2019_2020学年高中数学第五章三角函数5.2.2同角三角函数的基本关系课后篇巩固提升（含解析）新人教A版.docx

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cos =45,且322,则1tan的值为()A.34B.-34C.43D.-43解析因为cos=45,且322,所以sin=-1-cos2=-35.所以tan=-34,故1tan=-43.选D.答案D2.已知cos +sin =-12,则sin cos 的值为()A.-38B.38C.-34D.34解析由已知得(cos+sin)2=sin2+cos2+2sincos=1+2sincos=14,解得sincos=-38.答案A3.已知是第四象限角,tan =-512,则sin =()A.15B.-15C.513D.-513解析是第四象限角,sin0.由tan=-512,得sincos=-512,cos=-125sin.由sin2+cos2=1,得sin2+-125sin2=1,16925sin2=1,sin=513.sin0,sin=-513.答案D4.(多选题)化简cos1-sin2+2sin1-cos2的值为()A.-1B.1C.-3D.3解析原式=cos|cos|+2sin|sin|,当为第一象限角时,上式值为3;当为第二象限角时,上式值为1;当为第三象限角时,上式值为-3;当为第四象限角时,上式值为-1.答案ABCD5.化简11+tan2160的结果为()A.-cos 160B.cos 160C.1cos160D.1-cos160解析原式=11+sin2160cos2160=1cos2160+sin2160cos2160=11cos2160=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+4=13,02,则sin+4=.解析sin2+4+cos2+4=1,sin2+4=1-19=89.02,4+434.sin+4=223.答案2238.化简:1-2sin130cos130sin130+1-sin2130.解原式=sin2130-2sin130cos130+cos2130sin130+cos2130=|sin130-cos130|sin130+|cos130|=sin130-cos130sin130-cos130=1.9.证明:1+2sincoscos2-sin2=1+tan1-tan.证明左边=sin2+cos2+2sincos(cos+sin)(cos-sin)=(sin+cos)2(cos+sin)(cos-sin)=cos+sincos-sin=cos+sincoscos-sincos=1+tan1-tan=右边,原等式成立.能力提升1.若cos +2sin =-5,则tan 等于()A.12B.2C.-12D.-2解析(方法一)由cos+2sin=-5,cos2+sin2=1联立消去cos,得(-5-2sin)2+sin2=1.化简得5sin2+45sin+4=0,(5sin+2)2=0,sin=-255.cos=-5-2sin=-55.tan=sincos=2.(方法二)cos+2sin=-5,cos2+4sincos+4sin2=5.cos2+4sincos+4sin2cos2+sin2=5.1+4tan+4tan21+tan2=5,tan2-4tan+4=0.(tan-2)2=0,tan=2.答案B2.(一题多空题)已知tan ,1tan是关于x的方程x2-kx+k2-3=0的两个实根,且372,则tan =,cos +sin =.解析tan1tan=k2-3=1,k=2,而372,则tan+1tan=k=2,得tan=1,则sin=cos=-22,cos+sin=-2.答案1-23.若322,化简:1-cos1+cos+1+cos1-cos.解322,sin0.原式=(1-cos)2(1+cos)(1-cos)+(1+cos)2(1-cos)(1+cos)=(1-cos)2sin2+(1+cos)2sin2=|1-cos|sin|+|1+cos|sin|=-1-cossin-1+cossin=-2sin.4.已知(0,),且sin ,cos 是方程25x2-5x-12=0的两个根,求sin3+cos3和tan -1tan的值.解(方法一)由题意得sin+cos=15,sincos=-1225,易知2.sin3+cos3=(sin+cos)(sin2-sincos+cos2)=(sin+cos)(1-sincos)=151+1225=37125.tan-1tan=sincos-cossin=sin2-cos2sincos=(sin+cos)(sin-cos)sincos.(0,),sincos0,cos0.sin-cos=(sin-cos)2=1-2sincos=1+21225=4925=75.tan-1tan=1575-1225=-712.(方法二)方程25x2-5x-12=0的两根分别为45和-35.(0,),且sinco