2018届高三数学复习两角和与差的正弦余弦和正切公式及二倍角公式夯基提能作业本.docx

第五节两角和与差的正弦、余弦和正切公式及二倍角公式A组基础题组1.已知sin=12,-0,则cos的值是()A.12 B.23C.-12 D.12.已知cos=-,则cos x+cos=()A.-233B.233C.-1 D.13.已知sin=7210,cos 2=725,则sin =()A.45 B.-45C.35 D.-354.(2016江西九校联考)已知锐角,满足sin -cos =16,tan +tan +3tan tan =3,则,的大小关系是()A.B.C.D.5.已知cos =-513,则sin的值为.6.已知sin=13,则cos的值是.7.计算=.8.已知,tan =12,求tan 2和sin的值.9.已知,且sin+cos=62.(1)求cos 的值;(2)若sin(-)=-35,求cos 的值.B组提升题组10.cos cos cos=()A.-18 B.-116C.116 D.1811.定义运算=ad-bc.若cos =17,=3314,0,则等于()A. B.C. D.12.设,且tan =,则()A.3-=B.3+=C.2-=D.2+=13.(2015广东,16,12分)已知tan =2.(1)求tan的值;(2)求的值.14.已知coscos=-14,蟺3,蟺2.(1)求sin 2的值; (2)求tan -1tan伪的值.答案全解全析A组基础题组1.C由sin=12得cos =12,-0,.又tan +tan +3tan tan =3,tan(+)=3,由题意知0+,.5.答案5-12326解析由cos =-513,得sin =-=-1213,故sin=sin cos-cossin=-121332-51312=5-12326.6.答案-79解析sin=13,cos=cos2蟺6-伪=1-2sin2=79,cos=cos=cos=-cos=-79. 7.答案12解析=12.8.解析tan =12,tan 2=43,且=12,即cos =2sin ,又sin2+cos2=1,5sin2=1,而,sin =55,则cos =255.sin 2=2sin cos =255255=45,cos 2=cos2-sin2=45-15=35,sin=sin 2cos+cos 2sin=12+3532=4+3310.9.解析(1)将sin+cos=62两边同时平方,得1+sin =32,则sin =12.又,所以cos =-=-32.(2)因为,所以-.所以由sin(-)=-35,得cos(-)=45,所以cos =cos-(-)=cos cos(-)+sin sin(-)=-3245+12-35=-43+310.B组提升题组10.Acos cos cos=cos 20cos 40cos 100=-cos 20cos 40cos 80=-=-=-=-=-=-18.11.D依题意得sin cos -cos sin =sin(-)=3314.0,0-,cos(-)=1314.cos =17,0,sin =437.sin =sin-(-)=sin cos(-)-cos sin(-)=4371314-173314=32,=.故选D.12.C由tan =,得=,即sin cos =cos +sin cos ,所以sin(-)=cos ,又cos =sin,所以sin(-)=sin,又因为,所以-,0-,因此-=-,所以2-=,故选C.13.解析(1)因为tan =2,所以tan=-3.(2)因为tan =2,所以=1.14.解析(1)coscos=cossin=12sin2+=-14,即sin=-12.,2+,cos=-32,sin 2=sin=sincos-cos