1、第2课时诱导公式五、六课后训练巩固提升一、A组1.(多选题)下列各式中,正确的是()A.sin(-)=sin B.cos=sin C.cos=-sin D.sin=cos 解析:sin=sin ,故A正确.cos =-sin ,故B错误.cos=cos=-sin ,故C正确.sin=sin=cos ,故D正确.答案:ACD2.化简的结果是()A.1B.-1C.tan D.-tan 解析:原式=-1.答案:B3.已知sin 10=k,则cos 620的值为()A.kB.-kC.kD.不确定解析:cos 620=cos(360+260)=cos 260=cos(270-10)=-sin 10=-k
2、.答案:B4.若f(sin x)=3-cos 2x,则f(cos x)等于()A.3-cos 2xB.3-sin 2xC.3+cos 2xD.3+sin 2x解析:f(cos x)=f=3-cos 2=3-cos(-2x)=3+cos 2x.答案:C5.化简=.解析:原式=-1.答案:-16.已知角的终边过点P(1,-2),则tan =,=.解析:因为角的终边过点P(1,-2),所以tan =-2,所以.答案:-27.求值:sin2+sin2=.解析:-+=,sin2=sin2=cos2.sin2+sin2=sin2+cos2=1.答案:18.求证:=tan .证明:左边=tan =右边,原等
3、式成立.9.已知函数f()=.(1)化简f();(2)若f()f=-,求f()+f2的值.解:(1)由题意得f()=-cos .(2)由(1)知f=-cos=sin .f()f=-,cos sin =.=(sin -cos )2=1-2cos sin =.二、B组1.若cos,则sin=()A.B.C.-D.-解析:cos,sin=sin=cos,故选A.答案:A2.已知sin+3sin=0,则tan=()A.B.C.2D.3解析:sin+3sin=0,sin=-3sin=-3sin+(-)=3sin=3cos=3cos.tan=3.答案:D3.已知为锐角,2tan(-)-3cos=-5,ta
4、n(+)+6sin(+)=1,则sin =()A.B.C.D.解析:2tan(-)-3cos=-5,tan(+)+6sin(+)=1,-2tan +3sin +5=0,tan -6sin =1,解得tan =3.又是锐角,sin =.答案:C4.已知sin(3+)=2sin,则=()A.-B.C.D.解析:sin(3+)=2sin,-sin =-2cos ,即sin =2cos .原式=-.答案:A5.已知sin =,且,则=.解析:,cos =-=-.原式=-tan =-.答案:6.已知tan(3+)=2,求的值.解:tan(3+)=2,tan =2.原式=2.7.已知sin ,cos 是关于x的方程x2-ax+a=0(aR)的两个根.(1)求cos+sin的值;(2)求tan(-)-的值.解:由已知原方程判别式0,即(-a)2-4a0,则a4或a0.又(sin +cos )2=1+2sin cos ,所以a2-2a-1=0,解得a=1-或a=1+(舍去).所以sin +cos =sin cos =1-.(1)cos+sin=sin +cos =1-.(2)tan(-)-=-tan -=-(tan +)=-=-=-+1.
