin4-,得2(cos2-sin2)=22(cos -sin ),又2,则cos -sin 0,得cos +sin =12,两边平方得cos2+sin2+2sin cos =14,即sin 2=-34.7.已知为钝角,且sin +cos =15,则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.因为为钝角,所以2,所以2(,2),cos 2=-725,所以tan 2=247,故选B.8.(2019效实中学月考)已知2sin(+)-cos2-sin(-)-cos(+)=4,则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,则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,则cos =,tan =.答案12;3解析由sin2+cos 2=14,得sin2+1-2sin2=1-sin2=cos2=14,因为0,2,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宁波调研)已知cos(+)=-12,求sin+(2n+1)+sin(+)sin(-)cos(+2n)(nZ).解析因为cos(+)=-12,所以-cos =-12,cos =12.sin+(2n+1)+sin(+)sin(-)cos(+2n)=sin(+2n+)-sinsincos=sin(+)-sinsincos=-2sinsincos=-2cos=-4.B组提升题组1.已知2sin tan =3,则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),则sin4-cos4=(sin2+cos2)(sin2-cos2)=sin2-cos2=1-2cos2=12.2.若1sin+1cos=3,则sin cos =()A.-13B.13C.-13或1D.13或-1答案A1sin+1cos=sin+cossincos=3,sin +cos =3sin cos ,两边平方得1+2sin cos =3(sin cos )2,(sin cos -1)(3sin cos +1)=0,因为sin cos =12sin 212,所以sin cos =-13,故选A.3.(2017浙江镇海中学模拟)已知f(x)=x2+sin+cos-sincosx为偶函数,则sin 2的值为()A.2-22B.33-6C.32-5D.1-3答案A因为f(x)为偶函数,即f(-x)=f(x)恒成立,所以sin +cos =sin cos .设sin +cos =sin cos =t,则t2-1=2t,故t=1-2或t=1+2(舍).所以sin 2=2t=2-22,故选A.4.若sin3+=512,则cos6-=.答案512解析因为6-+3+=2,所以cos6-=sin3+=512.5.若sin +2cos =-25(0),则tan =;cos2+4=.答案-43;17250解析由sin +2cos =-25(0)可知,为钝角,结合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浙江镇海中学月考)已知x,y0,2,且2sin x=6sin y,tan x=3tan y,则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,则cos2x=14,又x0,2,故cos x=12.7.在ABC中,若sin(2+A)=