2018届高三数学复习三角函数解三角形第四节简单的三角恒等变换夯基提能作业本理.docx

第四节简单的三角恒等变换A组基础题组1.已知sin 2=13,则cos2=() A.13B.-13C.23D.-232.(2016河南八市重点高中质检)已知,tan=17,那么sin 2+cos 2的值为()A.-15B.75C.-75D.343.化简:=()A.1B.3C.2D.24.已知cos=-33,则cos x+cos=()A.-233B.233C.-1D.15.的值为()A.1B.-1C.12D.-126.(2016河北“五校联盟”质量检测)在ABC中,sin(C-A)=1,sin B=13,则sin A=.7.已知-0,cos(-)=35,sin =-513,则sin =.8.已知2tanx1+tan2x=35,则sin2蟺4+x=.9.已知tan =-13,cos =55,求tan(+)的值,并求出+的值.10.已知角的顶点是坐标原点,始边与x轴的正半轴重合,终边经过点P(-3,3).(1)求sin 2-tan 的值;(2)若函数f(x)=cos(x-)cos -sin(x-)sin ,求函数g(x)=3f 蟺2-2x-2f 2(x)在区间上的值域.B组提升题组11.已知R,sin +2cos =102,则tan 2=() A.43B.34C.-34D.-4312.cos cos cos=() A.-18B.-116C.116D.1813.=.14.(2016郑州模拟)已知直线l1l2,A是l1,l2之间的一定点,并且A点到l1,l2的距离分别为h1,h2,B是直线l2上一动点,作ACAB,且使AC与直线l1交于点C,则ABC面积的最小值为.15.(2014广东,16,12分)已知函数f(x)=Asin,xR,且f5蟺12=32.(1)求A的值;(2)若f()+f(-)=32,求f.16.(2014江西,16,12分)已知函数f(x)=sin(x+)+acos(x+2),其中aR,.(1)当a=2,=时,求f(x)在区间0,上的最大值与最小值;(2)若f蟺2=0, f()=1,求a,的值.答案全解全析A组基础题组1.Ccos2=,将sin 2=13代入,得原式=1+132=23,故选C.2.A由tan=17,知=17,tan 2=-34.2,sin 2=35,cos 2=-45.sin 2+cos 2=-15,故选A.3.C原式=2,故选C.4.Ccos x+cos=cos x+12cos x+32sin x=32cos x+32sin x=332cosx+12sinx=3cos,将cos=-33代入,得原式=-1.5.D原式=-12.6.答案33解析由题意得0C180,0A180,-180C-A180,sin(C-A)=1,C-A=90,即C=90+A,sin B=13,sin B=sin(A+C)=sin(90+2A)=cos 2A=13,即1-2sin2A=13,sin A=33.7.答案3365解析-0,0,0-.由cos(-)=35,sin =-513,可得sin(-)=45,cos =1213,sin =sin(-)+=sin(-)cos +cos(-)sin =451213+35-513=3365.8.答案45解析因为2tanx1+tan2x=35=2sinxcosxcos2x+sin2x=sin 2x,所以sin 2x=35,则sin2蟺4+x=1+sin2x2=45.9.解析由cos =55,得sin =255,则tan =2.tan(+)=-13+21+23=1.,+,+=.10.解析(1)角的终边经过点P(-3,3),sin =12,cos =-32,tan =-33.sin 2-tan =2sin cos -tan =-32+33=-36.(2)f(x)=cos(x-)cos -sin(x-)sin =cos x,g(x)=3cos蟺2-2x-2cos2x=3sin 2x-1-cos 2x=2sin-1,0x,-2x-.-12sin1,-22sin-11,故函数g(x)=3f 蟺2-2x-2f 2(x)在区间上的值域是-2,1.B组提升题组11.C因为sin +2cos =102,所以sin2+4cos2+4sin cos =104(sin2+cos2),整理得3sin2-3cos2-8sin cos =0,两边同时除以cos2,得3tan2-8tan -3=0,解得tan =3或tan =-13,代入tan 2=,得到tan 2=-34.12.Acos cos cos=cos 20cos 40cos 100=-cos 20cos 40cos 80=-=-=-=-=-=-18.13.答案-43解析原式=-43.14.答案h1h2解析如图,设ABD=,则CAE=,AB=h2sin伪,AC=h1cos伪.所以SABC=12ABAC=.易得当2=,即=时,SABC取最小值,且最小值为h1h2.15.解析(1)f5蟺12=Asin=32,A32=32,A=3.(2)f()+f(-)=3sin+3sin=32,322(sin +cos )+22(-sin +cos )=32,6cos =32,cos =64,又 ,sin =104,f34蟺-胃=3sin(-)=3sin =304.16.解析(1)当