考點(diǎn)規(guī)范練18同角三角函數(shù)的基本關(guān)系及誘導(dǎo)公式一、基礎(chǔ)鞏固1.已知sin(+)<0,cos(-)>0,則下列不等關(guān)系中必定成立的是()A.sin <0,cos >0B.sin >0,cos <0C.sin >0,cos >0D.sin <0,cos <0答案B解析sin(+)<0,-sin<0,即sin>0.cos(-)>0,-cos>0,即cos<0.故選B.2.若cos(3-x)-3cosx+2=0,則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,故選D.3.已知tan(-)=34,且2,32,則sin+2=()A.45B.-45C.35D.-35答案B解析tan(-)=34,tan=34.又2,32,為第三象限角.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,則tan 的值為()A.-2B.2C.2316D.-2316答案D解析由題意可知cos0,sin-2cos3sin+5cos=tan-23tan+5=-5,解得tan=-2316.6.已知sin(-)=-2sin2+,則sin cos 等于()A.25B.-25C.25或-25D.-15答案B解析sin(-)=-2sin2+,sin=-2cos,tan=-2.sin·cos=sin·cossin2+cos2=tan1+tan2=-25,故選B.7.已知cos512+=13,且-<<-2,則cos12-等于()A.223B.-13C.13D.-223答案D解析cos512+=sin12-=13,又-<<-2,712<12-<1312.cos12-=-1-sin212-=-223.8.若(0,),sin(-)+cos =23,則sin -cos 的值為()A.23B.-23C.43D.-43答案C解析由誘導(dǎo)公式得sin(-)+cos=sin+cos=23,平方得(sin+cos)2=1+2sincos=29,則2sincos=-79<0,所以(sin-cos)2=1-2sincos=169,又因?yàn)?0,),所以sin-cos>0,所以sin-cos=43.9.已知2,sin =45,則tan =. 答案-43解析2,cos=-1-sin2=-35.tan=sincos=-43.10.若f(cos x)=cos 2x,則f(sin 15°)=. 答案-32解析f(sin15°)=f(cos75°)=cos150°=cos(180°-30°)=-cos30°=-32.11.已知為第二象限角,則cos 1+tan2+sin 1+1tan2=. 答案0解析原式=cossin2+cos2cos2+sinsin2+cos2sin2=cos1|cos|+sin1|sin|.因?yàn)槭堑诙笙藿?所以sin>0,cos<0,所以cos1|cos|+sin1|sin|=-1+1=0,即原式等于0.12.已知kZ,則sin(k-)cos(k-1)-sin(k+1)+cos(k+)的值為. 答案-1解析當(dāng)k=2n(nZ)時(shí),原式=sin(2n-)cos(2n-1)-sin(2n+1)+cos(2n+)=sin(-)·cos(-)sin(+)·cos=-sin(-cos)-sin·cos=-1.當(dāng)k=2n+1(nZ)時(shí),原式=sin(2n+1)-·cos(2n+1-1)-sin(2n+1+1)+·cos(2n+1)+=sin(-)·cossin·cos(+)=sin·cossin(-cos)=-1.綜上,原式=-1.二、能力提升13.已知sin(-)=log814,且-2,0,則tan(2-)的值為()A.-255B.255C.±255D.52答案B解析sin(-)=sin=log814=-23.又因?yàn)?2,0,所以cos=1-sin2=53,所以tan(2-)=tan(-)=-tan=-sincos=255.14.已知2tan ·sin =3,-2<<0,則sin 等于()A.32B.-32C.12D.-12答案B解析2tan·sin=3,2sin2cos=3,即2cos2+3cos-2=0.又-2<<0,cos=12(cos=-2舍去),sin=-32.15.已知角和的終邊關(guān)于直線y=x對(duì)稱,且=-3,則sin 等于()A.-32B.32C.-12D.12答案D解析終邊在直線y=x上的角為k+4(kZ),因?yàn)榻呛偷慕K邊關(guān)于直線y=x對(duì)稱,所以+=2k+2(kZ).又=-3,所以=2k+56(kZ),即得sin=12.16.(2018山東日照期中聯(lián)考)已知sinx+6=14,則sin56-x+cos3-x的值為()A.0B.14C.12D.-12答案C解析因?yàn)閟inx+6=14,所以sin56-x+cos3-x=sin-x+6+cos2-x+6=2sinx+6=2×14=12.故選C.17.已知函數(shù)f(x)=asin5x+btan5x(a,b為常數(shù),xR).若f(1)=1,則不等式f(31)>log2x的解集為. 答案(0,2)解析由f(31)=asin5×31+btan5×31=asin5+btan5=f(1)=1,則f(31)>log2x,即1>log2x,解得0<x<2.三、高考預(yù)測(cè)18.(2018廣東佛山七校聯(lián)考)已知sin =2cos ,則sin cos =()A.-25B.-15C.25D.15答案C解析由題意得tan=2,所以sincos=sincos1=sincossin2+cos2=tantan2+1=25.6