1、8.2三角恒等变换8.2.1两角和与差的余弦课后篇巩固提升基础巩固1.cos 70cos 335+sin 110sin 25的值为()A.1B.22C.32D.12解析原式=cos 70cos 25+sin 70sin 25=cos(70-25)=cos 45=22.答案B2.化简sin4-3xcos3-3x-sin4+3xsin3-3x的结果为()A.cos512B.-cos512C.sin512D.-sin512答案B3.在ABC中,cos A=35,cos B=513,则cos C等于()A.-3365B.3365C.-6365D.6365答案B4.(多选)已知cos =55,则cos-
2、4可以取的值为()A.31010B.-1010C.255D.-31010答案AB5.若cos(+)=15,cos(-)=35,则tan tan 的值为()A.2B.12C.-2D.-12解析由cos(+)=15,cos(-)=35可得coscos-sinsin=15,coscos+sinsin=35,则sin sin =15,cos cos =25.故tan tan =sinsincoscos=1525=12.答案B6.向量a=(2cos ,2sin ),b=(3cos ,3sin ),a与b的夹角为60,则直线xcos -ysin =12与圆(x-cos )2+(y+sin )2=12的位置
3、关系是()A.相切B.相交C.相离D.随,的值而定答案B7.(双空)已知,均为锐角,且sin =55,cos =1010,则-的值为,cos(+)=.答案-4-2108.已知sin(-45)=-210,090,则cos =.解析因为090,所以-45-4545,所以cos(-45)=1-sin2(-45)=7210,所以cos =cos(-45)+45=cos(-45)cos 45-sin(-45)sin 45=45.答案459.已知,均为锐角,cos =17,cos(+)=-1114,求角.解因为,均为锐角,所以02,02,0+0,xR)的最小正周期为10,(1)求的值;(2)设,0,2,f
4、5+53=-65,f5-56=1617,求cos(+)的值.解(1)f(x)=2cosx+6,0的最小正周期T=10=2,=15.(2)f(x)=2cos15x+6,f5+53=2cos+3+6=-2sin .sin =35.f5-56=2cos-6+6=2cos ,cos =817.,0,2,cos =45,sin =1517,cos(+)=cos cos -sin sin =45817-351517=-1385.能力提升1.cos 80cos 20-sin(-80)sin 160的值是()A.12B.32C.-12D.-32解析cos 80cos 20-sin(-80)sin 160=co
5、s 80cos 20+sin 80sin 20=cos 60=12,故选A.答案A2.已知cos =210,(-,0),则cos-4=()A.-35B.-45C.35D.45解析cos =210,(-,0),sin =-1-cos2=-7210,cos-4=cos cos4+sin sin4=21022+-721022=-35.故选A.答案A3.(多选)若,均为第二象限角,满足sin =35,cos =-513,则cos(+)和cos(-)的值分别为()A.-3365B.-1665C.3356D.5665解析sin =35,cos =-513,均为第二象限角,cos =-1-sin2=-45,
6、sin =1-cos2=1213,cos(+)=cos cos -sin sin =-45-513-351213=-1665,cos(-)=cos cos +sin sin =-45-513+351213=5665,故选BD.答案BD4.若-202,cos4+=13,cos4-2=33,则cos+2=()A.539B.33C.-33D.-69解析-2035,所以,因此cos(-)=cos cos +sin sin =2245+2235=7210,sin(-)=1-cos(-)2=210,cos(+)=cos cos -sin sin =2245-2235=210,sin(+)=1-cos(+)
7、2=7210,cos 2=cos(+)-(-)=cos(-)cos(+)+sin(-)sin(+)=725.答案72107256.已知cos-6-sin =235,则cos+76的值是.解析由于cos-6-sin =235,整理得32cos +12sin -sin =235,即32cos -12sin =235,则cos+6=235,可得cos+76=-cos+6=-235.答案-2357.已知a=(cos ,sin ),b=(cos ,-sin ),均为锐角,且|a-b|=255.(1)求cos(+)的值;(2)若sin =35,求cos 的值.解(1)由题意得|a|=1,|b|=1,则|a
8、-b|2=(a-b)2=a2-2ab+b2=2-2(cos cos -sin sin )=2-2cos(+)=45,解得cos(+)=35.(2),0,2,+(0,),由sin =35,cos(+)=35可得cos =45,sin(+)=45,故cos =cos(+)-=cos(+)cos +sin(+)sin =3545+4535=2425.8.已知向量a=(sin ,5cos -sin ),b=(cos -5sin ,cos ),且ab=2.(1)求cos(+)的值;(2)若02,02,且sin =1010,求2a+的值.解(1)因为a=(sin ,5cos -sin ),b=(cos -
9、5sin ,cos ),所以ab=sin (cos -5sin )+(5cos -sin )cos =5cos cos -5sin sin =5cos(+),因为ab=2,所以5cos(+)=2,即cos(+)=255.(2)因为02,sin =1010,所以cos =31010.因为02,02,所以0+.因为cos(+)=255,所以sin(+)=55,所以cos(2+)=cos+(+)=cos cos(+)-sin sin(+)=22.因为02,02,所以02+32,所以2+=4.9.已知函数f(x)=Asinx+4(xR),且f(0)=1.(1)求A的值;(2)若f()=-15,是第二象限角,求cos .解(1)依题意得f(0)=Asin4=22A=1,故A=2.(2)由(1)得f(x)=2sinx+4,由f()=-15可得f()=2sin+4=-15,则sin+4=-210,是第二象限角,2k+22k+(kZ),2k+34+42k+54(kZ),又sin+4=-2100,+4是第三象限角,cos+4=-1-sin2(+4)=-7210,cos =cos(+4)-4=cos+4cos4+sin+4sin4=-721022-21022=-45.
