1、第2课时诱导公式五、六课后训练巩固提升A组1.已知sin52+=15,则cos 等于()A.-25B.-15C.15D.25解析:sin52+=cos ,cos =15,故选C.答案:C2.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)=fsin2-x=3-cos 22-x=3-cos(-2x)=3+cos 2x.答案:C3.化简cos(180+)sin(+360)cos(-270-)sin(-180)cos(-180-)sin(360-)的结果是()A.1B.-1C.tan
2、 D.-tan 解析:原式=-cossinsinsin(-cos)(-sin)=-1.答案:B4.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.答案:B5.化简sin152+cos-2sin92-cos32+=.解析:原式=sin32+cos2-sin2-sin=(-cos)sincossin=-1.答案:-16.若cos =15,且是第四象限角,则cos+2=.解析:cos =15,且是第四象限角,sin =-1-cos2=-1-152=-265.co
3、s+2=-sin =265.答案:2657.给出下列三个结论,其中正确结论的序号是.sin(+)=-sin 成立的条件是角是锐角;若cos(n-)=13(nZ),则cos =13;若k2(kZ),则tan2+=-1tan.解析:由诱导公式,当R时,都有sin(+)=-sin ,所以错误.当n=2k(kZ)时,cos(n-)=cos(-)=cos ,此时cos =13;当n=2k+1(kZ)时,cos(n-)=cos(2k+1)-=cos(-)=-cos ,此时cos =-13,所以错误.若k2(kZ),则tan2+=sin2+cos2+=cos-sin=-1tan,所以正确.答案:8.已知函数
4、f()=sin-2cos32+tan(2-)tan(+)sin(+).(1)化简f();(2)若f()f+2=-18,求f()+f+22的值.解:(1)由题意得f()=-cossin(-tan)tan(-sin)=-cos .(2)由(1)知f+2=-cos+2=sin .f()f+2=-18,cos sin =18.f()+f+22=(sin -cos )2=1-2cos sin =34.9.求证:tan(2-)cos32-cos(8-)sin-32cos+72=tan .证明:左边=tan(-)-cos2-cos(-)sin+2cos-2=-tan(-sin)coscossin=tan =
5、右边,原等式成立.B组1.若cos12-=13,则sin512+=()A.13B.223C.-13D.-223解析:cos12-=13,sin512+=sin2-12-=cos12-=13,故选A.答案:A2.已知sin125+3sin1110-=0,则tan25+=()A.13B.12C.2D.3解析:sin125+3sin1110-=0,sin25+=-3sin1110-=-3sin+10-=3sin10-=3cos2-10-=3cos25+.tan25+=sin25+cos25+=3.答案:D3.已知为锐角,2tan(-)-3cos2+=-5,tan(+)+6sin(+)=1,则sin
6、=()A.355B.377C.31010D.13解析:2tan(-)-3cos2+=-5,tan(+)+6sin(+)=1,-2tan +3sin +5=0,tan -6sin =1,解得tan =3.又是锐角,sin =31010.答案:C4.已知sin(3+)=2sin32+,则sin(-)-4sin2+5sin(2+)+2cos(2-)=()A.-16B.16C.13D.12解析:sin(3+)=2sin32+,-sin =-2cos ,即sin =2cos .原式=sin-4cos5sin+2cos=2cos-4cos10cos+2cos=-212=-16.答案:A5.已知sin =2
7、3,且2,则cos(-2)sin(2-)tan(-)cos32-sin2+=.解析:2,cos =-1-29=-73.原式=cos(-sin)(-tan)(-sin)cos=-tan =-sincos=147.答案:1476.已知tan(3+)=2,求sin(-3)+cos(-)+sin2-2cos2+-sin(-)+cos(+)的值.解:tan(3+)=2,tan =2.原式=-sin-cos+cos+2sinsin-cos=sinsin-cos=tantan-1=22-1=2.7.已知sin ,cos 是关于x的方程x2-ax+a=0(aR)的两个根.(1)求cos2-+sin2+的值;(2)求tan(-)-1tan的值.解:由已知原方程判别式0,即(-a)2-4a0,则a4或a0.又sin+cos=a,sincos=a,(sin +cos )2=1+2sin cos ,所以a2-2a-1=0,解得a=1-2或a=1+2(舍去).所以sin +cos =sin cos =1-2.(1)cos2-+sin2+=sin +cos =1-2.(2)tan(-)-1tan=-tan -1tan=-tan+1tan=-sincos+cossin=-1sincos=-11-2=2+1.
