考点规范练17同角三角函数的基本关系及诱导公式一、基础巩固1.若α∈(-π2,π2),sinα=-35,则cos(-α)=()A.-45B.45C.35D.-352.若cos(π2-α)=√23,则cos(π-2α)=()A.29B.59C.-29D.-593.已知tan(α-π)=34,且α∈(π2,3π2),则sin(α+π2)=()A.45B.-45C.35D.-354.sin29π6+cos(-29π3)-tan25π4=()A.0B.12C.1D.-125.若sin(π6-α)=13,则cos(2π3+2α)等于()A.-79B.-13C.13D.796.已知sin(π-α)=-2sin(π2+α),则sinα·cosα等于()A.25B.-25C.25或-25D.-157.已知cos(5π12+α)=13,且-π<α<-π2,则cos(π12-α)等于()A.2√23B.-13C.13D.-2√238.若tanα=34,则cos2α+2sin2α=()A.6425B.4825C.1D.16259.已知α∈(π2,π),sinα=45,则tanα=.10.若f(cosx)=cos2x,则f(sin15°)=.11.已知α为第二象限角,则cosα√1+tan2α+sinα√1+1tan2α=.12.若sinα是方程5x2-7x-6=0的一个根,则sin(-α-3π2)sin(3π2-α)tan2(2π-α)cos(π2-α)cos(π2+α)sin(π+α)=.二、能力提升13.已知sin(π-α)=log814,且α∈(-π2,0),则tan(2π-α)的值为()A.-2√55B.2√55C.±2√55D.√5214.已知2tanα·sinα=3,-π2<α<0,则sinα等于()A.√32B.-√32C.12D.-1215.已知f(x)=asin(πx+α)+bcos(πx+β)+4,若f(2018)=5,则f(2019)的值是()A.2B.3C.4D.516.已知cos(π6-θ)=a(|a|≤1),则cos(5π6+θ)+sin(2π3-θ)的值是.17.sin21°+sin22°+…+sin290°=.三、高考预测18.已知sin(π+α)=-12,则cos(32π+α)等于()A.-12B.12C.-√32D.√32考点规范练17同角三角函数的基本关系及诱导公式1.B解析因为α∈(-π2,π2),sinα=-35,所以cosα=45,即cos(-α)=45.2.D解析∵cos(π2-α)=√23,∴sinα=√23.∵cos(π-2α)=-cos2α=2sin2α-1=2×(√23)2-1=-59.3.B解析∵tan(α-π)=34,∴tanα=34.又α∈(π2,3π2),∴α为第三象限角.∴sin(α+π2)=cosα=-45.4.A解析原式=sin(4π+5π6)+cos(-10π+π3)-tan(6π+π4)=sin5π6+cosπ3-tanπ4=12+12-1=0.5.A解析∵(π3+α)+(π6-α)=π2,∴sin(π6-α)=sin[π2-(π3+α)]=cos(π3+α)=13.∴cos(2π3+2α)=2cos2(π3+α)-1=-79.6.B解析∵sin(π-α)=-2sin(π2+α),∴sinα=-2cosα,∴tanα=-2.∴sinα·cosα=sinα·cosαsin2α+cos2α=tanα1+tan2α=-25,故选B.7.D解析∵cos(5π12+α)=sin(π12-α)=13,又-π<α<-π2,∴7π12<π12-α<13π12.∴cos(π12-α)=-√1-sin2(π12-α)=-2√23.8.A解析(方法1)由tanα=34,得cos2α+2sin2α=cos2α+4sinαcosαcos2α+sin2α=1+4tanα1+tan2α=1+4×341+(34)2=42516=6425.故选A.(方法2)∵tanα=34,∴3cosα=4sinα,即9cos2α=16sin2α.又sin2α+cos2α=1,∴9cos2α=16(1-cos2α),∴cos2α=1625.∴cos2α+2sin2α=cos2α+4sinαcosα=cos2α+3cos2α=4cos2α=4×1625=6425,故选A.9.-43解析∵α∈(π2,π),∴cosα=-√1-sin2α=-35.∴tanα=sinαcosα=-43.10.-√32解析f(sin15°)=f(cos75°)=cos150°=cos(180°-30°)=-cos30°=-√32.11.0解析原式=cosα√sin2α+cos2αcos2α+sinα·√sin2α+cos2αsin2α=cosα|cosα|+sinα|sinα|.因为α是第二象限角,所以sinα>0,cosα<0,所以cosα|cosα|+sinα|sinα|=-1+1=0,即原式等于0.12.53解析方程5x2-7x-6=0的两根为x1=-35,x2=2,则sinα=-35.原式=cosα(-cosα)tan2αsinα(-sinα)(-sinα)=-1sinα=53.13.B解析sin(π-α)=sinα=log814=-23.因为α∈(-π2,0),所以cosα=√1-sin2α=√53,所以tan(2π-α)=tan(-α)=-tanα=-sinαcosα=2√55.14.B解析∵2tanα·sinα=3,∴2sin2αcosα=3,即2cos2α+3cosα-2=0.又-π2<α<0,∴cosα=12(cosα=-2舍去),∴sinα=-√32.15.B解析∵f(2018)=5,∴asin(2018π+α)+bcos(2018π+β)+4=5,即asinα+bcosβ=1.∴f(2019)=asin(2019π+α)+bcos(2019π+β)+4=-asinα-bcosβ+4=-1+4=3.16.0解析∵cos(5π6+θ)=cos[π-(π6-θ)]=-cos(π6-θ)=-a,sin(2π3-θ)=sin[π2+(π6-θ)]=cos(π6-θ)=a,∴cos(5π6+θ)+sin(2π3-θ)=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(π+α)=-12,得sinα=12,故cos(32π+α)=sinα=12.
