高考数学一轮复习考点规范练17同角三角函数的基本关系及诱导公式（含解析）新人教A版

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(-A.-45 B.45 C.35 D2.若cosπ2-α=23,则cos(π-2A.29 B.59 C.-29 D3.已知tan(α-π)=34,且α∈π2,3π2,则sinA.45 B.-45 C.35 D4.sin29π6+cos-29π3-tanA.0 B.12 C.1 D.-5.若sinπ6-α=1A.-79 B.-13 C.136.已知sin(π-α)=-2sinπ2+α,则sinα·cosαA.25 B.-25 C.25或-257.已知cos5π12+α=13,且-π<αA.223 B.-13 C.138.若tanα=34,则cos2α+2sin2α=(A.6425 B.4825 C.1 D9.已知α∈π2,π,sinα=45,则tanα10.若f(cosx)=cos2x,则f(sin15°)=.

11.已知α为第二象限角,则cosα1+tan2α+sinα1+112.若sinα是方程5x2-7x-6=0的一个根,则sin-α-3π二、能力提升13.已知sin(π-α)=log814,且α∈-π2,0,则tan(2πA.-255 B.255 C.±14.已知2tanα·sinα=3,-π2<α<0,则sinα等于(A.32 B.-32 C.12 D15.已知f(x)=asin(πx+α)+bcos(πx+β)+4,若f(2018)=5,则f(2019)的值是()A.2 B.3 C.4 D.516.已知cosπ6-θ=a(|a|≤1),则cos5π6+θ17.sin21°+sin22°+…+sin290°=.

三、高考预测18.已知sin(π+α)=-12,则cos32πA.-12 B.12 C.-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所以cosα=45,即cos(-α)=42.D解析∵cosπ2∴sinα=23∵cos(π-2α)=-cos2α=2sin2α-1=2×232-1=-3.B解析∵tan(α-π)=34,∴tanα=3又α∈π2,3π2∴sinα+π2=cosα4.A解析原式=sin4π+5π6+cos-10π+π3-tan6π+π45.A解析∵π3∴sinπ6-α=sinπ2∴cos2π3+2α=2cos2π36.B解析∵sin(π-α)=-2sinπ2∴sinα=-2cosα,∴tanα=-2.∴sinα·cosα=sinα·cos故选B.7.D解析∵cos5π12+α又-π<α<-π2,∴7π12<π∴cosπ12-α=-18.A解析(方法1)由tanα=34,得cos2α+2sin2α=co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=9.-43解析∵α∈π∴cosα=-1-sin∴tanα=sinαcosα10.-32解析f(sin15°)=f(cos75°)=cos150°=cos(180°-30°)=-cos30°=-311.0解析原式=cosαsin2α+cos因为α是第二象限角,所以sinα>0,cosα<0,所以cosα|cosα|+sin12.53解析方程5x2-7x-6=0的两根为x1=-35,x2=2,则sinα=-原式=cosα(-cos13.B解析sin(π-α)=sinα=log814=-2因为α∈-π所以cosα=1-所以tan(2π-α)=tan(-α)=-tanα=-sinα14.B解析∵2tanα·sinα=3,∴2sin2即2cos2α+3cosα-2=0.又-π2<α<∴cosα=12(cosα=-∴sinα=-3215.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∵cos5π6+θ=cosπ-sin2π3-θ=sinπ2∴cos5π6+θ+sin17.912解析sin21°+sin22°+…+sin2=sin21°+sin22°+…+sin244°+sin245°+cos244°+cos243°+…+cos21°+sin290°=(sin21°+cos21°