（新课标）2020版高考数学总复习第四章第六节简单的三角恒等变换练习文新人教A版.docx

第六节简单的三角恒等变换A组基础题组1.(2019江西临川模拟)已知cos6-=33,则sin56-2的值为()A.13B.-13C.23D.-23答案Bsin56-2=sin2+3-2=cos3-2=cos26-=2cos26-1=2332-1=-13.故选B.2.若2,且3cos 2=sin4-,则sin 2的值为()A.-118B.118C.-1718D.1718答案C由3cos 2=sin4-可得3(cos2-sin2)=22(cos -sin ),又由2,可知cos -sin 0,于是3(cos +sin )=22,所以1+2sin cos =118,故sin 2=-1718.3.(2019安徽淮南一模)设0,2,0,4,且tan =1+sin2cos2,则下列结论正确的是()A.-=4B.+=4C.2-=4D.2+=4答案Atan =1+sin2cos2=(sin+cos)2cos2-sin2=sin+coscos-sin=1+tan1-tan=tan+4.因为0,2,+44,2,所以=+4,即-=4.4.(2019湖北八校联考)已知34,且1+cos2+1-cos2=62,则=()A.103或113B.3712或4712C.134或154D.196或236答案D34,3222,cos20,sin20,则1+cos2+1-cos2=cos22+sin22=cos2-sin2=2cos2+4=62,cos2+4=32.2+4=6+2k或2+4=-6+2k,kZ,即=-6+4k或=-56+4k,kZ.34,=196或236,故选D.5.4cos 50-tan 40等于()A.2B.2+32C.3D.22-1答案C4cos 50-tan 40=4sin 40-sin40cos40=4cos40sin40-sin40cos40=2sin80-sin40cos40=2sin(120-40)-sin40cos40=3cos40+sin40-sin40cos40=3cos40cos40=3.6.cos9cos29cos-239=.答案-18解析cos9cos29cos-239=cos 20cos 40cos 100=-cos 20cos 40cos 80=-sin20cos20cos40cos80sin20=-12sin40cos40cos80sin20=-14sin80cos80sin20=-18sin160sin20=-18sin20sin20=-18.7.2sin 50+sin 10(1+3tan 10)2sin280=.答案6解析原式=2sin50+sin10cos10+3sin10cos102sin 80=2sin50+2sin1012cos10+32sin10cos102cos 10=22sin 50cos 10+sin 10cos(60-10)=22sin(50+10)=2232=6.8.已知0,2,且2sin2-sin cos -3cos2=0,则sin+4sin2+cos2+1=.答案268解析0,2,且2sin2-sin cos -3cos2=0,则(2sin -3cos )(sin +cos )=0,2sin =3cos ,又sin2+cos2=1,cos =213,sin =313,sin+4sin2+cos2+1=22(sin+cos)(sin+cos)2+(cos2-sin2)=222cos=268.9.已知sin+4=210,2,.求:(1)cos 的值;(2)sin2-4的值.解析(1)解法一:因为2,所以+434,54.又sin+4=210,所以cos+4=-1-sin2+4=-1-2102=-7210.所以cos =cos+4-4=cos+4cos4+sin+4sin 4=-721022+21022=-35.解法二:由sin+4=210得sin cos4+cos sin4=210,即sin +cos =15,结合sin2+cos2=1,得cos =-35或cos =45.因为2,所以cos =-35.(2)因为2,又由(1)知cos =-35,所以sin =1-cos2=1-352=45.所以sin 2=2sin cos =245-35=-2425,cos 2=2cos2-1=2-352-1=-725.所以sin2-4=sin 2cos4-cos 2sin4=-242522-72522=-17250.10.已知函数f(x)=Acosx4+6,xR,且f3=2.(1)求A的值;(2)设,0,2, f4+43=-3017, f4-23=85,求cos(+)的值.解析(1)因为f3=Acos12+6=Acos4=22A=2,所以A=2.(2)由f4+43=2cos+3+6=2cos+2=-2sin =-3017,得sin =1517,又0,2,所以cos =817.由f4-23=2cos-6+6=2cos =85,得cos =45,又0,2,所以sin =35,所以cos(+)=cos cos -sin sin =81745-151735=-1385.B组提升题组1.(2019河南平顶山模拟)已知sin =-4532,2,若sin(+)cos=2,则tan(+)=()A.613B.136C.-613D.-136答案Bsin =-45,32,2,cos =35.由sin(+)cos=2,得sin(+)=2cos(+)-,即65cos(+)=135sin(+),故tan(+)=613.2.若函数f(x)=5cos x+12sin x在x=时取得最小值,则cos 等于()A.513B.-513C.1213D.-1213答案Bf(x)=5cos x+12sin x=13513cosx+1213sinx=13sin(x+),其中sin =513,cos =1213,由题意知+=2k-2(kZ),得=2k-2-(kZ),所以cos =cos2k-2-=cos2+=-sin =-513.3.已知函数f(x)=(2cos2x-1)sin 2x+12cos 4x.(1)求f(x)的最小正周期及最大值;(2)若2,且f()=22,求的值.解析(1)因为f(x)=(2cos2x-1)sin 2x+12cos 4x=cos 2xsin 2x+12cos 4x=12(sin 4x+cos 4x)=2