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

第三节两角和与差的正弦、余弦、正切公式及二倍角公式A组基础题组1.sin 163sin 223+sin 253sin 313=() A.-12B.12C.-32D.322.若=12,则tan 2等于()A.54B.-54C.43D.-433.(2016宿州模拟)若sin=13,则cos等于()A.429B.-429C.79D.-794.(2016青岛模拟)化简12sin 2-2cos2=()A.cos2B.sin2C.cos 2D.-cos 25.设tan(+)=25,tan=14,则tan的值是()A.16B.322C.1322D.13186.sin 15+sin 75的值是.7.(2016广州高考模拟)已知cos(+)=-13,则sin=.8.设tan=14,则tan=.9.已知,且sin +cos =62.(1)求cos 的值;(2)若sin(-)=-35,求cos 的值.10.已知函数f(x)=Asin,xR,且f5蟺12=322.(1)求A的值;(2)若f()-f(-)=3,求f.B组提升题组11.(2016烟台模拟)若cos 2+cos =0,则sin 2+sin =() A.0B.3C.0或3D.0或312.已知sin=7210,cos 2=725,则sin =()A.45B.-45C.35D.-3513.计算=.14.(2016合肥质检)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.Bsin 163sin 223+sin 253sin 313=-sin 17cos 47+cos 17sin 47=sin(47-17)=12.2.D因为=12,所以tan =2,所以tan 2=41-4=-43.3.D因为sin=13,所以cos=sin 2=-cos=-cos 2=-=2sin2-1=-79.4.D原式=sin伪cos伪+cos伪sin伪sin cos -2cos2=(sin2+cos2)-2cos2=1-2cos2=-cos 2.5.Btan=tan=,将tan(+)=25,tan=14代入,得原式=322.6.答案62解析sin 15+sin 75=sin 15+cos 15=2sin(15+45)=2sin 60=62.7.答案-79解析因为cos(+)=-13,所以cos =13,则sin=cos 2=2cos2-1=-79.8.答案-4解析tan=14,tan =53,tan=-4.9.解析(1)sin +cos =62,两边同时平方,得sin =12.又,所以cos =-=-32.(2)因为,所以-.则由sin(-)=-35,得cos(-)=45.所以cos =cos-(-)=cos cos(-)+sin sin(-)=-3245+12-35=-43+310.10.解析(1)由f 5蟺12=Asin=Asin =22A=322可得A=3.(2)f()-f(-)=3,则3sin-3sin=3,3-3=3,sin =33.因为,所以cos =63,所以f =3sin=3sin=3cos =6.B组提升题组11.D由cos 2+cos =0得2cos2-1+cos =0,所以cos =-1或12.当cos =-1时,有sin =0;当cos =12时,有sin =32.于是sin 2+sin =sin (2cos +1)=0或3或-3.12.C由sin=7210得sin -cos =75,由cos 2=725得cos2 -sin2=725,所以(cos -sin )(cos +sin )=725,由可得cos +sin =-15,由可得sin =35.13.答案12解析=12.14.解析(1)coscos=cossin=12sin=-14,即sin=-.蟺3,蟺2,2+,cos=-32,sin 2=sin=sin