广西2020版高考数学一轮复习 考点规范练18 同角三角函数的基本关系及诱导公式 文

考点规范练18同角三角函数的基本关系及诱导公式一、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,-sin0.cos(-)0,-cos0,即cos0.故选B.2.若cos(3-x)-3cosx+2=0,则tan x等于()A.-12B.-2C.12D.13答案D解析cos(3-x)-3cosx+2=0,-cosx+3sinx=0,tanx=13,故选D.3.已知tan(-)=34,且2,32,则sin+2=()A.45B.-45C.35D.-35答案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cos0,sin-2cos3sin+5cos=tan-23tan+5=-5,解得tan=-2316.6.已知sin(-)=-2sin2+,则sin cos 等于()A.25B.-25C.25或-25D.-15答案B解析sin(-)=-2sin2+,sin=-2cos,tan=-2.sincos=sincossin2+cos2=tan1+tan2=-25,故选B.7.已知cos512+=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-=-223.8.若(0,),sin(-)+cos =23,则sin -cos 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+2sincos=29,则2sincos=-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(sin15)=f(cos75)=cos150=cos(180-30)=-cos30=-32.11.已知为第二象限角,则cos 1+tan2+sin 1+1tan2=.答案0解析原式=cossin2+cos2cos2+sinsin2+cos2sin2=cos1|cos|+sin1|sin|.因为是第二象限角,所以sin0,cos0,所以cos1|cos|+sin1|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=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=-1.二、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.12D.-12答案B解析2tansin=3,2sin2cos=3,即2cos2+3cos-2=0.又-2log2x的解集为.答案(0,2)解析由f(31)=asin531+btan531=asin5+btan5=f(1)=1,则f(31)log2x,即1log2x,解得0x2.三、高考预测18.(2018广东佛山七校联考)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sincos=sincos1=sincossin2+cos2=tantan2+1=25.6