2018-2019学年高中数学 第一章 三角函数 1.2 任意角的三角函数2课后习题 新人教A版必修4.doc

1.2.2同角三角函数的基本关系课后篇巩固探究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.若为第三象限角,则cos1-sin2+2sin1-cos2的值为()A.3B.-3C.1D.-1解析因为为第三象限角,所以原式=cos-cos+2sin-sin=-3.答案B3.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,sin 0.由tan =-512,得sincos=-512,cos =-125sin .由sin2+cos2=1,得sin2+-125sin2=1,16925sin2=1,sin =513.sin 0,sin =-513.答案D4.(2018全国高考)若sin =13,则cos 2=()A.89B.79C.-79D.-89解析cos 2=1-2sin2=1-2132=79.答案B5.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+2sin cos =1+2sin cos =14,解得sin cos =-38.答案A6.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=|cos 160|=-cos 160.故选A.答案A7.导学号68254013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+4sin cos +4sin2=5.cos2+4sincos+4sin2cos2+sin2=5.1+4tan+4tan21+tan2=5,tan2-4tan +4=0.(tan -2)2=0,tan =2.答案B8.若tan2x-sin2x=165,则tan2xsin2x=.解析tan2xsin2x=tan2x(1-cos2x)=tan2x-tan2xcos2x=tan2x-sin2x=165.答案1659.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22310.已知tan ,1tan是关于x的方程x2-kx+k2-3=0的两个实根,且372,则cos +sin =.解析tan 1tan=k2-3=1,k=2,而372,则tan +1tan=k=2,得tan =1,则sin =cos =-22,cos +sin =-2.答案-211.化简:1-2sin130cos130sin130+1-sin2130.解原式=sin2130-2sin130cos130+cos2130sin130+cos2130=|sin130-cos130|sin130+|cos130|=sin130-cos130sin130-cos130=1.12.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.13.若322,化简:1-cos1+cos+1+cos1-cos.解322,sin 0.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.14.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.解法一由题意得sin +cos =15,sin cos =-1225,易知2.sin3+cos3=(sin +cos )(sin2-sin cos +cos2)=(sin +cos )(1-sin cos )=151+1225=37125.tan -1tan=sincos-cossin=sin2-cos2sincos=(sin+cos)(sin-cos)sincos.(0,),sin cos 0,cos 0.sin -cos =(sin-cos)2=1-2sincos=1+21225=4925=75.tan -1tan=1575-1225=-712.解法二方程25x2-5x-12=0的两根分别为45和-35.(0,),且sin cos