1、第2课时两角和与差的正弦、余弦、正切公式课后篇巩固提升合格考达标练1.(2021黑龙江哈尔滨香坊高一期末)化简cos 16cos 44-cos 74sin 44的值为()A.32B.-32C.12D.-12答案C解析cos16cos44-cos74sin44=cos16cos44-sin16sin44=cos(16+44)=cos60=12,故选C.2.化简:sinx+3+sinx-3=()A.-sin xB.sin xC.-cos xD.cos x答案B解析sinx+3+sinx-3=12sinx+32cosx+12sinx-32cosx=sinx.3.若sin6-=cos6+,则tan =
2、()A.-1B.0C.12D.1答案A解析由已知得12cos-32sin=32cos-12sin,因此1-32sin=3-12cos,于是tan=-1.4.(2021新疆维吾尔自治区哈密伊州高一期末)已知tan-34=23,则tan =()A.15B.-15C.5D.-5答案B解析tan-34=tan-tan341+tantan34=tan+11-tan=23,解得tan=-15,故选B.5.(2021天津和平高一期末)已知tan A=2tan B,sin(A+B)=14,则sin(A-B)=()A.13B.14C.112D.-112答案C解析由tanA=2tanB得sinAcosA=2sin
3、BcosB,即sinAcosB=2cosAsinB,sin(A+B)=14,sinAcosB+cosAsinB=14,得sinAcosB=16,cosAsinB=112.则sin(A-B)=sinAcosB-cosAsinB=16-112=112.故选C.6.已知cos(+)=45,cos(-)=-45,则cos cos =.答案0解析由已知得coscos-sinsin=45,coscos+sinsin=-45,两式相加得2coscos=0,故coscos=0.7.化简:sin(-150)+cos(-120)cos=.答案-1解析原式=sincos150-cossin150+coscos120
4、+sinsin120cos=-32sin-12cos-12cos+32sincos=-1.8.化简求值:(1)sin(+)cos(-)+cos(+)sin(-);(2)cos(70+)sin(170-)-sin(70+)cos(10+);(3)cos 21cos 24+sin 159sin 204.解(1)原式=sin(+-)=sin2.(2)原式=cos(70+)sin(10+)-sin(70+)cos(10+)=sin(10+)-(70+)=sin(-60)=-32.(3)原式=cos21cos24+sin(180-21)sin(180+24)=cos21cos24-sin21sin24=
5、cos(21+24)=cos45=22.等级考提升练9.若tan(+)=25,tan(-)=14,则tan 2=()A.16B.2213C.322D.1318答案D解析tan2=tan(+)+(-)=tan(+)+tan(-)1-tan(+)tan(-)=25+141-2514=1318.10.设0,2,0,2,且tan =1+sincos,则()A.3-=2B.3+=2C.2-=2D.2+=2答案C解析由tan=1+sincos,得sincos=1+sincos,得sincos-cossin=cos,sin(-)=sin2-.又0,2,0,2,故-=2-,即2-=2.11.(2021北京朝阳
6、高一期末)已知tan-6=2,tan(+)=-3,则tan+6=()A.1B.2C.3D.4答案A解析因为-6+6=+,所以tan(+)=tan-6+6=tan(-6)+tan(+6)1-tan(-6)tan(+6)=2+tan(+6)1-2tan(+6)=-3,整理解得tan+6=1.故选A.12.在ABC中,tan A+tan B+tan C=33,tan2B=tan Atan C,则角B等于()A.30B.45C.120D.60答案D解析由公式变形得tanA+tanB=tan(A+B)(1-tanAtanB)=tan(180-C)(1-tanAtanB)=-tanC(1-tanAtanB
7、)=-tanC+tanAtanBtanC,tanA+tanB+tanC=-tanC+tanAtanBtanC+tanC=tanAtanBtanC=33.tan2B=tanAtanC,tan3B=33,tanB=3,则B=60.故选D.13.(多选题)下面各式中,正确的是()A.sin4+3=sin4cos3+32cos4B.cos512=22sin3-cos4cos3C.cos-12=cos4cos3+64D.cos12=cos3-cos4答案ABC解析sin4+3=sin4cos3+cos4sin3=sin4cos3+32cos4,A正确;cos512=-cos712=-cos3+4=22s
8、in3-cos4cos3,B正确;cos-12=cos4-3=cos4cos3+64,C正确;cos12=cos3-4cos3-cos4,D不正确.14.(多选题)在ABC中,C=120,tan A+tan B=233,下列各式正确的是()A.A+B=2CB.tan(A+B)=-3C.tan A=tan BD.cos B=3sin A答案CD解析C=120,A+B=60,2(A+B)=C,tan(A+B)=3,选项A,B错误;tanA+tanB=3(1-tanAtanB)=233,tanAtanB=13,又tanA+tanB=233,联立解得tanA=tanB=33,cosB=3sinA,故选
9、项C,D正确.15.已知锐角,满足(tan -1)(tan -1)=2,则tan(+)=,+=.答案-134解析因为(tan-1)(tan-1)=2,所以tan+tan=tantan-1.因此tan(+)=tan+tan1-tantan=-1,因为+(0,),所以+=34.16.若cos =-13,sin =-33,2,32,2,则sin(+)的值为.答案539解析cos=-13,2,sin=1-cos2=223.sin=-33,32,2,cos=1-sin2=63.sin(+)=sincos+cossin=22363+-13-33=539.17.已知,均为锐角,且tan =cos-sincos+sin,求tan(+)的值.解tan=cos-sincos+sin=1-tan1+tan=tan4-,因为,均为锐角,所以-44-4,00,即x22x,即x8,或x0(舍去),x的最小值为8.当且仅当tanB=2+2,tanC=2-2,tanA=4(或tanB,tanC互换)时取等号,此时A,B,C均为锐角.可得tanAtanBtanC的取值范围是8,+).
