（天津專用）2020屆高考數學一輪復習 考點規(guī)范練17 同角三角函數的基本關系及誘導公式（含解析）新人教A版

考點規(guī)范練17同角三角函數的基本關系及誘導公式一、基礎鞏固1.若-2,2,sin =-35,則cos(-)=()A.-45B.45C.35D.-352.若cos2-=23,則cos(-2)=()A.29B.59C.-29D.-593.已知tan(-)=34,且2,32,則sin+2=()A.45B.-45C.35D.-354.sin296+cos-293-tan254=()A.0B.12C.1D.-125.若sin6-=13,則cos23+2等于()A.-79B.-13C.13D.796.已知sin(-)=-2sin2+,則sin ·cos 等于()A.25B.-25C.25或-25D.-157.已知cos512+=13,且-<<-2,則cos12-等于()A.223B.-13C.13D.-2238.若tan =34,則cos2+2sin 2=()A.6425B.4825C.1D.16259.已知2,sin =45,則tan =. 10.若f(cos x)=cos 2x,則f(sin 15°)=. 11.已知為第二象限角,則cos 1+tan2+sin 1+1tan2=. 12.若sin 是方程5x2-7x-6=0的一個根,則sin-32sin32-tan2(2-)cos2-cos2+sin(+)=. 二、能力提升13.已知sin(-)=log814,且-2,0,則tan(2-)的值為()A.-255B.255C.±255D.5214.已知2tan ·sin =3,-2<<0,則sin 等于()A.32B.-32C.12D.-1215.已知f(x)=asin(x+)+bcos(x+)+4,若f(2 018)=5,則f(2 019)的值是()A.2B.3C.4D.516.已知cos6-=a(|a|1),則cos56+sin23-的值是. 17.sin21°+sin22°+sin290°=. 三、高考預測18.已知sin(+)=-12,則cos32+等于()A.-12B.12C.-32D.32考點規(guī)范練17同角三角函數的基本關系及誘導公式1.B解析因為-2,2,sin=-35,所以cos=45,即cos(-)=45.2.D解析cos2-=23,sin=23.cos(-2)=-cos2=2sin2-1=2×232-1=-59.3.B解析tan(-)=34,tan=34.又2,32,為第三象限角.sin+2=cos=-45.4.A解析原式=sin4+56+cos-10+3-tan6+4=sin56+cos3-tan4=12+12-1=0.5.A解析3+6-=2,sin6-=sin2-3+=cos3+=13.cos23+2=2cos23+-1=-79.6.B解析sin(-)=-2sin2+,sin=-2cos,tan=-2.sin·cos=sin·cossin2+cos2=tan1+tan2=-25,故選B.7.D解析cos512+=sin12-=13,又-<<-2,712<12-<1312.cos12-=-1-sin212-=-223.8.A解析(方法1)由tan=34,得cos2+2sin2=cos2+4sincoscos2+sin2=1+4tan1+tan2=1+4×341+342=42516=6425.故選A.(方法2)tan=34,3cos=4sin,即9cos2=16sin2.又sin2+cos2=1,9cos2=16(1-cos2),cos2=1625.cos2+2sin2=cos2+4sincos=cos2+3cos2=4cos2=4×1625=6425,故選A.9.-43解析2,cos=-1-sin2=-35.tan=sincos=-43.10.-32解析f(sin15°)=f(cos75°)=cos150°=cos(180°-30°)=-cos30°=-32.11.0解析原式=cossin2+cos2cos2+sin·sin2+cos2sin2=cos|cos|+sin|sin|.因為是第二象限角,所以sin>0,cos<0,所以cos|cos|+sin|sin|=-1+1=0,即原式等于0.12.53解析方程5x2-7x-6=0的兩根為x1=-35,x2=2,則sin=-35.原式=cos(-cos)tan2sin(-sin)(-sin)=-1sin=53.13.B解析sin(-)=sin=log814=-23.因為-2,0,所以cos=1-sin2=53,所以tan(2-)=tan(-)=-tan=-sincos=255.14.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.B解析f(2018)=5,asin(2018+)+bcos(2018+)+4=5,即asin+bcos=1.f(2019)=asin(2019+)+bcos(2019+)+4=-asin-bcos+4=-1+4=3.16.0解析cos56+=cos-6-=-cos6-=-a,sin23-=sin2+6-=cos6-=a,cos56+sin23-=0.17.912解析sin21°+sin22°+sin290°=sin21°+sin22°+sin244°+sin245°+cos244°+cos243°+cos21°+sin290°=(sin21°+cos21°)+(sin22°+cos22°)+(sin244°+cos244°)+sin245°+sin290°=44+12+1=912.18.B解析由sin(+)=-12,得sin=12,故cos32+=sin=12.5