2019版高考数学一轮复习第四章三角函数解三角形第五节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文243103.doc

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.(2017北京东城一模)“sin +cos =0”是“cos 2=0”的()A.充分而不必要条件B.必要而不充分条件C.充分必要条件D.既不充分也不必要条件2.已知sin2+=12,-20,则cos-3的值是()A.12B.23C.-12D.13.已知sin-4=7210,cos 2=725,则sin =()A.45B.-45C.35D.-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.4B.4C.4D.40,xR)的最小正周期为2.(1)求的值;(2)如果0,2,且f()=85,求cos 的值.10.已知2,且sin2+cos2=62.(1)求cos 的值;(2)若sin(-)=-35,2,求cos 的值.B组提升题组11.cos 9cos 29cos-239=()A.-18B.-116C.116D.1812.定义运算abcd=ad-bc.若cos =17,sinsincoscos=3314,02,则等于()A.12B.6C.4D.313.设0,2,0,2,且tan =1+sincos,则()A.3-=2B.3+=2C.2-=2D.2+=214.已知tan =2.(1)求tan+4的值;(2)求sin2sin2+sincos-cos2-1的值.15.已知cos6+cos3-=-14,3,2.(1)求sin 2的值; (2)求tan -1tan的值.答案精解精析A组基础题组1.Acos 2=cos2-sin2=(cos +sin )(cos -sin )=0cos +sin =0或cos -sin =0,“sin +cos =0”是“cos 2=0”的充分而不必要条件,故选A.2.C由sin2+=12得cos =12,-20,4.又tan +tan +3tan tan =3,tan(+)=tan+tan1-tantan=3,由题意知0+4,40,所以=12.(2)由(1)可知f()=cos+6+1=85,所以cos+6=35,因为0,2,所以+66,23,所以sin+6=45.所以cos =cos+6-6=cos+6cos6+sin+6sin 6=3532+4512=33+410.10.解析(1)将sin2+cos2=62两边同时平方,得1+sin =32,则sin =12.又2,所以cos =-1-sin2=-32.(2)因为2,2,所以-2-2.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11.Acos 9cos 29cos-239=cos 20cos 40cos 100=-cos 20cos 40cos 80=-sin20cos20cos40cos80sin20=-12sin40cos40cos80sin20=-14sin80cos80sin20=-18sin160sin20=-18sin20sin20=-18.12.D依题意得sin cos -cos sin =sin(-)=3314.02,0-2,cos(-)=1314.cos =17,02,sin =437.sin =sin-(-)=sin cos(-)-cos sin(-)=4371314-173314=32,=3.故选D.13.C由tan =1+sincos,得sincos=1+sincos,即sin cos =cos +sin cos ,所以sin(-)=cos ,又cos =sin2-,所以sin(-)=sin2-,又因为0,2,0,2,所以-2-2,02-2,因此-=2-,所以2-=2,故选C.14.解析(1)因为tan =2,所以tan+4=tan+tan41-tantan4=2+11-21=-3.(2)因为tan =2,所以sin2sin2+sincos-cos2-1=2sincossin2+sincos-(cos2-sin2)-(sin2+cos2)=2sincossin2+sincos-2cos2=2tantan2+tan-2=2222+2-2=1.15.解析(1)cos6+cos3-=cos6+sin6+=12sin2+3=-14,即sin2+3=-12.3,2,2+3,43,cos2+3=-32,sin 2=sin2+3-3=sin2+3cos3-cos2+3sin3=12.(