2019届高考数学一轮复习夯基提能作业：第四章三角函数解三角形第六节简单的三角恒等变换 .doc
第六节简单的三角恒等变换A组基础题组1.若cos2sin+74=-22,则sin +cos 的值为()A.-22B.-12 C.12D.722.已知cos4-x=35,则sin 2x=()A.1825 B.725C.-725D.-16253.已知sin6-=cos6+,则cos 2=()A.1B.-1C.12D.04.2cos10-sin20sin70的值是()A.12B.32C.3D.25.已知sin-4=7210,cos 2=725,则sin =()A.45B.-45C.35D.-356.已知sin 2=352<2<,tan(-)=12,则tan(+)等于.7.已知cos(+)=16,cos(-)=13,则tan tan 的值为.8.若tan =3,则sin2+4的值为.9.已知tan =-13,cos =55,2,0,2,求tan(+)的值,并求出+的值.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.B组提升题组1.已知cos3-2x=-78,则sinx+3的值为()A.14B.78C.14D.782.cos9cos29cos-239=.3.已知函数f(x)=cos2x+sin xcos x,xR.(1)求f6的值;(2)若sin =35,且2,求f2+24.4.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.答案精解精析A组基础题组1.Ccos2sin+74=cos2-sin222(sin-cos)=(cos+sin)(cos-sin)22(sin-cos)=-22,整理得sin +cos =12.2.Ccos4-x=cos4cos x+sin4sin x=22(cos x+sin x)=35,sin x+cos x=325,1+2sin xcos x=1825,即sin 2x=1825-1=-725.3.Dsin6-=cos6+,12cos -32sin =32cos -12sin ,即12-32sin =-12-32cos ,tan =sincos=-1,cos 2=cos2-sin2=cos2-sin2sin2+cos2=1-tan2tan2+1=0.4.C原式=2cos(30-20)-sin20sin70=2(cos30cos20+sin30sin20)-sin20sin70=3cos20cos20=3.5.C由sin-4=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 =35.6.答案-2解析由题意,可得cos 2=-45,则tan 2=-34,故tan(+)=tan2-(-)=tan2-tan(-)1+tan2tan(-)=-2.7.答案13解析因为cos(+)=16,所以cos cos -sin sin =16.因为cos(-)=13,所以cos cos +sin sin =13.+得cos cos =14.-得sin sin =112.所以tan tan =sinsincoscos=13.8.答案-210解析sin 2=2sin cos =2sincossin2+cos2=2tantan2+1=35,cos 2=cos2-sin2=cos2-sin2cos2+sin2=1-tan21+tan2=-45,sin2+4=22sin 2+22cos 2=2235+-45=-210.9.解析由cos =55,0,2,得sin =255,则tan =2.tan(+)=tan+tan1-tantan=-13+21+23=1.2,0,2,2<+<32,+=54.10.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. C因为cos-3-2x=cos2x+23=78,所以有sin2x+3=121-78=116,从而求得sinx+3的值为14,故选C.2.答案-18解析cos9cos29cos-239=cos 20cos 40cos 100=-cos 20cos 40cos 80=-sin20cos20cos40cos80sin20=-12sin40cos40cos80sin20=-14sin80cos80sin20=-18sin160sin20=-18sin20sin20=-18.3.解析(1)f6=cos26+sin6cos6=322+1232=3+34.(2)因为f(x)=cos2x+sin xcos x=1+cos2x2+12sin 2x=12+12(sin 2x+cos 2x)=12+22sin2x+4,所以f2+24=12+22sin+12+4=12+22sin+3=12+2212sin+32cos.又因为sin =35,且2,所以cos =-45,所以f2+24=12+221235-3245=10+32-4620.4.解析(1)因为f(x)=(2cos2x-1)sin 2x+12cos 4x=cos 2xsin 2x+12cos 4x=12(sin 4x+cos 4x)=22sin4x+4,所以f(x)的最小正周期为2,最大值为22.(2)解法一:因为f()=22,所以sin4+4=1.因为2,所以4+494,174.所以4+4=52.故=916.解法二:f()=22,sin4+4=1.4+4=2+2k,kZ.=16+k2,kZ.又2,当k=1,即=916时,符合题意.故=916.版权所有：高考资源网(www.ks5u.com)
