1、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 =,且2,则的值为()A.B.-C.D.-解析因为cos =,且2,所以sin =-=-.所以tan =-,故=-.选D.答案D2.若为第三象限角,则的值为()A.3B.-3C.1D.-1解析因为为第三象限角,所以原式=-3.答案B3.已知是第四象限角,tan =-,则sin =()A.B.-C.D.-解析是第四象限角,sin 0.由tan =-,得=-,cos =-sin .由sin2+cos2=1,得sin2+=1,sin2=1,sin =.sin 0,sin =-.答案D4.(2018全国高考)若sin =,则cos 2=()
2、A.B.C.-D.-解析cos 2=1-2sin2=1-2.答案B5.已知cos +sin =-,则sin cos 的值为()A.-B.C.-D.解析由已知得(cos +sin )2=sin2+cos2+2sin cos =1+2sin cos =,解得sin cos =-.答案A6.化简的结果为()A.-cos 160B.cos 160C.D.解析原式=|cos 160|=-cos 160.故选A.答案A7.导学号68254013若cos +2sin =-,则tan 等于()A.B.2C.-D.-2解析(方法一)由联立消去cos ,得(-2sin )2+sin2=1.化简得5sin2+4si
3、n +4=0,(sin +2)2=0,sin =-.cos =-2sin =-.tan =2.(方法二)cos +2sin =-,cos2+4sin cos +4sin2=5.=5.=5,tan2-4tan +4=0.(tan -2)2=0,tan =2.答案B8.若tan2x-sin2x=,则tan2xsin2x=.解析tan2xsin2x=tan2x(1-cos2x)=tan2x-tan2xcos2x=tan2x-sin2x=.答案9.已知cos,0,则sin=.解析sin2+cos2=1,sin2=1-.0,+.sin.答案10.已知tan ,是关于x的方程x2-kx+k2-3=0的两个
4、实根,且3,则cos +sin =.解析tan =k2-3=1,k=2,而3,则tan +=k=2,得tan =1,则sin =cos =-,cos +sin =-.答案-11.化简:.解原式=1.12.证明:.证明左边=右边,原等式成立.13.若2,化简:.解2,sin 0.原式=-=-.14.已知(0,),且sin ,cos 是方程25x2-5x-12=0的两个根,求sin3+cos3和tan -的值.解法一由题意得sin +cos =,sin cos =-,易知.sin3+cos3=(sin +cos )(sin2-sin cos +cos2)=(sin +cos )(1-sin cos )=1+=.tan -=.(0,),sin cos 0,cos 0.sin -cos =.tan -=-.解法二方程25x2-5x-12=0的两根分别为和-.(0,),且sin cos =-0,cos 0,则sin =,cos =-,sin3+cos3=3+-3=,tan -=-=-.5
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