1、考点规范练20两角和与差的正弦、余弦与正切公式一、基础巩固1.(2020全国,理9)已知2tan -tan+4=7,则tan =()A.-2B.-1C.1D.2答案:D解析:由已知得2tan-1+tan1-tan=7,即tan2-4tan+4=0,解得tan=2.2.已知角的顶点与原点重合,始边与x轴的正半轴重合,终边在直线y=2x上,则cos 2=()A.-45B.-35C.35D.45答案:B解析:由题意知tan=2,故cos2=cos2-sin2cos2+sin2=1-tan21+tan2=1-221+22=-35.3.已知,32,且cos =-45,则tan4-等于()A.7B.17C
2、.-17D.-7答案:B解析:因为,32,且cos=-45,所以sin=-35,所以tan=34.所以tan4-=1-tan1+tan=1-341+34=17.4.若tan =2tan5,则cos-310sin-5=()A.1B.2C.3D.4答案:C解析:因为tan=2tan5,所以cos-310sin-5=sin-310+2sin-5=sin+5sin-5=sincos5+cossin5sincos5-cossin5=tan+tan5tan-tan5=3tan5tan5=3.5.已知cos-6+sin =435,则sin+76的值为()A.12B.32C.-45D.-12答案:C解析:co
3、s-6+sin=32cos+32sin=435,12cos+32sin=45,即sin+6=45.sin+76=-sin+6=-45.6.若0yx2,且tan x=3tan y,则x-y的最大值为()A.4B.6C.3D.2答案:B解析:0yx2,x-y0,2.又tanx=3tany,tan(x-y)=tanx-tany1+tanxtany=2tany1+3tan2y=21tany+3tany33=tan6.当且仅当3tan2y=1时取等号,x-y的最大值为6,故选B.7.函数f(x)=sin 2xsin6-cos 2xcos56在区间-2,2上的单调递增区间为.答案:-512,12解析:f(
4、x)=sin2xsin6-cos2xcos56=sin2xsin6+cos2xcos6=cos2x-6.当2k-2x-62k(kZ),即k-512xk+12(kZ)时,函数f(x)单调递增.取k=0,得-512x12,故函数f(x)在区间-2,2上的单调递增区间为-512,12.8.已知tan(3-)=-12,tan(-)=-13,则tan =.答案:17解析:由题意得tan=12,所以tan=tan(-)+=tan(-)+tan1-tan(-)tan=-13+121-1312=17.9.在ABC中,C=60,tanA2+tanB2=1,则tanA2tanB2=.答案:1-33解析:由C=60
5、,则A+B=120,即A2+B2=60.根据tanA2+B2=tanA2+tanB21-tanA2tanB2,tanA2+tanB2=1,得3=11-tanA2tanB2,解得tanA2tanB2=1-33.10.已知,均为锐角,且sin =35,tan(-)=-13.(1)求sin(-)的值;(2)求cos 的值.解:(1),0,2,-2-2.又tan(-)=-130,-2-bcB.bacC.cabD.acb答案:D解析:a=sin40cos127+cos40sin127=sin(40+127)=sin167=sin13,b=22(sin56-cos56)=22sin56-22cos56=s
6、in(56-45)=sin11,c=1-tan2391+tan239=cos239-sin239cos239cos239+sin239cos239=cos239-sin239=cos78=sin12.sin13sin12sin11,acb.故选D.13.已知sin+4=14,-32,-,则cos+712的值为.答案:-15+38解析:由-32,-,得+4-54,-34.因为sin+4=14,所以cos+4=-154.cos+712=cos+4+3=cos+4cos3-sin+4sin3=-15412-1432=-15+38.14.设,0,2,且tan =1+sincos,则2-=.答案:2解析
7、:,0,2,且tan=1+sincos,sincos=1+sincos,sincos=cos+cossin.sincos-cossin=cos.sin(-)=cos=sin2-.,0,2,-2,2,2-0,2.函数y=sinx在-2,2内单调递增,由sin(-)=sin2-可得-=2-,即2-=2.15.已知角的顶点与原点O重合,始边与x轴的非负半轴重合,它的终边过点P-35,-45.(1)求sin(+)的值;(2)若角满足sin(+)=513,求cos 的值.解:(1)由角的终边过点P-35,-45,得sin=-45,所以sin(+)=-sin=45.(2)由角的终边过点P-35,-45,得cos=-35,由sin(+)=513,得cos(+)=1213.由=(+)-,得cos=cos(+)cos+sin(+)sin,所以cos=-5665或cos=1665.三、高考预测16.在平面直角坐标系中,点P的坐标为35,45,Q是第三象限内一点,|OQ|=1,且POQ=34,则点Q的横坐标为()A.-7210B.-325C.-7212D.-8213答案:A解析:设xOP=,则cos=35,sin=45.因为是第三象限内一点,所以xQ=cos+34=35-22-4522=-7210,故选A.
