5.2.2同角三角函数的基本关系课后篇巩固提升基础巩固1.已知cosθ=45,且3π2<θ<2π,则1tanθ的值为()A.34B.-34C.43D.-43解析因为cosθ=45,且3π2<θ<2π,所以sinθ=-❑√1-cos2θ=-35.所以tanθ=-34,故1tanθ=-43.选D.答案D2.已知cosα+sinα=-12,则sinαcosα的值为()A.-38B.±38C.-34D.±34解析由已知得(cosα+sinα)2=sin2α+cos2α+2sinαcosα=1+2sinαcosα=14,解得sinαcosα=-38.答案A3.已知α是第四象限角,tanα=-512,则sinα=()A.15B.-15C.513D.-513解析∵α是第四象限角,∴sinα<0.由tanα=-512,得sinαcosα=-512,∴cosα=-125sinα.由sin2α+cos2α=1,得sin2α+(-125sinα)2=1,∴16925sin2α=1,sinα=±513.∵sinα<0,∴sinα=-513.答案D4.(多选题)化简cosα❑√1-sin2α+2sinα❑√1-cos2α的值为()A.-1B.1C.-3D.3解析原式=cosα|cosα|+2sinα|sinα|,当α为第一象限角时,上式值为3;当α为第二象限角时,上式值为1;当α为第三象限角时,上式值为-3;当α为第四象限角时,上式值为-1.答案ABCD5.化简1❑√1+tan2160°的结果为()A.-cos160°B.cos160°C.1cos160°D.1-cos160°解析原式=1❑√1+sin2160°cos2160°=1❑√cos2160°+sin2160°cos2160°=1❑√1cos2160°=❑√cos2160°=|cos160°|=-cos160°.故选A.答案A6.若tan2x-sin2x=165,则tan2xsin2x=.解析tan2xsin2x=tan2x(1-cos2x)=tan2x-tan2xcos2x=tan2x-sin2x=165.答案1657.已知cos(α+π4)=13,0<α<π2,则sin(α+π4)=.解析∵sin2(α+π4)+cos2(α+π4)=1,∴sin2(α+π4)=1-19=89.∵0<α<π2,∴π4<α+π4<3π4.∴sin(α+π4)=2❑√23.答案2❑√238.化简:❑√1-2sin130°cos130°sin130°+❑√1-sin2130°.解原式=❑√sin2130°-2sin130°cos130°+cos2130°sin130°+❑√cos2130°=|sin130°-cos130°|sin130°+|cos130°|=sin130°-cos130°sin130°-cos130°=1.9.证明:1+2sinθcosθcos2θ-sin2θ=1+tanθ1-tanθ.证明∵左边=sin2θ+cos2θ+2sinθcosθ(cosθ+sinθ)(cosθ-sinθ)=(sinθ+cosθ)2(cosθ+sinθ)(cosθ-sinθ)=cosθ+sinθcosθ-sinθ=cosθ+sinθcosθcosθ-sinθcosθ=1+tanθ1-tanθ=右边,∴原等式成立.能力提升1.若cosα+2sinα=-❑√5,则tanα等于()A.12B.2C.-12D.-2解析(方法一)由{cosα+2sinα=-❑√5,cos2α+sin2α=1联立消去cosα,得(-❑√5-2sinα)2+sin2α=1.化简得5sin2α+4❑√5sinα+4=0,∴(❑√5sinα+2)2=0,∴sinα=-2❑√55.∴cosα=-❑√5-2sinα=-❑√55.∴tanα=sinαcosα=2.(方法二)∵cosα+2sinα=-❑√5,∴cos2α+4sinαcosα+4sin2α=5.∴cos2α+4sinαcosα+4sin2αcos2α+sin2α=5.∴1+4tanα+4tan2α1+tan2α=5,∴tan2α-4tanα+4=0.∴(tanα-2)2=0,∴tanα=2.答案B2.(一题多空题)已知tanα,1tanα是关于x的方程x2-kx+k2-3=0的两个实根,且3π<α<72π,则tanα=,cosα+sinα=.解析∵tanα·1tanα=k2-3=1,∴k=±2,而3π<α<72π,则tanα+1tanα=k=2,得tanα=1,则sinα=cosα=-❑√22,∴cosα+sinα=-❑√2.答案1-❑√23.若3π2<α<2π,化简:❑√1-cosα1+cosα+❑√1+cosα1-cosα.解∵3π2<α<2π,∴sinα<0.∴原式=❑√(1-cosα)2(1+cosα)(1-cosα)+❑√(1+cosα)2(1-cosα)(1+cosα)=❑√(1-cosα)2sin2α+❑√(1+cosα)2sin2α=|1-cosα||sinα|+|1+cosα||sinα|=-1-cosαsinα−1+cosαsinα=-2sinα.4.已知θ∈(0,π),且sinθ,cosθ是方程25x2-5x-12=0的两个根,求sin3θ+cos3θ和tanθ-1tanθ的值.解(方法一)由题意得sinθ+cosθ=15,sinθcosθ=-1225,易知θ≠π2.∴sin3θ+cos3θ=(sinθ+cosθ)(sin2θ-sinθcosθ+cos2θ)=(sinθ+cosθ)(1-sinθcosθ)=15×1+1225=37125.tanθ-1tanθ=sinθcosθ−cosθsinθ=sin2θ-cos2θsinθcosθ=(sinθ+cosθ)(sinθ-cosθ)sinθcosθ.∵θ∈(0,π),sinθcosθ<0,∴sinθ>0,cosθ<0,则sinθ-cosθ>0.∴sinθ-cosθ=❑√(sinθ-cosθ)2=❑√1-2sinθcosθ=❑√1+2×1225=❑√4925=75.∴tanθ-1tanθ=15×75-1225=-712.(方法二)方程25x2-5x-12=0的两根分别为45和-35.∵θ∈(0,π),且sinθcosθ=-1225<0,∴sinθ>0,cosθ<0,则sinθ=45,cosθ=-35,∴sin3θ+cos3θ=453+-353=64125−27125=37125,tanθ-1tanθ=sinθcosθ−cosθsinθ=45-35−-3545=-43+34=-712.
