2021高考数学（理）导学大一轮人教A广西专用考点规范练19　同角三角函数的基本关系及诱导公式 WORD版含解析.docx

1、考点规范练19同角三角函数的基本关系及诱导公式考点规范练A册第12页基础巩固1.已知sin(+)0,则下列不等关系中必定成立的是()A.sin 0B.sin 0,cos 0,cos 0D.sin 0,cos 0答案:B解析:sin(+)0,-sin 0.cos(-)0,-cos 0,即cos 0.故选B.2.若cos2-=23,则cos(-2)=()A.29B.59C.-29D.-59答案:D解析:cos2-=23,sin =23.cos(-2)=-cos 2=2sin2-1=2232-1=-59.3.已知tan(-)=34,且2,32,则sin+2=()A.45B.-45C.35D.-35答

2、案:B解析:tan(-)=34,tan =34.又2,32,为第三象限角.sin+2=cos =-45.4.sin296+cos-293-tan254=()A.0B.12C.1D.-12答案:A解析:原式=sin4+56+cos-10+3-tan6+4=sin56+cos3-tan4=12+12-1=0.5.已知sin-2cos3sin+5cos=-5,则tan 的值为()A.-2B.2C.2316D.-2316答案:D解析:由题意可知cos 0,sin-2cos3sin+5cos=tan-23tan+5=-5,解得tan =-2316.6.(2019山东济宁一模)若sin x=3sinx-2

3、,则cos xcosx+2=()A.310B.-310C.34D.-34答案:A解析:由sin x=3sinx-2,得sin x=-3cos x,即tan x=-3,所以cos xcosx+2=-cos xsin x=-cosxsinxsin2x+cos2x=-tanxtan 2x+1=310.7.已知sinx+6=14,则sin56-x+cos3-x的值为()A.0B.14C.12D.-12答案:C解析:因为sinx+6=14,所以sin56-x+cos3-x=sin-x+6+cos2-x+6=2sinx+6=214=12.故选C.8.若(0,),sin(-)+cos =23,则sin -c

4、os 的值为()A.23B.-23C.43D.-43答案:C解析:由诱导公式得sin(-)+cos =sin +cos =23,平方得(sin +cos )2=1+2sin cos =29,则2sin cos =-790,所以sin -cos =43.9.已知2,sin =45,则tan =.答案:-43解析:2,cos =-1-sin2=-35.tan =sincos=-43.10.若f(cos x)=cos 2x,则f(sin 15)=.答案:-32解析:f(sin 15)=f(cos 75)=cos 150=cos(180-30)=-cos 30=-32.11.已知为第二象限角,则cos

5、 1+tan2+sin 1+1tan2=.答案:0解析:原式=cos sin2+cos2cos2+sin sin2+cos2sin2=cos 1|cos|+sin 1|sin|.因为是第二象限角,所以sin 0,cos 0,所以cos 1|cos|+sin 1|sin|=-1+1=0,即原式等于0.12.已知kZ,则sin(k-)cos(k-1)-sin(k+1)+cos(k+)的值为.答案:-1解析:当k=2n(nZ)时,原式=sin(2n-)cos(2n-1)-sin(2n+1)+cos(2n+)=sin(-)cos(-)sin(+)cos=-sin(-cos)-sincos=-1.当k=

6、2n+1(nZ)时,原式=sin(2n+1)-cos(2n+1-1)-sin(2n+1+1)+cos(2n+1)+=sin(-)cossincos(+)=sincossin(-cos)=-1.综上,原式=-1.能力提升13.已知sin(-)=log814,且-2,0,则tan(2-)的值为()A.-255B.255C.255D.52答案:B解析:sin(-)=sin =log814=-23.又因为-2,0,所以cos =1-sin2=53,所以tan(2-)=tan(-)=-tan =-sincos=255.14.已知2tan sin =3,-20,则sin 等于()A.32B.-32C.12

7、D.-12答案:B解析:2tan sin =3,2sin2cos=3,即2cos2+3cos -2=0.又-20,cos =12(cos =-2舍去),sin =-32.15.在ABC中,3sin2-A=3sin(-A),且cos A=-3cos(-B),则ABC为()A.等腰三角形B.直角三角形C.等腰直角三角形D.等边三角形答案:B解析:由3sin2-A=3sin(-A),得3cos A=3sin A,所以tan A=33,所以A=6.由cos A=-3cos(-B),得cos 6=3cos B,所以cos B=12,所以B=3.所以C=-A-B=2.故ABC为直角三角形.16.已知cos

8、512+=13,且-2,则cos12-等于()A.223B.-13C.13D.-223答案:D解析:cos512+=sin12-=13,又-2,71212-1312.cos12-=-1-sin212-=-223.17.(2019湖北武昌调研)若tan =cos ,则1sin+cos4=.答案:2解析:tan =cos ,sincos=cos ,sin =cos2,1sin+cos4=sin2+cos2sin+cos4=sin +cos2sin+cos4=sin +sinsin+sin2=sin2+sin +1=sin2+cos2+1=1+1=2.高考预测18.已知sin =2cos ,则sin cos =()A.-25B.-15C.25D.15答案:C解析:由题意得tan =2,所以sin cos =sincos1=sincossin2+cos2=tantan2+1=25.