（浙江专用）高考数学大一轮复习 第四章 三角函数、解三角形 考点规范练16 同角三角函数的基本关系及诱导公式-人教版高三全册数学试题VIP免费

考点规范练16同角三角函数的基本关系及诱导公式基础巩固组1.sin600°的值为()A.-12B.-√32C.12D.√32答案B解析sin600°=sin(360°+240°)=sin240°=sin(180°+60°)=-sin60°=-√32.2.已知sin(π2+α)=-35,α∈(π2,π),则tanα=()A.34B.-34C.-43D.43答案C解析 sin(π2+α)=-35,sin(π2+α)=cosα,∴cosα=-35,又α∈(π2,π),∴sinα=√1-cos2α=45,∴tanα=sinαcosα=-43.故选C.3.若cos(3π-x)-3cos(x+π2)=0,则tanx等于()A.-12B.-2C.12D.13答案D解析 cos(3π-x)-3cos(x+π2)=0,∴-cosx+3sinx=0.∴tanx=13.故选D.4.√1+2sin(π-3)cos(π+3)化简的结果是()A.sin3-cos3B.cos3-sin3C.±(sin3-cos3)D.以上都不对答案A解析 sin(π-3)=sin3,cos(π+3)=-cos3,∴原式=√1-2sin3·cos3=√(sin3-cos3)2=|sin3-cos3|. π2<3<π,∴sin3>0,cos3<0.∴原式=sin3-cos3.故选A.15.已知tanα=3,则1+2sinαcosαsin2α-cos2α的值是()A.12B.2C.-12D.-2答案B解析原式=sin2α+cos2α+2sinαcosαsin2α-cos2α=(sinα+cosα)2(sinα+cosα)(sinα-cosα)=sinα+cosαsinα-cosα=tanα+1tanα-1=3+13-1=2.6.sin(-1071°)sin99°+sin(-171°)sin(-261°)+tan(-1089°)tan(-540°)=.答案0解析原式=(-sin1071°)·sin99°+sin171°·sin261°+tan1089°·tan540°=-sin(3×360°-9°)sin(90°+9°)+sin(180°-9°)sin(270°-9°)+tan(3×360°+9°)tan(360°+180°)=sin9°cos9°-sin9°cos9°+tan9°tan180°=0.7.(2018浙江绍兴3月模拟)已知sin(30°+α)=35,60°<α<150°,则cos(30°+α)=;cosα=.答案-453-4√310解析 60°<α<150°,∴90°<30°+α<180°. sin(30°+α)=35,∴cos(30°+α)=-√1-sin2(30°+α)=-√1-(35)2=-45.cosα=cos(30°+α-30°)=cos(30°+α)cos30°+sin(30°+α)·sin30°=-45×√32+35×12=3-4√310,故答案为-45,3-4√310.8.已知α∈R,sin2α+4sinαcosα+4cos2α=52,则tanα=.答案3或-13解析由sin2α+4sinαcosα+4cos2α=52可得52=sin2α+4sinαcosα+4cos2αsin2α+cos2α=tan2α+4tanα+4tan2α+1,解得tanα=3或tanα=-13.能力提升组29.已知:①sin(-1000°);②cos(-2200°);③tan(-10);④sin7π10cosπtan17π9.其中符号为负的是()A.①B.②C.③D.④答案C解析sin(-1000°)=sin80°>0;cos(-2200°)=cos(-40°)=cos40°>0,tan(-10)=tan(3π-10)<0;sin7π10·cosπtan17π9=-sin7π10tan17π9,sin7π10>0,tan17π9<0.故选C.10.已知倾斜角为θ的直线与直线x-3y+1=0垂直,则23sin2θ-cos2θ=()A.103B.-103C.1013D.-1013答案C解析直线x-3y+1=0的斜率为13,因此与此直线垂直的直线的斜率k=-3,∴tanθ=-3,∴23sin2θ-cos2θ=2(sin2θ+cos2θ)3sin2θ-cos2θ=2(tan2θ+1)3tan2θ-1,把tanθ=-3代入得,原式=2×[(-3)2+1]3×(-3)2-1=1013.故选C.11.已知cos(5π12+α)=13,且-π<α<-π2,则cos(π12-α)等于()A.2√23B.13C.-13D.-2√23答案D解析因为(512π+α)+(π12-α)=π2,所以cos(π12-α)=sin[π2-(π12-α)]=sin(5π12+α).因为-π<α<-π2,所以-7π12<α+5π12<-π12.又cos(5π12+α)=13>0,所以-π2<α+5π12<-π12,所以sin(5π12+α)=-√1-cos2(5π12+α)3=-√1-(13)2=-2√23.12.若1sinα+1cosα=√3,则sinαcosα=()A.-13B.13C.-13或1D.13或-1答案A解析由1sinα+1cosα=√3,可得sinα+cosα=√3sinαcosα,两边平方,得1+2sinαcosα=3sin2αcos2α,解得sinαcosα=-13或sinαcosα=1.由题意,知-1