考点规范练18同角三角函数的基本关系及诱导公式一、基础巩固1.已知sin(θ+π)<0,cos(θ-π)>0,则下列不等关系中必定成立的是()A.sinθ<0,cosθ>0B.sinθ>0,cosθ<0C.sinθ>0,cosθ>0D.sinθ<0,cosθ<0答案B解析∵sin(θ+π)<0,∴-sinθ<0,即sinθ>0.∵cos(θ-π)>0,∴-cosθ>0,即cosθ<0.故选B.2.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.3.已知tan(α-π)=34,且α∈(π2,3π2),则sin(α+π2)=()A.45B.-45C.35D.-35答案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.sin29π6+cos(-29π3)-tan25π4=()A.0B.12C.1D.-12答案A解析原式=sin(4π+5π6)+cos(-10π+π3)-tan(6π+π4)=sin5π6+cosπ3-tanπ4=12+12-1=0.5.已知sinα-2cosα3sinα+5cosα=-5,则tanα的值为()A.-2B.2C.2316D.-2316答案D解析由题意可知cosα≠0,∴sinα-2cosα3sinα+5cosα=tanα-23tanα+5=-5,解得tanα=-2316.6.已知sin(π-α)=-2sin(π2+α),则sinαcosα等于()A.25B.-25C.25或-25D.-15答案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.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∵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.若α∈(0,π),sin(π-α)+cosα=❑√23,则sinα-cosα的值为()A.❑√23B.-❑√23C.43D.-43答案C解析由诱导公式得sin(π-α)+cosα=sinα+cosα=❑√23,平方得(sinα+cosα)2=1+2sinαcosα=29,则2sinαcosα=-79<0,所以(sinα-cosα)2=1-2sinαcosα=169,又因为α∈(0,π),所以sinα-cosα>0,所以sinα-cosα=43.9.已知α∈(π2,π),sinα=45,则tanα=.答案-43解析∵α∈(π2,π),∴cosα=-❑√1-sin2α=-35.∴tanα=sinαcosα=-43.10.若f(cosx)=cos2x,则f(sin15°)=.答案-❑√32解析f(sin15°)=f(cos75°)=cos150°=cos(180°-30°)=-cos30°=-❑√32.11.已知α为第二象限角,则cosα❑√1+tan2α+sinα❑√1+1tan2α=.答案0解析原式=cosα❑√sin2α+cos2αcos2α+sinα❑√sin2α+cos2αsin2α=cosα1|cosα|+sinα1|sinα|.因为α是第二象限角,所以sinα>0,cosα<0,所以cosα1|cosα|+sinα1|sinα|=-1+1=0,即原式等于0.12.已知k∈Z,则sin(kπ-α)cos[(k-1)π-α]sin[(k+1)π+α]cos(kπ+α)的值为.答案-1解析当k=2n(n∈Z)时,原式=sin(2nπ-α)cos[(2n-1)π-α]sin[(2n+1)π+α]cos(2nπ+α)=sin(-α)·cos(-π-α)sin(π+α)·cosα=-sinα(-cosα)-sinα·cosα=-1.当k=2n+1(n∈Z)时,原式=sin[(2n+1)π-α]·cos[(2n+1-1)π-α]sin[(2n+1+1)π+α]·cos[(2n+1)π+α]=sin(π-α)·cosαsinα·cos(π+α)=sinα·cosαsinα(-cosα)=-1.综上,原式=-1.二、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.❑√52答案B解析sin(π-α)=sinα=log814=-23.又因为α∈(-π2,0),所以cosα=❑√1-sin2α=❑√53,所以tan(2π-α)=tan(-α)=-tanα=-sinαcosα=2❑√55.14.已知2tanα·sinα=3,-π2<α<0,则sinα等于()A.❑√32B.-❑√32C.12D.-12答案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.已知角α和β的终边关于直线y=x对称,且β=-π3,则sinα等于()A.-❑√32B.❑√32C.-12D.12答案D解析终边在直线y=x上的角为kπ+π4(k∈Z),因为角α和β的终边关于直线y=x对称,所以α+β=2kπ+π2(k∈Z).又β=-π3,所以α=2kπ+5π6(k∈Z),即得sinα=12.16.(2018山东日照期中联考)已知sin(x+π6)=14,则sin(5π6-x)+cos(π3-x)的值为()A.0B.14C.12D.-12答案C解析因为sin(x+π6)=14,所以sin(5π6-x)+cos(π3-x)=sin[π-(x+π6)]+cos[π2-(x+π6)]=2sin(x+π6)=2×14=12.故选C.17.已知函数f(x)=asin(π5x)+btan(π5x)(a,b为常数,x∈R).若f(1)=1,则不等式f(31)>log2x的解集为.答案(0,2)解析由f(31)=asin(π5×31)+btan(π5×31)=asinπ5+btanπ5=f(1)=1,则f(31)>log2x,即1>log2x,解得0
